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Case Study Questions Class 10 Science Electricity

Case study questions class 10 science chapter 12 electricity.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

Case study: 1

We can see that, as the applied voltage is increased the current through the wire also increases. It means that, the potential difference across the terminals of the wire is directly proportional to the electric current passing through it at a given temperature.

Thus, V= IR

Where R is the proportionality constant called as resistance of the wire. Thus, we can say that the resistance of the wire is inversely proportional to the electric current. As the resistance increases current through the wire decreases. The resistance of the conductor is directly proportional to length of the conductor, inversely proportional to the area of cross section of the conductor and also depends on the nature of the material from which conductor is made. Thus R= qL/A, where q is the resistivity of the material of conductor. According resistivity of the material they are classified as conductors, insulators and semiconductors. It is observed that the resistance and resistivity of the material varies with temperature. And hence there are vast applications of these materials based on their resistivity.

The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

1) What is SI unit of resistivity?

2) What is variable resistance?

3) Why tungsten is used in electric bulbs?

4) 1M ohm = ?

1) The SI unit of resistivity is ohm meter.

2) The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance.

3) Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high.

4) 1M ohm = 10 6 ohm

Case study: 2

Resistance is the opposition offered by the conductor to the flow of electric current. When two or more resistors are connected in series then electric current through each resistor is same but the electric potential across each resistor will be different. If R1, R2 and R3 are the resistance connected in series then current through each resistor will be I but potential difference across each resistor is V1, V2 and V3 respectively.

Thus, the total potential difference is equal to the sum of potential difference across each resistor. Hence, V= V1 + V2 + V3

Again, IR = IR1 + IR2 + IR3

Thus, R = R1 + R2 + R3

Hence in case of series combination of resistors, the total resistance is the sum of resistance of each resistor in a circuit.

Now, in case of parallel combination of resistors electric current through each resistor is different but the potential difference across each resistor is same. If resistors R1, R2 and R3 are connected in parallel combination then potential difference across each resistor will be V but current through each resistor is I1, I2 and I3 respectively.

Thus, total current through the circuit is the sum of current flowing through each resistor.

I = I1 + I2 + I3

Again, V/R= V/R1 + V/R2 + V/R3

Thus, 1/R = 1/R1 + 1/R2 + 1/R3

Hence, in case of parallel combination of resistors, the reciprocal of total resistance is the sum of reciprocal of each resistance connected in parallel.

 Questions:

1) In which case the equivalent resistance is more and why?

2) In our home, which type of combination of electric devices is preferred? Why?

3) If n resistors of resistance R are connected in parallel then what is the equivalent resistance?

1) In case of parallel combination of resistors the equivalent resistance is less than the individual resistance connected in parallel.

Since, 1/R = 1/R1 + 1/R2 + 1/R3 +….

2) At our home, we are connecting electrical devices in parallel combination because in parallel combination equivalent resistance is less and also we can draw an electric current according to the need of electric devices.

3) If n resistors of resistance R are connected in parallel then equivalent resistance is given by,

1/Re = 1/R + 1/R + 1/R +….n times 1/R

Thus, 1/Re = n/R

Hence, Re= R/n is the required equivalent resistance of the given combination.

Case study:3

When electric current flows through the circuit this electrical energy is used in two ways, some part is used for doing work and remaining may be expended in the form of heat. We can see, in mixers after using it for long time it become more hot, fans also become hot after continuous use. This type of effect of electric current is called as heating effect of electric current. If I is the current flowing through the circuit then the amount of heat dissipated in that resistor will be H = VIt

This effect was discovered by Joule, hence it is called as Joule’s law of heating.

Also, we can write, H = I 2 Rt

Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point.

In case of electric circuit, this heating effect is used to protect the electric circuit from damage.

The rate of doing work  or rate of consumption of energy is called as power. Here, the rate at which electric energy dissipated or consumed in an electric circuit is called as electric power. And it is given by P= VI

The SI unit of electric power is watt.

1) What is the SI unit of electric energy?

2) How heating effect works to protect electric circuit?

3) 1KW h = ?

4) If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb?

1) The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h.

2) In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

3) 1kW h = 3.6*10 6 joule

4) Given that,

V = 200V, I = 1A

Then, P = VI = 200*1 = 200 J/s = 200 W

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Case Study Questions Class 10 Science Chapter 12 Electricity

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Case study Questions Class 10 Science Chapter 12  are very important to solve for your exam. Class 10 Science Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 10 Science Chapter 12 Electricity

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 10 Science Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electricity Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Science  Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser, etc. all are based on the heating effect of current.

(i) What are the properties of heating elements? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point.

Answer: (b) Low resistance, high melting point

(ii) What are the properties of an electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

Answer: (a) doubled ​

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

Answer: (b) 2 times ​

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

Answer: (c) 750J ​

Question 2:

The relationship between potential difference and the current was first established by George Simon Ohm. This relationship is known as Ohm’s law. According to this law, the current passed through a conductor is proportional to the potential difference applied between its ends provided the temperature remains constant i.e. I ∝ V or V = IR where R is the constant for the conductor and it is known as the resistance of the conductor. Although Ohm’s law has been found valid over a large class of materials, there are some materials that do not hold Ohm’s law.

