Please refer to Chapter 12 Electricity Class 10 Science Important Questions with solutions provided below. These questions and answers have been provided for Class 10 Science based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 10 Science for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 10.

## Class 10 Science Important Questions Electricity Chapter 12

**Very Short Answer:**

**Question: Draw a schematic diagram of an electric circuit consisting of a battery of two cells each of 1.5 V, 5 Ω, 10 Ω and 15 Ω resistors and a** **plug key, all connected in series. Answer:** The required circuit diagram is as below :

**Question: Out of 60 W and 40 W lamps, which one has a higher electrical resistance when in use?Answer:** We know that : P= V

^{2}/R

Power is inversely proportional to the resistance.

∴ 40 W has a higher electrical resistance.

**Question: Why is a series arrangement not used for connecting domestic electrical appliances in a circuit?Answer: (1) In series arrangement, same current will **

**flow through all the appliances, which is not**

**required for domestic electric circuit.**

**(2) Total resistance of domestic circuit will be**

**sum of the resistance of all appliances and hence**

**current drawn by the circuit will be less.**

**(3) We cannot use independent on/off switches**

**with individual appliances.**

**(4) All appliances are to be used simultaneously**

**even if we do not need them.**

**Question: What is an electric circuit? Distinguish between an open and a closed circuit.Answer: **Electric circuit : Closed and continuous path of an electric current is called an electric circuit.

**Short Answer: I **

**Question: The electrical resistivity of silver is 1.60 × 10– ^{6} Ω m. What will be the resistance of a silver wire of length 10 m and crosssectional**

**area 2 × 10–**

Answer:Given, electrical resistivity of silver,

^{3}m^{2}?Answer:

P = 1.6 × 10–

^{6}Ω m

Length of silver wire, l = 10 m

Area of cross-section, A = 2 × 10–

^{3}m

^{2}

As we know,

resistance of wire is given by

R= P,l/A

Putting the values of l, P and A, we get

R= 1.6X10-

^{6}X10/2X10-

^{3}= 8.0X10-

^{3}Ω

**Short Answer: II **

**Question: (a) Define the term “volt”.****(b) State the relation between work, charge and potential difference for an electric circuit.****Calculate the potential difference between the two terminals of a battery if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to the other.Answer:** (a) ‘Volt’ is the SI unit of electric potential difference, it is defined as the amount of work done in moving a unit positive charge from one point to another.

(b) The relationship between work (W), charge (Q) and potential dfference (V) for an electric circuit is given by V =W/Q

Given : W = 100 J, Q = 20 C, V = ?

Using the relation, V=W/Q= = 100/20 =5 V So, 5 V of potential difference between the two terminals to transfer 20 C of charge when work done is 100 J.

**Question: Two resistors, with resistances 5 Ω and 10 Ω respectively are to be connected to a battery** **of emf 6 V so as to obtain :****(i) minimum current flowing****(ii) maximum current flowing****(a) How will you connect the resistances in each case?****(b) Calculate the strength of the total current in the circuit in the two cases.Answer:** (a) In order to make the flow of minimum current in the circuit, we can connect 5 Ω and 10 Ω resistors in series. And to get maximum flow of current in the circuit,we can connect 5 Ω and 10 Ω resistors in parallel. 15

**Long Answer:**

**Question: Name an instrument that measures electric current in a circuit, define the unit of electric current.Answer:** Ammeter is an instrument that measures electric current in the circuit. One ampere is defined as the flow of one coulomb of charge per second. i.e., 1 A = 1 C s–1.

**Question: What is meant by saying that the potential difference between two points is 1 volt? Name a device that helps to measure the potential difference across a conductor?Answer:** When we say that the potential difference between two points is 1 V, it means that 1 J of work is being done to move a unit charge between that two points. Voltmeter is a device that helps to measure the potential difference.

**Question: What do the following symbols mean in circuit diagrams?Answer: **

**Question: An electric circuit consisting of a 0.5 m long nichrome wire XY, and an ammeter, a voltmeter, four cells of 1.5 V each and a plug** **key was set up.****(i) Draw a diagram of this electric circuit to study the relation between the potential difference maintained between the points ‘X’ and ‘Y’ and the electric current flowing through XY,****(ii) Following graph was plotted between V and I values : 9What would be the values of V/1 ratios when the potential difference is 0.8 V, 1.2 V and 1.6 V respectively? What conclusion do you draw from these values?Answer:** (i) Electric circuit to study the relation between potential difference and the current.

**Question: Electrical resistivities of some substances at****20°C are given below:****Silver 1.60 × 10– ^{8} Ω m**

**Copper 1.62 × 10–**

^{8}Ω m**Tungsten 5.20 × 10–**

^{8}Ω m**Iron 10.0 × 10–**

^{8}Ω m**Mercury 94.0 × 10–**

^{8}Ω m**Nichrome 100 × 10–**

^{8}Ω m**Answer the following questions in relation to them :**

**(i) Among silver and copper, which one is a better conductor? Why?**

**(ii) Which material would you advise to be used in electrical heating devices? Why?**

Answer:(i) Silver is the best conductor of electricity because of low resistivity.

Answer:

(ii) Nichrome should be used in electrical heating devices, due to very high resistivity. It has a high resistance and produces a lot of heat on passing current.

**Question: (A) Two lamps rated 100 W, 220 V and 10 W, 220 V are connected in parallel to 220 V supply. Calculate the total current through the circuit.****(B) Two resistors X and Y of resistances 2 Ω and 3 W respectively are first ??oined in parallel and then in series. In each case the voltage supplied is 5 V.****(i) Draw circuit diagrams to show the combination of resistors in each case.****(ii) Calculate the voltage across the 3 W resistor in the series combination of resistors. ****Answer:** (A) Resistance of first lamp R_{1} = V^{2}/P

= 220 x220/100= 484 Ω

Resistance of second lamp R_{2} = V^{2}/P

= 220 x220/10 = 4840 Ω

Since the two lamps are connected in parallel, the equivalent resistance is given by:

(B) Resistance of Resistor X(R_{1}) = 2Ω

Resistance of resistor Y(R_{2}) = 3 Ω

V = 5V

**Question: Draw a schematic diagram of a circuit consisting of a battery of 3 cells of 2V each, a combination of three resistors of 10 W, 20****W and 30 W connected in parallel, a plug key and an ammeter, all connected in series. Use this circuit to find the value of the following:****(A) Current through each resistor****(B) Total current in the circuit****(C) Total effective resistance of the circuit****Answer:** Schematic diagram of a circuit Battery of 3 cells of 2V each so the potential difference

V = 3 × 2 = 6 V

**Question:(A) A 6 Ω resistance wire is doubled on itself.Calculate the new resistance of the wire.****(B) Three 2 Ω resistors A, B and C are connected in such a way that the total resistance of the combination is 3 Ω. Show the arrangement of the three resistors and justify your answer. ****Answer:** (A) R = 6 Ω

When the wire of length l and cross-sectional area A is doubled on itself, its length becomes ½ and cross-section area becomes 2 A.

R = P l/A

R’ =P(l/2)

=1/4[pl/A]

=1/4x6x6 as R=6Ω=[pl/A]

R’ =1.5 Ω

New resistance of the wire will be 1.5 W

(B) To get a total resistance of 3 W from three resistors A, B and C, two resistors of resistances 2 W each should be connected in parallel. Their equivalent resistance is :

1/R=1/R1+1/R2

1/R=1/2+1/2=2/2=1 Ω

R=1 Ω

This combination of equivalent resistance 1 Ω should be connected in series with the resistor of resistance 2 W. So that the equivalent resistance becomes 1 Ω + 2 Ω =3 Ω

**Question:** **How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery? ****Answer:** That the same current flows through every part of the circuit containing three resistances in series connected to a battery can be concluded with the help of the following experiment:

(1) Take three resistors R_{1}, R_{2} and R_{3} and connect these resistances in series with an ammeter, key and a battery of known voltage as shown in the figure given below.** **

(2) Plug the key. Note the ammeter reading.

(3) Switch the position of the ammeter between R_{1} and R_{2}. Note the reading by closing and opening the key.

(4) Switch the position of the ammeter between R_{2} and R_{3} and note the reading by closing and opening the key.