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm​

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

Case Study 3

3.1) The current passing through an electric kettle has been doubled. The heat produced will become : (a) half (b) double (c) four times (d) one fourth

Answer(c) four times

3.2) The heat produced in a wire of resistance ‘a’ when a current ‘b’ flows through it in time ‘c’ is given by : (a) a 2 bc (b) abc 2 (c) ab 2 c (d) abc

Answer(c) ab2c

3.3) What are the properties of heating element ? (a) high resistance, high melting point (b) low resistance, high melting point (c) low resistance, high melting point (d) low resistance, low melting point

Answer (a) high resistance, high melting point

3.4) Calculate the heat produced when 96,000 coulombs of charge is transferred in one hour through a potential difference of 50 volts. (a) 4788 J (b) 4788 kJ (c) 478 kJ (d) 478 J

Answer (b) 4788 kJ

3.5) Which of the following characteristic is not suitable for a fuse wire ? (a) thin and short (b) low melting point (c) thick and short (d) high resistance

Answer (c) thick and short

Case Study 4

Substance through which charges cannot pass is called insulators. Glass, pure water, and all gases are insulators. Insulators are also called dielectrics. In insulators, the electrons are strongly bound to their atoms and cannot get themselves freed. Thus, free electrons are absent in insulators. Insulators can easily be charged by friction. This is due to the reason that when an electric charge is given to an insulator, it is unable to move freely and remains localized. But this does not mean that conductors cannot be charged. A metal rod can be charged by rubbing it with silk if it is held in a handle of glass or amber

4.1) Calculate the current in a wire if a 1500 C charge is passed through it in 5 minutes. (a) 2 A (b) 5 A (c) 3 A (d) 4 A

Answer (b) 5 A

4.2) Electrons and conventional current flows in : (a) The same direction (b) The opposite direction (c) Any direction (d) Can’t say

Answer (b) The opposite direction

4.3) If the current passing through a lamp is 5 A, what charge passes in 10 second ? (a) 0.5 C (b) 3 C (c) 5 C (d) 50 C

Answer (d) 50 C

4.4) One-coulomb charge is equivalent to the charge contained in : (a) 6.2 × 10 19  electrons (b) 2.6 × 10 18  electrons (c) 2.65 × 10 19  electrons (d) 6.25 × 10 18  electrons

Answer (d) 6.25 × 1018 electrons

4.5) When an electric lamp is connected to 12 V battery, it draws a current of 0.5 A. The power of the lamp is :  (a) 0.5 W (b) 6 W (c) 12 W (d) 24 W

Answer (b) 6 W

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Science Chapter 12 Electricity with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 10 Science Electricity Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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CBSE 12th Standard Physics Subject Current Electricity Chapter Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams 

QB365 - Question Bank Software

Cbse 12th standard physics subject current electricity case study questions 2021.

12th Standard CBSE

Final Semester - June 2015

(ii) SI unit of electric current is

(iii) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor?

(v) When a current of 40 A flows through a conductor of area 10 m 2 , then the current density is

According to Ohm's law, the current flowing through a conductor is directly proportional to the potential difference across the ends of the conductor i.e  \(I \propto V \Rightarrow \frac{V}{I}=R\)  where R is resistance of the conductor Electrical resistance of a conductor is the obstruction posed by the conductor to the flow of electric current through it. It depends upon length, area of cross-section, nature of material and temperature of the conductor We can write  \(R \propto \frac{l}{A} \text { or } R=\rho \frac{l}{A}\)  where \(\rho\)  is electrical resistivity of the material of the conductor. (i) Dimensions of electric resistance is

(ii) If  \(1 \mu \mathrm{A}\)  current flows through a conductor when potential difference of2 volt is applied across its ends, then the resistance of the conductor is

(iii) Specific resistance of a wire depends upon

(iv) The slope of the graph between potential difference and current through a conductor is

(v) The resistivity of the material of a wire 1.0 m long, 0.4 mm in diameter and having a resistance of 2.0 ohm is

(ii) A cell of emf E and internal resistance r gives a current of 0.5 A with an external resistance of  \(12 \Omega\)  and a current of 0.25 A with an external resistance of  \(25 \Omega\)  .What is the value of internal resistance of the cell?

(iii) Choose the wrong statement.

(iv) An external resistance R is connected to a cell of internal resistance r, the maximum current flows in the external resistance, when

(v) IF external resistance connected to a cell has been increased to 5 times, the potential difference across the terminals of the cell increases from 10 V to 30 V. Then, the emf of the cell is

(ii) The total emf of the cells when n identical cells each of emf e are connected in parallel is

(iii) 4 cells each of emf 2 V and internal resistance of  \(1 \Omega\)  are connected in parallel to a load resistor of  \(2 \Omega\) . Then the current through the load resistor is

(iv) If two cells out of n number of cells each of internal resistance 'r' are wrongly connected in series, then total resistance of the cell is

(v) Two identical non-ideal batteries are connected in parallel. Consider the following statements. (i). The equivalent emf is smaller than either of the two emfs. (ii) The equivalent internal resistance is smaller than either of the two internal resistances

(ii) A current of 1.0 mA is flowing through a potentiometer wire of length 4 cm and of resistance  \(4 \Omega\)  .The potential gradient of the potentiometer wire is

(iii) Sensitivity of a potentiometer can be increased by

(iv) A potentiometer is an accurate and versatile device to make electrical measurements of EMF because the method involves

(v) In a potentiometer experiment, the balancing length is 8 m, when the two cells E l and E 2 are joined in series. When the two cells are connected in opposition the balancing length is 4 m. The ratio of the e. m. f. of two cells (E l /E 2 ) is

*****************************************

Cbse 12th standard physics subject current electricity case study questions 2021 answer keys.

(I) (b):  \(q=10^{6} \times 1.6 \times 10^{-19} \mathrm{C}=1.6 \times 10^{-13} \mathrm{C}\) t = 10 -3 s \(I=\frac{q}{t}=\frac{1.6 \times 10^{-13}}{10^{-3}}=1.6 \times 10^{-10} \mathrm{~A}\) (ii) (C): C S -1 (iii) (C): The current flowing through a conductor of non-uniform cross-section remain same in the whole of the conductor. (iv) (a): When a constant current is flowing through a conductor of non-uniform cross-section, electron density does not depend upon the area of cross section, while current density, drift velocity and electric field all vary inversely with area of cross-section. (v) (a): Given, I = 40 A ;A = 10m 2 \(\therefore\)  Current density,  \(J=\frac{I}{A} \text { or } J=\frac{40}{10}=4 \mathrm{~A} / \mathrm{m}^{2}\)