(5) Switch the position of the ammeter after R_{3} and note the reading by closing and opening the key.**Observations:**

We will observe that the value of the current in the ammeter is the same, independent of its position in the electric circuit. It means that in a series combination of resistors, the current is the same in every part of the circuit or the same current through each resistor. We will find that the reading in each case will be the same in all cases.

**Question:** **(A) Define electrical energy with S.I. unit?****(B) A household uses the following electric appliance;****(i) Refrigerator of rating 400w for ten hour each day.****(ii) Two electric fans of rating 80w each for twelve hours each day.****(iii) Six electric tubes of rating 18w each for six hours each day.****Calculate the electricity bill of the household for the month of ??une if the cost per unit of electric energy is ` 3.00. ****Answer:** (A) The work done by a surce of electricity to maintain current in a circuit is known as electrical energy. Its S.I. unit is Joule.

(B) (i) Electricity consumed by refrigerator

= Power × Time

= 400 × 10

= 4000 Wh

= 4 kWh

(ii) Electricity consumed by two fans

= Power × Time

= 80 × 2 × 12

= 1920 Wh

= 1.92 kWh

(iii) Electricity consumed by six electric tubes

= 6 × 18 × 6

= 648 h

= 0.648 kWh

Total energy consumed in one day

= 4 + 1.92 + 0.648

= 6.548 kWh

Total energy consumed in one month

= 6.548 × 30

= 197.04 kWh

Cost of 1 unit (kWh) = Rs. 3.00

Cost of 197.04 kWh = 197.04 × 3.00

Total electricity bill = `Rs.591.12

**Question:****Two identical resistors, each of resistance 10W, are connected in (i) series and then (ii) in parallel, in line to a battery of 6 volts.****Calculate the ratio of power consumed in the combination of resistor in the two case.****Answer:** (i) Total resistance in series

= R_{1} + R_{2}

= 10 + 10

= 20 Ohm

Let us now calculate the power

PS=V^{2}/R

=6X6/20

=1.8 W

Similarly, we calculate power consumed in parallel connection.

(ii) Total resistance in parallel

=1/R_{1}+1/R_{2}

1/R_{P}=1/10+1/10

R_{P}=1/10+1/10

R_{P}=5 ohm

Let us now calculate the power

R_{P} = V^{2}/R

36/5

=7.2

Ratio of power consumed

= Power is Series/Power inParallel

1.8/7.2

= 0.25

Or = 1:4

**Question: Consider the scale of a voltmeter shown in the diagram and answer the following questions: **

**(A) What is the least count of the voltmeter?****(B) What is the reading shown by the voltmeter?****(C) If this voltmeter is connected across a resistor of 20 W, how much current is** **flowing through the resistor?****Answer:** (A) Least count of the voltmeter = 1 5/10 V = 0.15 V

(B) Reading shown by the voltmeter = 1.5 + 2 × 0.15 = 1.8 V

(C) If R = 20 Ω V = 1.8 V, current I = V/R=1.8/20 A = 0.09 A.

**Question:(A) What are the advantages of connecting electrical devices in parallel with the** **battery instead of connecting them in series?****(B) Two resistors of 20W and 40W are connected in parallel in an electric circuit.****Compare the current passing through the two resistors?****Answer:** (A) Advantages of connecting electrical devices in parallel are:

(i) There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.

(ii) The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.

(iii) If one appliance fails to work, others will continue to work properly if they are connected in parallel combination.

(B) Here, for parallel combination the potential difference will be same.

Let the potential difference =V

Then the current passing through the resistance 20 ohm will be,

I_{1} = V/20.

The current passing through the resistance 40 ohm will be,

I_{2} = V/40

Clearly we see that, I_{1} = 2I_{2}

Hence current in 20 Ω is double as compared to the current in 40Ω resistor.

**Question: What would you suggest to a student if while performing an experiment he finds****that the pointer??needle of the ammeter and voltmeter do not coincide with the zero****marks on the scales when circuit is open? No extra ammeter??voltmeter is available in the laboratory.****Answer:** As pointer of both the ammeter and voltmeter do not coincide with the zero marks on the scales when circuit is open, it indicates zero error in both the instruments. This zero error should be subtracted from the readings taken

when circuit is closed.

To get correct readings using these instruments, first we should find out the Least counts of both the instruments, i.e., the minimum value which can be accurately measured.