(i) (b) (ii) (a):   \(R=\frac{V}{I}=\frac{2}{10^{-6}}=2 \times 10^{6} \Omega\) (iii) (d): Specific resistance depends upon the nature of material and is independent of mass and dimensions of the material (iv) (a) (v) (d) : l = 1.0 m; D = 0.4 mm = 4 x 10 -4 m \(R=2 \Omega\) \(A=\frac{\pi D^{2}}{4}=\frac{\pi \times\left(4 \times 10^{-4}\right)^{2}}{4}=4 \pi \times 10^{-8} \mathrm{~m}^{2}\) Now,  \(\rho=\frac{R A}{l}=\frac{2 \times 4 \pi \times 10^{-8}}{1}=2.55 \times 10^{-7} \Omega \mathrm{m}\)

(i) (b) (ii) (b): As  \(I=\frac{\varepsilon}{R+r}\)   In first case, \(I=0.5 \mathrm{~A} ; R=12 \Omega\) \(0.5=\frac{\varepsilon}{12+r} \Rightarrow \varepsilon=6.0+0.5 r\)  ....(i) In second case  \(I=0.25 \mathrm{~A} ; R=25 \Omega\) \(\varepsilon=6.25+0.25 r\)  ...(ii) From equation (i) and (ii), r =  \(1 \Omega\) (iii) (b) (iv) (a): Current in the circuit  \(I=\frac{E}{R+r}\) Power delivered to the resistance R is \(P=I^{2} R=\frac{E^{2} R}{(R+r)^{2}}\) It is maximum when  \(\frac{d P}{d R}=0\) \(\frac{d P}{d R}=E^{2}\left[\frac{(r+R)^{2}-2 R(r+R)}{(r+R)^{4}}\right]=0\) \(\text { or } \quad(r+R)^{2}=2 R(r+R) \text { or } R=r\) (v) (b): For first case,  \(\frac{\varepsilon}{R+r}=\frac{10}{R}\)  ...(i)  For second case,  \(\frac{\varepsilon}{5 R+r}=\frac{30}{5 R}\) Dividing (i) by (ii), we get r = 5R From (i),  \(\frac{E}{R+5 R}=\frac{10}{R}\) E = 60 V

(i) (b) (ii) (a): Given,  \(I=1.0 \mathrm{~mA}=10^{-3} \mathrm{~A} ; R=4 \Omega ; L=4 \mathrm{~m}\) Potential drop across potentiometer wire V = IR = 10 -3 X 4 V Potential gradient,  \(k=\frac{V}{L}=\frac{4 \times 10^{-3}}{4}\) = 10 -3 V m -1 (iii) (a) (iv) (b): A potentiometer is an accurate and versatile device to make electrical measurements of EMF because the method involves a condition of no current flow through the galvanometer. It can be used to measure potential difference, internal resistance of a cell and compare EMF's of two sources. (v) (d): \(\frac{E_{1}}{E_{2}}=\frac{l_{1}+l_{2}}{l_{1}-l_{2}}=\frac{8+4}{8-4}=\frac{12}{4}=\frac{3}{1}\)

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Class 10 Science Chapter 11 Case Based Questions - Electricity

Case study - 1.

When electric current flows through the circuit this electrical energy is used in two ways, some part is used for doing work and remaining may be expended in the form of heat. We can see, in mixers after using it for long time it become more hot, fans also become hot after continuous use. This type of effect of electric current is called as heating effect of electric current. If I is the current flowing through the circuit then the amount of heat dissipated in that resistor will be H = VIt This effect was discovered by Joule, hence it is called as Joule’s law of heating. Also, we can write, H = I 2 Rt Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point. In case of electric circuit, this heating effect is used to protect the electric circuit from damage. The rate of doing work  or rate of consumption of energy is called as power. Here, the rate at which electric energy dissipated or consumed in an electric circuit is called as electric power. And it is given by P= VI The SI unit of electric power is watt.

Q1: What is the SI unit of electric energy? Ans:  The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h. Q2: How heating effect works to protect electric circuit? Ans:  In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

Q3: 1KW h = ? Ans: 1kW h = 3.6*10 6  joule   Q4: If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb? Ans:  Given that, V = 200V, I = 1A Then, P = VI = 200*1 = 200 J/s = 200 W

Case Study - 2

Resistance is the opposition offered by the conductor to the flow of electric current. When two or more resistors are connected in series then electric current through each resistor is same but the electric potential across each resistor will be different. If R1, R2 and R3 are the resistance connected in series then current through each resistor will be I but potential difference across each resistor is V1, V2 and V3 respectively. Thus, the total potential difference is equal to the sum of potential difference across each resistor. Hence, V= V1 + V2 + V3 Again, IR = IR1 + IR2 + IR3 Thus, R = R1 + R2 + R3 Hence in case of series combination of resistors, the total resistance is the sum of resistance of each resistor in a circuit. Now, in case of parallel combination of resistors electric current through each resistor is different but the potential difference across each resistor is same. If resistors R1, R2 and R3 are connected in parallel combination then potential difference across each resistor will be V but current through each resistor is I1, I2 and I3 respectively. Thus, total current through the circuit is the sum of current flowing through each resistor. I = I1 + I2 + I3 Again, V/R= V/R1 + V/R2 + V/R3 Thus, 1/R = 1/R1 + 1/R2 + 1/R3 Hence, in case of parallel combination of resistors, the reciprocal of total resistance is the sum of reciprocal of each resistance connected in parallel.

Q1: In which case the equivalent resistance is more and why? Ans: In case of parallel combination of resistors the equivalent resistance is less than the individual resistance connected in parallel. Since, 1/R = 1/R1 + 1/R2 + 1/R3 +…. Q2: In our home, which type of combination of electric devices is preferred? Why? Ans:  At our home, we are connecting electrical devices in parallel combination because in parallel combination equivalent resistance is less and also we can draw an electric current according to the need of electric devices. Q3: If n resistors of resistance R are connected in parallel then what is the equivalent resistance? Ans:  If n resistors of resistance R are connected in parallel then equivalent resistance is given by, 1/Re = 1/R + 1/R + 1/R +….n times 1/R Thus, 1/Re = n/R Hence, Re= R/n is the required equivalent resistance of the given combination.

Case study - 3

We can see that, as the applied voltage is increased the current through the wire also increases. It means that, the potential difference across the terminals of the wire is directly proportional to the electric current passing through it at a given temperature. Thus, V= IR Where R is the proportionality constant called as resistance of the wire. Thus, we can say that the resistance of the wire is inversely proportional to the electric current. As the resistance increases current through the wire decreases. The resistance of the conductor is directly proportional to length of the conductor, inversely proportional to the area of cross section of the conductor and also depends on the nature of the material from which conductor is made. Thus R= qL/A, where q is the resistivity of the material of conductor. According resistivity of the material they are classified as conductors, insulators and semiconductors. It is observed that the resistance and resistivity of the material varies with temperature. And hence there are vast applications of these materials based on their resistivity. The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

Q1: What is SI unit of resistivity? Ans:  The SI unit of resistivity is ohm meter. Q2: What is variable resistance? Ans: The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance. Q3: Why tungsten is used in electric bulbs? Ans:  Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high. Q4: 1M ohm = ? Ans: 1M ohm = 10 6  ohm 

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Electricity Case Study Based Questions Class 10

Students who are studying in CBSE class 10 board, need to get the knowledge about the Electricity Case Study Based Questions. Case based questions are generally based on the seen passages from the chapter Electricity. Through solving the case based questions, students can understand each and every concept. 

 With the help of Electricity Case Study Based Questions, students don’t need to memorise each answer. As answers for these case studies are already available in the given passage. Questions are asked through MCQs so student’s won’t take time to mark the answers. These multiple choice questions can help students to score the weightage of Electricity. 

Electricity Case Study Based Questions with Solutions 

Selfstudys provides case studies for the Class 10 Science chapter Electricity with solutions. The Solutions can be helpful for students to refer to if there is a doubt in any of the case studies problems. The solutions from the Selfstudys website are easily accessible and free of cost to download. This accessibility can help students to download case studies from anywhere with the help of the Internet. 

Electricity Case Study Based Questions with solutions are in the form of PDF. Portable Document Format (PDF) can be downloaded through any of the devices: smart phone, laptop. Through this accessibility, students don't need to carry those case based questions everywhere. 

Features of Electricity Case Study Based Questions

Before solving questions, students should understand the basic details of Electricity. Here are the features of case based questions on Electricity are:

• These case based questions start with short or long passages. In these passages some concepts included in the chapter can be explained.
• After reading the passage, students need to answer the given questions. These questions are asked in the Multiple Choice Questions (MCQ). 
• These case based questions are a type of open book test. These case based questions can help students to score well in the particular subject. 
•  These Electricity Case Study Based Questions can also be asked in the form of CBSE Assertion and Reason .

Benefits of Solving Electricity Case Study Based Questions

According to the CBSE board, some part of the questions are asked in the board exam question papers according to the case studies. As some benefits of solving Electricity Case Study Based Questions can be obtained by the students. Those benefits are: 

• Through solving case studies students will be able to understand every concept included in the chapter Electricity
• Passages included in the case study are seen passages, so students don’t need to struggle for getting answers. As these questions and answers can be discussed by their concerned teacher. 
• Through these students can develop their observation skills. This skill can help students to study further concepts clearly. 
• Case studies covers all the concepts which are included in the Electricity

How to Download Electricity Case Based Questions? 

Students studying in CBSE class 10 board, need to solve questions based on case study. It is necessary for students to know the basic idea of Electricity Case Study Based Questions. Students can obtain the basic idea of case based questions through Selfstudys website. Easy steps to download it are:

• Open Selfstudys website. 

• Bring the arrow towards CBSE which is visible in the navigation bar. 

• A pop-up menu will appear, Select case study from the list. 

• New page will appear, select 10 from the list of classes. 

• Select Science from the subject list. 

• And in the new page, you can access the Electricity Case Study Based Questions. 

Tips to solve Electricity Case Study Questions-

Students should follow some basic tips to solve Electricity Case Study Based Questions. These tips can help students to score good marks in CBSE Class 10 Science. 

• Generally, the case based questions are in the form of Multiple Choice Questions (MCQs). 
• Students should start solving the case based questions through reading the given passage. 
• Identify the questions and give the answers according to the case given. 
• Read the passage again, so that you can easily answer the complex questions. 
• Answer according to the options given below the questions provided in the Electricity Case Study Based Questions. 

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Case Study Chapter 12 Electricity

Please refer to Chapter 12 Electricity Case Study Questions with answers provided below. We have provided Case Study Questions for Class 10 Science for all chapters as per CBSE, NCERT and KVS examination guidelines. These case based questions are expected to come in your exams this year. Please practise these case study based Class 10 Science Questions and answers to get more marks in examinations.

Case Study Questions Chapter 12 Electricity

Case/Passage – 1

Two tungston lamps with resistances R1 and R2 respectively at full incandescence are connected first in parallel and then in series, in a lighting circuit of negaligible internal resistance. It is given that: R 1  > R 2 .

Question: Which lamp will glow more brightly when they are connected in parallel? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs 

Question: Which lamp will glow more brightly when they are connected in series? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs   

Question: If the lamp of resistance R 2 now burns out and the lamp of resistance R1 alone is plugged in, will the illumination increase or decrease? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) None 

Question: If the lamp of resistance R 1 now burns out, how will the illumination produced change? (a) Net illumination will increase (b) Net illumination will decrease (c) Net illumination will remain same (d) Net illumination will reduced to zero   

Question: Would physically bending a supply wire cause any change in the illumination? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) It is not possible to predict from the given datas 

Case/Passage – 2

The rate at which electric energy is dissipated or consumed in an electric circuit. This is termed as electric power,  P = IV, According to Ohm’s law V = IR  We can express the power dissipated in the alternative forms P =I 2 R=V 2 /R

If 100W – 220V is written on the bulb then it means that the bulb will consume 100 joule in one second if used at the potential difference of 220 volts. The value of electricity consumed in houses is decided on the basis of the total electric energy used. Electric power tells us about the electric energy used per second not the total electric energy. The total energy used in a circuit = power of the electric circuit × time.

Question: Which of the following terms does not represent electrical power in a circuit? (a) I 2 R (b) IR 2 (c) VI (d) V 2 /R 

Question: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in sereis and then in parallel in an electric circuit. The ratio of heat produced in series and in parallel combinations would be– (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1   

Question: In an electrical circuit, two resistors of 2Ω and 4Ω respectively are connected in series to a 6V battery. The heat dissipated by the 4Ω resistor in 5s will be (a) 5 J (b) 10 J (c) 20 J (d) 30 J   

Question: In an electrical circuit three incandescent bulbs. A, B and C of rating 40 W, 60 W and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness? (a) Brightness of all the bulbs will be the same (b) Brightness of bulb A will be the maximum (c) Brightness of bulb B will be more than that of A (d) Brightness of bulb C will be less than that of B     

Question: An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be– (a) 100 W (b) 75 W (c) 50 W (d) 25 W   

Case/Passage – 3

Answer the following questions based on the given circuit.

Question: The equivalent resistance between points A and B is (a) 7Ω (b) 6Ω (c) 13Ω (d) 5Ω 

Question: The potential drop across the 3Ω resistor is (a) 1 V (b) 1.5 V (c) 2 V (d) 3 V   

Question: The current flowing through in the given circuit is (a) 0.5 A (b) 1.5 A (c) 6 A (d) 3 A   

Case/Passage – 4

Answer the following questions based on the given circuit. 

Question: The current through each resistor is (a) 1 A (b) 2.3 A (c) 0.5 A (d) 0.75 A 

Question: The equivalent resistance between points A and B, is (a) 12 Ω (b) 36 Ω (c) 32 Ω (d) 24 Ω   

Question: The potential drop across the 12Ω resistor is (a) 12 V (b) 6 V (c) 8 V (d) 0.5 V 

Case/Passage – 5

Question: The equivalent resistance between points A and B (a) 6.2 Ω (b) 5.1 Ω (c) 13.33 Ω (d) 1.33 Ω 

Question: The current through the 4.0 ohm resistor is (a) 5.6 A (b) 0.98 A (c) 0.35 A (d) 0.68 A   

Question: The current through the battery is (a) 2.33 A (b) 3.12 A (c) 4.16 A (d) 5.19 A   

Case/Passage – 6

Question: The total resistance of the circuit is (a) 2 Ω (b) 4 Ω (c) 1.5 Ω (d) 0.5 Ω   

Question: The current flowing through 6Ω resistor is (a) 0.50 A (b) 0.75 A (c) 0.80 A (d) 0.25 

Question: The current flowing through 0.5Ω resistor is (a) 1 A (b) 1.5 A (c) 3 A (d) 2.5 A 

Case/Passage – 7

Ohm’s law gives the relationship between current flowing through a conductor with potential difference across it provided the physical conditions and temperature remains constant. The electric current flowing in a circuit can be measured by an ammeter. Potential difference is measured by voltmeter connected in parallel to the battery or cell. Resistances can reduce current in the circuit. A variable resistor or rheostat is used to vary the current in the circuit.

Question. Which type of conductor is represented by the graph given alongside?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these 

Question. What is the slope of graph in (i) equal to? (a) V (b) I (c) R (d) VI

Question. Which of the following is the factor on which resistance of a conductor does not depend? (a) Length (b) Area (c) Temperature (d) Pressur

Question. What type of conductor is represented by the following graph?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these

Question. What type of conductors are represented by the following graph?

Study this table related to material and their resistivity and answer the questions that follow.

Question. Which of the following is used in transmission wires? (a) Cr (b) Al (c) Zn (d) Fe

Question. Which is the best conducting metal? (a) Cu (b) Ag (c) Au (d) Hg

Question. Which of the following is used as a filament in electric bulbs? (a) Nichrome (b) Tungsten (c) Manganese (d) Silver

Question. What is the range of resistivity in metals, good conductors of electricity? (a) 10–8 to 10–6 Wm (b) 10–6 to 10–4 Wm (c) 1010 to 1014 Wm (d) 1012 to 1014 Wm

Question. Which property of the alloy makes it useful in heating devices like electric iron, toasters, immersion rods, etc.? (a) Higher resistivity (b) Do not oxidise at low temperature (c) Do not reduce at high temperature (d) Oxidise at high temperature

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Class 12 Physics Chapter 3 Case Study Question Current Electricity PDF Download

In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Physics Chapter 3 Current Electricity Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Current Electricity  to know their preparation level.

In CBSE Class 12 Physics Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Current Electricity Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 12 Physics  Chapter 3 Current Electricity

Case Study/Passage-Based Questions

Question 1:

(i) The terminal potential difference of two electrodes of a cell is equal to emf of the cell when

Answer: (b) I=0

(ii) A cell of emf E and internal resistance r gives a current of 0.5 A with an external resistance of 12Ω and a current of 0.25 A with an external resistance of 25Ω . What is the value of the internal resistance of the cell?

Answer:  (b) 1Ω 

(iii) Choose the wrong statement.

Answer: (b) Internal resistance of a cell decrease with the decrease in temperature of the electrolyte.

(iv) An external resistance R is connected to a cell of internal resistance r, the maximum current flows in the external resistance, when

Answer: (a) R = r

(v) IF external resistance connected to a cell has been increased to 5 times, the potential difference across the terminals of the cell increases from 10 V to 30 V. Then, the emf of the cell is

Answer: (b) 60V

Question 2:

Answer: (b) It measures the emf of a cell very accurately

(iii) Sensitivity of a potentiometer can be increased by

Answer: (a) decreasing potential gradient along the wire.

(iv) A potentiometer is an accurate and versatile device to make electrical measurements of EMF because the method involves

Answer: (b) a condition of no current flow through the galvanometer

(v) In a potentiometer experiment, the balancing length is 8 m, when the two cells E l  and E 2  are joined in series. When the two cells are connected in opposition the balancing length is 4 m. The ratio of the e. m. f. of two cells (E l /E 2 ) is

Answer: (d) 3: 1

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Physics Chapter 3 Current Electricity with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 12 Physics Current Electricity Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible.

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6th Class Science Electricity and Circuits Question Bank

Done electricity and circuits total questions - 20.

A) Brighter than normal               done clear

B) Normal             done clear

C) Dimmer than normal              done clear

D) Very dim (cannot be seen)         done clear

A) PQRS              done clear

B) SRQP                          done clear

C) PRQS              done clear

D) SQRP.             done clear

question_answer 3) In which of the following circuits will the bulb glow?                              

A) P                                  done clear

B) Q                     done clear

C) R                                  done clear

D) S                      done clear

A) P only              done clear

B) R only              done clear

C) R and S        done clear

D)        Q and S                       done clear

A) Wool                done clear

B) Graphite          done clear

C)        Iron                  done clear

D) Silver                           done clear

A) (i) Carbon and Manganese dioxide mixture, (ii) Carbon rod, (iii) Ammonium chloride                done clear

B) (i) Carbon rod, (ii) Ammonium chloride, (iii) Zinc container                  done clear

C) (i) Ammonium chloride, (ii) Zinc container, (iii) Carbon rod           done clear

D) (i) Zinc container, (ii) Carbon rod, (iii) Carbon and Manganese dioxide mixture                           done clear

A) It is used to connect the bulb to the other components in the circuit. done clear

B) It provides the energy for the bulb to glow.                              done clear

C) It can break a circuit, interrupting the current or diverting it from one conductor to another.               done clear

D) It measures the current in a circuit. done clear

A) (i) and (ii)                    done clear

B) (iii) and (iv)     done clear

C) (i) and (iii)       done clear

D) (ii) and (iii) done clear

A) You will get electrocuted if you handle live wires. done clear

B) The rubber glove is a good insulator to protect us from electric shocks. done clear

C) The rubber gloves control the electric current flow. done clear

D) Electricity cannot flow through rubber. done clear

A) Replace one cell with a piece of chalk. done clear

B) Replace one cell with a piece of wire. done clear

C) Replace one bulb with a piece of wire. done clear

D) Replace one bulb with another cell. done clear

question_answer 13) Fuse is the most important safety device, used for protecting the circuits due to   

A) Short circuiting and overloading     done clear

B) Overloading and earthing           done clear

C) Earthing only                       done clear

D) Short circuiting, overloading and earthing,                          done clear

A) Crayon                          done clear

B) Comb               done clear

C)         Key                   done clear

D)          Piece of stone                     done clear

A) A similar battery is connected between the probes in same order. done clear

B) A similar battery is connected between the probes in reverse order. done clear

C) A similar bulb is connected between the probes. done clear

D) None of these done clear

A) R will glow bright but Q and P will be dim. done clear

B) P, Q and R all will glow equally bright. done clear

C) Q and R will immediately burnout. done clear

D) P will glow bright, but Q and R will be dim. done clear

A) Only bulb X will glow   done clear

B) Only bulb Y will glow done clear

C) Both bulbs X and Y will glow done clear

D) Neither bulb X nor bulb Y will glow. done clear

A) The bulb X will glow first done clear

B) The bulb Y will glow first done clear

C) The bulbs Z and X will glow first done clear

D) All the bulbs will glow simultaneously. done clear

A) Only bulb R will glow. done clear

B) Only bulb P will glow. done clear

C) Both bulbs P and R will glow. done clear

D) None of the bulbs will glow. done clear

A) Only (i) and (ii) done clear

B) Only (ii) and (iii) done clear

C) Only (i) and (iii) done clear

D) All (i), (ii) and (iii) done clear

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Class 6 Science Chapter 12 Electricity and Circuits Extra Questions

CBSE Class 6 Science Chapter 12 Electricity and Circuits Extra Questions and Answers is available here. Students can learn and download the PDF of these questions for free. These extra questions and answers are prepared by our expert teachers as per the latest NCERT textbook and guidelines. Learning these extra questions will help you to score excellent marks in the final exams.

Electricity and Circuits Class 6 Science Extra Questions and Answers

Very short answer questions.

1. What is the direction of flow of current in a dry cell? Answer: The current flows in closed circuit from +ve to -ve terminal of cell.

2. Name the +ve terminal of dry cell. Answer: Carbon rod with a metal cap on it.

3. Name the -ve terminal of a dry cell. Answer: Zinc metal plate.

4. What is dry cell? Answer: It is a device which converts chemical energy into electrical energy.

5. What is solar cell? Answer: A device which converts solar energy into electrical energy.

6. What is open circuit? Answer: An electric circuit in which electrical contact at any point is broken is called open circuit.

7. Write one use of insulators. Answer: Insulators are used in making switchboard, handles of testers, screw drivers.

8. What is the name of thin wire in the electric bulb? Answer: Filament.

9. In which of the following circuits A, B and C given below, the cell will be used up very rapidly?  

Answer: I circuit A the cell will get used up rapidly.

10. Figure given below shows a bulb with its different parts marked as 1, 2, 3, 4 and 5. Which of them label the terminals of the bulb?  

Answer: Labels 3 and 4 mark the terminals of the bulb.

11. What is inside the glass case of a bulb? Answer: Filament

12. Why does an electric bulb not glow when both the wires are connected to the same terminal of a cell?

Answer: The current flows from one terminal to the other. When both wires are connected to same terminal, current will not flow.

13. Give one difference between a cell and a battery.

Answer: A cell produces electricity by chemical reactions taking place in it whereas battery is made up of two or more cells joined together.

14. Write any two uses of electric cells. Answer: It is used in alarm clocks and wrist watches

Short Answer Type Questions

1. You are provided with a bulb, a cell, a switch and some connecting wires. Draw a diagram to show the connections between them to make the bulb glow.

2. Will the bulb glow in the circuit shown below? Explain.  

Answer: No. The bulb will not glow in this circuit because the switch is open and the circuit is incomplete. Due to this there will be no current flow.

3. An electric bulb is connected to a cell through a switch as shown in figure below. When the switch is brought in ‘ON’ position, the bulb does not glow. What could be the possible reason/s for it? Mention any two of them.

Answer: The reasons could be

• the bulb is fused.
• the cell is a used one.
• break in connecting wire.
• loose connections. (Any two)

4. A torch requires 3 cells. Show the arrangement of the cells, with a diagram, inside the torch so that the bulb glows. 

5. When the chemicals in the electric cell are used up, the electric cell stops producing electricity. The electric cell is then replaced with a new one. In case of rechargeable batteries (such as the type used in mobile phones, camera and inverters), they are used again and again. How?

Answer: Rechargeable batteries can be recharged by providing them appropriate current.

6. Paheli connected two bulbs to a cell as shown in the figure given below.

She found that filament of bulb B is broken. Will the bulb A glow in this circuit? Give reason.

Answer: No, bulb A will not glow as the circuit is not complete.

7. Why do bulbs have two terminals?

Answer: Bulbs have two terminals to connect the filament with the circuit so that the current can pass through it and get accumulated.

8. Which of the following arrangements A, B, C and D given below should not be set up? Explain, why.

Answer: Arrangement A is not desirable and should not be set up. This will exhaust the cell very quickly.

9. A fused bulb does not glow. Why?

Answer: In a fused bulb the filament is broken and the circuit is incomplete. The current will not flow and the bulb will not glow.

10. Paheli wanted to glow a torch bulb using a cell. She could not get connecting wires, instead, she got two strips of aluminium foil. Will she succeed? Explain, how.

Answer: Yes. Aluminium foils can act as connecting wires as it is a good conductor of electricity.

11. What are the essential components of an electric circuit?

Answer: Connecting wires, bulb, switch and cell are the essential components of an electric circuit.

12. Why should you not touch electric appliances and switches with wet hands?

Answer: Water is a good conductor of electricity and current passes very quickly through wet hands. Therefore, it can give us an electric shock.

13. Write two precautions that you must follow while handling electricity.

• Wear rubber gloves or slippers.
• Never touch switches with wet hands.

14. When a bulb is fused, it does not light up. Explain why?

Answer: When a bulb is fused its filament breaks, which in turn breaks the electric circuit. Thus, current does not pass through it.

15. Why does a cell stop producing electricity after sometime?

Answer: When the chemicals in the electric cell are used up, the electric cell stops producing electricity. Since all the chemicals are used, no chemical reaction takes place which will produce energy.

16. Write any two uses of electricity?

Answer: Two uses of electricity are:

• To operate pumps that left water from wells or from ground level to the roof top tank.
• Electricity makes it possible to light our homes, that makes our tasks easier.

17: What will happen if we join two terminals of electric cell directly through a wire?

Answer: If we join two terminals of electric cell directly through a wire, the chemical in an electric cell get used up very fast and the cell will stop working.

18: What is an electric circuit?

Answer: Arrangement that provides a complete path for electricity to pass (current to flow) is known as electric circuit.

19: Explain how the bulb glows in circuit when it is connected to an electric cell?

Answer: When the terminals of the bulb are connected with that of electric cell by wires, a current pass through the filament of the bulb and it makes the bulb glow.

20: How can an electric bulb get fused?

Answer: An electric bulb may get fused due to break in its filament. A break in the filament of bulb means break in the path of current between the terminals of electric cell.

21: Why rubbers and plastic used to cover electric wires and plug tops?

Answer: Rubbers and plastic are used to cover electric wires and plug tops because they are insulators and do not allow electric current to pass when we touch plugs and switches.

22: Why is distilled water used in the batteries and not the tap water?

Answer: Because distilled water acts as an insulator as it is purest water. Whereas tap water has salts and impurities and acts as a conductor.

23: Why is handle of tools like screw driver, pliers are covered with plastic or rubber?

Answer: Because they are insulators and avoid direct contact with electric current while touching electric wires.

24: Generally what can be components of an electric circuit?

Answer: Components of electric circuit can be following:

a. A cell or a battery

b. Connecting wires made of copper or aluminium

d. Bulb or other electric device

25: What is the function of an electric switch?

Answer: It is a simple device that either breaks the circuit or completes it. For example-in a microwave or toy cars we have switches to turn it ON or OFF.

26: What is an electric cell? How many terminals it has?

Answer: An electric cell is a device that converts chemical energy into electrical energy. It has two metal plates indicating two terminals-negative and positive. It has chemical inside it.

27: Statement is true or false: a. Bulb has one terminal. b. Electrolyte is present in the cell. c. Copper is good conductor of electricity. d. Circuit shows the path of current.

Answer: a. False b. True c. True d. True.

28: Classify the following as conductor or insulator: Eraser, coin, glass, pencil, needle, key, iron nail, plastic scale

Answer: Conductor-coin, needle, key, iron nail Insulator-eraser, glass, pencil, plastic scale

29. Mention two advantages of a dry cell. Answer:

1. It converts chemical energy into electrical energy. 2. It is light and small in size.

30. Draw the circuit diagram for operating a bulb with the help of a dry cell. Answer:

31. Define conductors and insulators. Give one example of each.

Answer: A conductor is that which easily allows the passage of current through it. Example: Aluminium or any metal. An insulator is that which does not allow the passage of current through it. Example: Rubber.

32. Identify conductors and insulators from the following: Eraser, paper, matchstick, copper wire, pencil lead, polythene Answer: Conductors: Copper wire, pencil lead. Insulator. Eraser, paper, matchstick, polythene.

33. Name the scientist who invented electric cell and the scientist who invented electric bulb. Answer: Electric cell: Alessandro Volta. Electric bulb: Thomas Alva Edison.

34. Give one activity to prove that air is an insulator.

Answer: Take an electric circuit, keep the terminals unconnected in the air. The bulb will do not glow, as air is an insulator and does not allow the current to flow through it.

35. In any electric circuit, when the switch is on and the current flows through it why do the wire, switches, bulb or devices become hot?

Answer: This is because electric energy changes into heat energy.

36.The headlights of a car have reflectors behind the bulb. What is the function of reflectors?

Answer: The reflector helps in reflecting the light into a focussed area.

Long Answer Type Questions

1: (a)What is electric circuit? (b) How many types of electric circuit are there? Define them. (c)Draw a diagram to show the closed circuit for switch, bulb and dry cell.

Answer: (a) An electrical circuit is a path or line through which an electrical current flows.

(b)There are two types of electric circuit:

• Open electric circuit
• Closed electric circuit

Open electric circuit:  The circuit in which electrical contact at any point is broken is called open electric circuit.

Closed electric circuit:  The circuit in which electric current flows from one terminal of a cell or battery to the other is called a closed circuit.

2. How can you explain that the human body is a good conductor of electricity?

Answer: If we stand barefoot on the ground and touch an electric wire, we will get an electric shock. This is because human body is a good conductor of electricity. Without slippers, current can easily pass through.

3. Why should we take care while handling electricity?

Answer: Carelessness in handling electricity and electric devices can cause servers injuries and sometimes even death, so we should take proper care while handling electricity.

4: What is the difference between conductor and an insulator? Explain with examples.

Answer: Materials that allow electric current to pass through them are called conductors. For example iron, copper etc. Metals are good conductors. Materials that do not allow electric current to pass through them are called insulators. For example- rubber, plastic etc

5. If you touch an electric wire carrying current you get a shock, but if on the same wire the birds sit they do not get any shock/current. Explain why?

Answer: When we hold the wire carrying current then the circuit is closed and the current flows from our body and enters earth but the birds sitting on the same wire do not get any current as the circuit is not complete. If the bird touches the earth wire, it will also die due to electric shock.

6. Boojho has a cell and a single piece of connecting wire. Without cutting the wire in two, will he be able to make the bulb glow? Explain with the help of a circuit diagram.

Answer: Yes, using the arrangement given below he can succeed in getting the bulb glow.

7. Figures A and B, show a bulb connected to a cell in two different ways. What will be the direction of the current through the bulb in both the cases. (Q to P or P to Q)

Answer: In Fig. (A) Q to P and in Fig. (B) P to Q.

8. A torch is not functioning, though contact points in the torch are in working condition. What can be the possible reasons for this? Mention any three.

Answer: The possible reasons could be

• the bulb may be fused.
• the cells may have been used up.
• the cells are not placed in the correct order.
• the switch is faulty. ( Any three )

9. Distinguish between the following.

(i) Insulator and Conductor Answer:

(ii) Open circuit and Closed circuit Answer:

(iii) Open switch and Closed switch Answer:

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Case Study Questions for Class 7 Science Chapter 14 Electric Current and Its Effects

• Last modified on: 10 months ago
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[Download] Case Study Questions for Class 7 Science Chapter 14 Electric Current and Its Effects

Here we are providing case study or passage-based questions for class 7 science chapter 14 Electric Current and Its Effects.

Case Study/Passage Based Questions

We use various appliances based on heating effect of current in our daily life. Some such appliances are electric room heater, electric heater, hot plates, electric iron, hair dryers etc. All these contain a coil of wire called element.

• The coil of wire in an electric bulb is known as (a) element (b) filament (c) conducting wire (d) None of these
• Which part of an electrical iron becomes hot and gives out heat? (a) the switch (b) the handle (c) the element (d) the connecting wire
• The amount of heat produced depends upon which of the following? (a) material of element (b) length of wire (c) thickness of wire (d) All of these

Related Posts

What is case study question for class 7 science.

Case study or passage-based questions in class 7 Science typically require students to read a given scenario or passage and answer questions based on the information provided. These questions assess students’ comprehension, analytical thinking, and application of scientific concepts. 

Best Ways to Prepare for Case Study Questions

To develop a strong command on class 6 Science case study questions, you can follow these steps:

• Read the textbook and study materials:  Familiarize yourself with the concepts and topics covered in your class 6 Science curriculum. Read the textbook thoroughly and take notes on important information.
• Practice analyzing case studies:  Look for case studies or passages related to class 6 Science topics. Analyze the given information, identify key details, and understand the context of the situation.
• Develop comprehension skills:  Focus on improving your reading comprehension skills. Practice reading passages or articles and try to summarize the main points or extract relevant information. Pay attention to details, vocabulary, and the overall structure of the passage.
• Understand scientific concepts:  Ensure that you have a solid understanding of the scientific concepts discussed in class. Review the fundamental principles and theories related to each topic.
• Make connections:  Try to connect the information provided in the case study to the concepts you have learned in class. Identify any cause-effect relationships, patterns, or relevant scientific principles that apply to the situation.
• Practice critical thinking:  Develop your critical thinking skills by analyzing and evaluating the information given in the case study. Think logically, consider multiple perspectives, and draw conclusions based on the evidence provided.
• Solve practice questions:  Look for practice questions or sample case study questions specifically designed for class 6 Science. Solve these questions to apply your knowledge, practice your analytical skills, and familiarize yourself with the format of case study questions.
• Seek clarification:  If you come across any challenging concepts or have doubts, don’t hesitate to ask your teacher for clarification. Understanding the underlying principles will help you tackle case study questions effectively.

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