Zero error = Initial reading (in open circuit) = Pointer reading (in open circuit) X Least count To get the actual reading, subtract zero error or the initial reading of the open circuit to the reading when you perform the experiment.

Actual reading = Final reading – Initial reading (zero error)

**Question: (A) What is meant by the series combination and parallel combination of resistances? **

**(B) In the circuit diagram given below five resistances of 5 Ω, 20 Ω, 15 Ω, 20 Ω and** **10 Ω, are connected as given in figure to a 6 V battery:****Calculate total resistance in the circuit.****Answer:** (A) Resistors connected in series: In a series combination of resistors the current is the same in every part of the circuit or the same current flows through each resistor, i.e., there is only one path for the flow of current.

When several resistors are ??oined in series, the resultant resistance of the combination RS equals the sum of their individual

resistances, R_{1}, R_{2}, R_{3}.

R_{S} = R_{1} + R_{2} + R_{3}**Parallel Combination of resistors:** In a parallel circuit each resistor is placed in its own separate branch. A parallel circuit provides multiple paths for the current to flow.

The reciprocal of the equivalent resistance of a group of resistors joined in parallel is equal to the sum of the reciprocals of the individual resistors.

1/R_{P}=1/R_{1}+1/R_{2}+1/R_{3 }

**Question:While studying the dependence of potential difference (V) across a resistor on the current (I) passing through it, in order to determine the resistance of the resistor, a student took 5 readings for different values of current and plotted a graph between V and I. He got a straight line graph passing through the origin. What does the straight line signify? Write the method of determining resistance of the resister using this graph.****Answer:** The graph obtained after plotting the values of V and I is a straight line passing through the origin which shows that the potential difference (V) and the current flowing (I) vary linearly with each other thereby verifying Ohm’s Law.**Further,** I = 0 when V = 0, as the graph passes through the origin.

Resistance of the resistor can be determined by finding the slope of the VI graph as we know from Ohm’s Law that V = IR, where R is the resistance of the resistor. \ R = V/I

**Question: Consider the following circuit: **

**What would be the readings of the ammeter and the voltmeter when key is closed?****Give reason to Justify your answers.****Answer:** The three resistors are connected in series. The effective resistance of 3 resistors R_{1}, R_{2} and R_{3 }connected in series is given by.

The effective resistance = 5+ 8 + 12 = 25 Ohm

Total current flowing I in the circuit is given by

Ohm’s law = I = V/R=6/25 A.

As the current flowing through the resistors connected in series is same, the current flowing through the Ammeter = I = 0.24 A.

Reading of ammeter = 0.24 A.

The potential difference V is equal to the sum of the potential difference V_{1}, V_{2} and V_{3} across the individual resistors R_{1}, R_{2} and R_{3} respectively.

That is, V = V_{1} + V_{2} + V_{3}

As the voltmeter is connected across the 12 Ohm resistor or R_{3}, we will calculate the potential difference V_{3} across R_{3}.

Using Ohm’s law, V_{3} = IR_{3} = 0.24 × 12V = 2.88V.

Reading of Voltmeter = 2.88 V

**Question: (A) On what factors does the resistance of a conductor depend?****(B) Give one example to show how the resistance depends on the nature of material of the conductor.****(C) Calculate the resistance of an aluminium cable of length 10 km and diameter 2.0 mm if the resistivity of aluminium is 2.7 x 10 ^{–8} Ω m.**

**Answer:**(A) Resistance of a conductor depends on the following factors:

(1) Length of the conductor

(2) Area of cross section of the conductor

(3) Nature of material of the conductor

(4) Temperature of the conductor

(B) If we take two similar wires of same length and same diameter, one of copper metal and other of nichrome alloy, we will find that the resistance of nichrome wire is about 60 times more than that of the copper wire. This shows that the resistance depends on the nature of material of the conductor.

(C) l = 10 km = 10000 m; d = 2 mm; r = 1 mm = 10

^{–3}m; r = 2.7 × 10

^{–8}W m

R = r l/A = 2.7 × 10

^{–8}× 10000/(3.14×(10

^{−3})

^{2})

= 0.859 × 102 W

= 86 W

**Question: Determine the following quantities for the circuits shown below: **

**(A) the equivalent resistance****(B) the total current from the power supply****(C) the current through each resistor****(D) the voltage drop across each resistor and****(E) the** power dissipated in each resistor**Answer:** Given circuit diagram:

(A) **Equivalent resistance:** Resistances in series.

R = R_{1} + R_{2} + R_{3}

= 20 Ω + 30 Ω + 50 Ω

= 100 Ω

(B)** Total current from the power supply:** Total current is determined by the voltage of the power supply and the equivalent resistance

of the circuit.

I = V/R = 125/100

125 = 1.25 A

(C) Current through each resistor: Current is constant through resistors connected in series.

(D) Voltage drop across each resistor

V1 = IR_{1} = 1.25 A × 20 Ω = 25 V

V2 = IR2 = 1.25 A × 30 Ω = 37.5V

V3 = IR3 = 1.25 A × 50Ω = 62.5 V

In a series circuit they should equal the voltage increase of the power supply.

V_{total} = V_{1} + V_{2} + V_{3}

125 V = 25 V + 37.5 V +62.5 V

125 V = 125 V

(E) The power dissipated in each resistor

P1 = V_{1} × I_{1}

= 25 V X 1.25 A = 31.250 Ω

P2 = V_{2} × I_{2}

= 37.5V X 1.25 A = 46.875 Ω

P_{3} = V_{3} × I_{3}

= 62.5 V X1.25 A = 78.125 Ω

In a series circuit, the element with the greatest resistance consumes the most power.

**Question: (i) Consider a conductor of resistance ‘R; length ‘L;, thickness‘d?? and resistivity ;5 Now this conductor is cut into four equal parts. What will be the new resistivity of** **each of these parts? Why?****(ii) Find the resistance if all of these parts are connected in:****(A) Parallel****(B) Series****(iii) Out of the combinations of resistors mentioned above in the previous part, for** **a given voltage which combination will consume more power and why?****Answer: **(i) Resistivity will not change as it depends on the nature of the material of the conductor. (ii) The length of each part becomes L/4. p,A constant.** **

**Question: (A) In a given ammeter, a student saw that needle indicates 12th division in ammeter while performing an experiment to verify Ohm??s law. If ammeter has 10 divisions between 0 to 0.5 A, then what is the ammeter reading corresponding to 12th division?****(B) How do you connect an ammeter and a voltmeter in an electric circuit?****Answer:** (A) Least count of ammeter = 0.5/10 = 0.05 A

Thus, value corresponding to 12 divisions = 0.05 ×12 = 0.6 A

(B) An ammeter is connected in series and a voltmeter is connected in parallel in an electric circuit.

**Question: (A) Name an instrument that measures electric current in a circuit. Define unit of electric current.****(B) How is static electricity different from current electricity?****(C) Mention some important uses of electricity in our daily life.****Answer:** (A) Ammeter is used to measure electric current in a circuit.

The SI unit of current is Ampere.

Current through a conductor is said to be one ampere, when one coulomb of charge flows through any section of a conductor in one second.

(B) Electricity is basically categorised into static and current electricity. The difference between the two is as follows:

**(C) Electricity has a wide range of applications:**

(1) Electricity is used in heating, lighting, cooking, operating fans, geysers etc. in our houses, hospitals, hotels, offce buildings, food storage plants.

(2) Electricity is used in industries to run various types of machines.

(3) In transportation, to pull electric trains, trams etc.

(4) Electricity is used for irrigation purposes in agriculture.

**Question: In the circuit given below, the resistors R _{1}, R_{2 }and R_{3} have the values 10 W, 20 W and 30 Ω respectively, which have been connected to a battery of 12 V. Calculate:**

**(A) the current through each resistor,**

**(B) the total circuit resistance, and**

**(C) the total current in the circuit.**

**Answer:**

**Question: You have following material:****An ammeter (0-1A), a voltmeter (0-3V), a resistor of 20 W, a key, a rheostat, a battery of 3 V and seven connecting wires.****Using this material draw a labelled circuit diagram to study the dependence of** **potential difference (V) across a resistor on the current (I) passing through it.****Answer:** Labelled circuit diagram to study the dependence of potential difference (V) across a resistor on the current (I) is drawn below: