Worksheets Chapter 12 Electricity Class 10 Science

Students should refer to Worksheets Class 10 Science Electricity Chapter 12 provided below with important questions and answers. These important questions with solutions for Chapter 12 Electricity have been prepared by expert teachers for Class 10 Science based on the expected pattern of questions in the class 10 exams. We have provided Worksheets for Class 10 Science for all chapters on our website. You should carefully learn all the important examinations questions provided below as they will help you to get better marks in your class tests and exams.

Electricity Worksheets Class 10 Science

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

Question. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is −
(a) 1/25 
(b) 1/ 5
(c) 5
(d) 25 

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

Question. Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/ R 

Question. How many electrons constitute a current of 1 ampere?
(a) 6*10¹⁵
(b) 6*10¹⁸
(c) 1.6*10^-19 
(d) 6*10²⁰   

Question. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
(a) 1/5 Ω
(b) 10 Ω
(c) 5 Ω
(d) 1 Ω 

Question. Electric current is measured by
(a) A voltmeter
(b) An ammeter
(c) A rheostat
(d) A potentiometer   

Question. The resistivity does not change if
(a) The material is changed
(b) The temperature is changed
(c) The shape of the resistor is changed
(d) Both material and temperature are changed 

Question. Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have
(a) Same current flowing through them when connected in parallel
(b) Same current flowing through them when connected in series
(c) Same potential difference across them when connected in series
(d) Different potential difference across them when connected in parallel   

Question. Assertion (A): A cell converts chemical energy into electrical energy.
Reason(R): A cell maintains a potential difference across its terminals due to chemical reactions. 
(a) Both A and R are true and R is the correct explanation for A
(b) Both A and R are true and R is not the correct explanation for A
(c) A is true but R is false
(d) A is false but R is true.

Question. Assertion (A): If a graph is plotted between potential difference and current a linear graph is obtained.
Reason(R): current is directly proportional to the potential difference.
(a) Both A and R are true and R is the correct explanation for A
(b) Both A and R are true and R is not the correct explanation for A
(c) A is true but R is false
(d) A is false but R is true.

Very Short Answer Type Questions :

Question. Mention one reason why tungsten is used for making filament of electric lamp.
Answer :  Tungsten is used for making filament because of its high melting point and low resistivity.

Question. Why is tungsten metal selected for making filaments of incandescent lamps?
Answer :  Tungsten has high resistance and high melting point.

Question. Which device helps to maintain a potential difference across a conductor?
Ans: Cell or battery

Question. Why are heating elements made of alloys rather than metals?
Ans: High resistivity, does not oxidise at high temperatures

Question. Calculate the amount of charge flowing in a wire if it draws a current of 2A in 10 minutes.
Ans: 2 x 10 x 60
= 1200 C

Question. Draw a circuit diagram having the following components
a. Bulb
b. A two cell battery
c. Ammeter
d. A closed key
Ans: Refer diagram 12.1

Question. What happens to resistance of a conductor if area of cross-section is doubled?
Ans: It halves

Question. a)What is the function of fuse in an electric circuit?
b) What would be the rating of the fuse for an electric kettle which is operated at 220V and consumes 500 W power?
c) How is the SI unit of electric energy related to its commercial unit? 
Ans. b) 2.2A flows through the circuit, fuse should be rated 3A.
c) 1 KWh = 3.6 X 106 J

Question. What do we mean when we say that potential difference between two points is 1volt?
Ans: Definition

Question. What are the advantages of connecting electrical devices in parallel with the battery rather than in series?
Ans: (1) The current required by each device is different which is possible only in parallel.
(2)Potential difference is constant for all devices.
(3) Total resistance in the circuit is decreased

Question. If three resistors of 6Ω, 9Ω and 21Ω are connected in series to a 12V battery, find
a) The total resistance of the circuit.
b) The current flowing through the circuit.
c) The potential difference across the 21 Ω resistor.
Ans: a) 36 Ω
b) 0.33A
c) 6.93V

Question. a) State Ohms law. Give the graphical relation between V & I.
b) An electric oven rated at 500W is connected to a 220V line and used for 2 hours daily. Calculate the cost of electric energy per month at the rate of Rs.5 per KWh.
Ans: a) Ohms law state that current flowing in aconductoris directly proportional tothe applied potentialprovided that temperature and physical conditions remainssame.
b) Energy consumed per day = 1 KWh (P x t) cost for 30 days = 1 X 5 X 30 = Rs.150.00

Question. A wire of resistance 5 W is bent in the form of a closed circle. What is the resistance between two points at the ends of any diameter of the circle?
Answer :  R = 5 W
It can be assumed as two semi circular elements are connected across diameter AB.
Total resistances of couple wire 5 W . Resistance of two semicircular elements = 5/2 Ω 
Both the elements are connected in parallel. 
1/R , = /R1 + 1/R2 + = 2/5 + 2/5 = 4/5
or R’ = 5/4 Ω

Question is much less, heat generated in long electric cables than in filaments of electric bulbs?
Answer :  As heat produced = I2Rt
Electric cables are made of thick good conductor where as filaments are made of thin tungsten wire whose resistances is high. The resistance of electric cables is very less. So heat produced in cable is much lesser than filaments.

Question. State which has a higher resistance a 50 W or a 25 W lamp bulb and how many times?
Answer : 
We have P1 = 50W
P2 = 25W
Let their resistances are R1 and R2 respectively.

i.e. 50 W bulbs has 1/2 resistance than 25 W bulb.

Question. Define resistivity of a material.
Answer :  We have R =P,l/A
Resistivity, P = R , A/l
If
A = 1, l = 1 then r = R i.e. resistivity of a material is the resistance of a conductor of this material whose length and area of cross- section both are unity.
or
Resistivity of a material is the resistance of a conductor of this material whose volume is unity.

Question. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer. The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal. 

Question. What does an electric circuit mean?
Answer. An electric circuit consists of electric devices, switching devices, source of electricity, etc. that are connected by conducting wires.

Question. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer. The energy given to each coulomb of charge is equal to the amount of work required to move it. The amount of work is given by the expression,

Where,
Charge = 1 C
Potential difference = 6 V
Work Done = Potential difference X 
Therefore, 6 J of energy is given to each coulomb of charge passing through a battery of
6 V.

Question. A cylinder of a material is 10 cm long and has a crosssection of 2 cm². If its resistance along the length be 20 ohm, what will be its resistivity in number and units?
Answer :  l = 10 cm, A = 2 cm²
R = 20 ohm
R = p,L/A
or P = R,A/l = 20 x 2/10 Ω cm
r = 4 W – cm

Question. (a) Name the instrument/device used to measure electric current in a circuit.
(b) How is an ammeter connected in a circuit to measure current flowing through it?
Answer : 
a. Ammeter is used to measure electric current.
b. Ammeter is connected in series in an electric circuit.

Question. In an electric circuit, state the relationship between the direction of conventional current and the direction of flow of electrons.
Answer :  Electrons flows from negative terminal to positive terminal where as current flows from +ve terminal to -ve terminal in external circuit i.e. Conventional current and electrons flow are opposite to each other.

Question. Calculate the number of electrons that would flow per second through the cross- section of a wire when 1 A current flows in it.
Answer :  Given: I = 1A, t = 1 s
I = Q/t ⇒ Q = I x t
Q = 1A#1 s (Q = 1 coulomb)
But Q = ne
n = Q/e = 1C/1.6 X 10^-19
= 6.25 X 10¹⁸ electrons

Question. (a) What material is used in making the filament of an electric bulb?
(b) Name the characteristics which make it suitable for this.
Answer : 
a. Tungsten is used in making filament.
b. Its high resistivity and high melting point. 

Question. In the circuit diagram shown, the two resistance wires A and B are of same length and same material, but A is thicker than B. Which ammeter A1 or A2 will indicate higher reading for current? Give reason.

Answer :  Length of A and B is same. A is thicker than B.
Hence RA < RB [∴ R ∝ 1/A]
Current in A1 is more than current in A2 i.e., reading in A1 is higher than reading in A2.

Question. Tungsten is used almost exclusively for filaments of electric bulb. List two reasons.
Answer :  (1) high resistivity (2) high melting point and does not oxidise at very high temperature.

Question. How does the resistivity of alloys compare with those of pure metals from which they may have been formed?
Answer :  The resistivity of pure metals is lesser than resistivity of alloys with which these alloys are made.

Question. Write SI unit of resistivity.
Answer :  Ohm-m

Short Answer Type Questions :

Question. Out of the two wires X and Y shown below, which one has greater resistance? img
Answer: Wire ‘Y’ has greater resistance as it has more length than wire ‘X’. It is because resistance of wire is directly proportional to the length of wire.

Question. A 9Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.
Answer:

Question. An electric iron has a rating of 750 W, 220 V. Calculate the (i) current flowing through it, and (ii) its resistance when in use.
Answer:

Question. (a) What do the following circuit symbols represent?

(b) The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. Find the resistance of heater when in use.
Answer:
(a) (i) Wires crossing without touching each other.
(ii) Rheostat/Variable resistor
(b) Given: V = 60 V, I = 4 A, R = ?
From Ohm’s law, V = IR
60 = 4 x R = 15 Ω

Question. Find the current flowing through the following electric circuit.

Answer:

Question. Draw a schematic diagrams of an electric circuit comprising of 3 cells and an electric bulb, ammeter, plug-key in the ON mode and another with same components but with two bulbs in parallel and a voltmeter across the combination.
Answer:

Question. Two identical wires one of nichrome and other of copper are connected in series and a current (I) is passed through them. State the change observed in the temperatures of the two wires. Justify your answer. State the law which explains the above observation.
Answer: The resistivity of nichrome is more than that of copper so its resistance is also high. Therefore, large amount of heat is produced in the nichrome wire for the same current as compared to that of copper wire. Accordingly, more change in temperature is observed in the nichrome wire. This is explained by Joule’s law of heating.
Joule’s law of heating: It states that the amount of heat produced in a conductor is

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 J of work is required to transfer 20 C of charge from one terminal of the battery to the other.
Answer:
(a) When 1 joule of work is done in carrying 1 coulomb of charge, from infinity to a point in the electric field, then potential at that point is called 1 volt.
(b) Potential difference, V = Work done on unit charge =W/q Work is 100 J,q=20C
Potential difference,V=W/q=100/20=5V

Question. An electric bulb is rated at 60 W, 240 V. Calculate its resistance. If the voltage drops to 192 V, calculate the power consumed and the current drawn by the bulb. (Assume that the resistance of the bulb remain unchanged.)
Answer:

Question. A torch bulb is rated 2.5 V and 750 mA. Calculate (i) its power, (ii) its resistance and (iii) the energy consumed, if this bulb is lighted for four hours.
Answer:

Question. Series arrangements are not used for domestic circuits. List any three reasons.
Answer: Series arrangements are not used for domestic circuit because
1. The electrical appliances need current of widely different values to operate properly.
2. In series arrangement, when one component fails, the circuit is broken and none of the components works.
3. All electrical appliances work at a constant voltage. But in series circuit, the current is constant throughout the electric circuit and potential is different across the different components. So, series arrangement is not suitable for domestic circuits.

Question. Name the physical quantity which is (i) same (ii) different in all the bulbs when three bulbs of:
(a) same wattage are connected in series.
(b) same wattage are connected in parallel.
(c) different wattage are connected in series.
(d) different wattage are connected in parallel.
Answer: (a) For identical bulbs in series- same current, same potential difference.
(b) For identical bulbs in parallel- same potential difference, different current.
(c) For unidentical bulbs in series- same current, different potential difference.
(d) For unidentical bulbs in parallel- different current, same potential difference.

Question.(a) How is the direction of electric current related to the direction of flow of electrons in a wire?
(b) Calculate the current in a circuit if 500 C of charge passes through it in 10 minutes.
Answer:
(a) Conventional direction of electric current is opposite to the direction of flow of electrons in a wire.
(b) q = 500 C, t = 10 X 60 = 600 s
I = 500/600 = 5/6A

Question. An electric bulb of resistance 200Ω draws a current of 1 Ampere. Calculate the power of the bulb the potential difference at its ends and the energy in kWh consumed burning it for 5h. 
Answer: Power of the bulb,

Question. A circuit is shown in the diagram given below.
(a) Find the value of R.
(b) Find the reading of the ammeter.
(c) Find the potential difference across the terminals of the battery.

Answer:
(a) Potential difference across 6Ω = 12 V
.’. Current through 6 Ω,

Question. Consider the circuit shown in the diagram. Find the current in 3Ω resistor.
Answer:

Question. Two resistors with resistances 5Ω and 10 Ω are to be connected to a battery of emf 6 V so as to obtain:
(i) minimum current
(ii) maximum current
(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) As current is inversely proportional to resistance for the same voltage. So, to get maximum current, the equivalent resistance has to be less. This means the resistors must be connected in parallel. To get minimum current, the equivalent resistance has to be greater as

Question. (a) Define the term ‘coulomb’.
(b) State the relationship between the electric current, the charge moving through a conductor and the time of flow.
Calculate the charge passing through an electric bulb in 20 minutes if the value of current is 200 mA.
Answer:

(a) When 1 A current flows across the wire in 1 second, the charge transfer across its ends is said to be 1 coulomb.
(b) The relationship between the electric current I, the charge q and time t is

Question. Two devices of rating 44 W, 220 V and 11 W, 220 V are connected in series. The combination is connected across a 440 V mains. The fuse of which of the two devices is likely to burn when the switch is ON? Justify your answer.
Answer:

Question. Five resistors are connected in a circuit as shown. Find the ammeter reading when circuit is closed.

Answer:

Question. Study the circuit shown in which three identical bulbs B1, B2 and B3 are connected in parallel with a battery of 4.5 V.
(i) What will happen to the glow of other two bulbs if the bulb B3 gets fused?
(ii) If the wattage of each bulb is 1.5 W, how much reading will the ammeter A show when all the three bulbs glow simultaneously?
(iii) Find the total resistance of the circuit.

Answer:
(i) The glow of other two bulbs remains the same.
(ii) In parallel combination, potential difference across each resistance will remain same.|

Wattage of each bulb = 1.5 W
Therefore, when all the three bulbs glow simultaneously, the ammeter A shows 1 A reading.
(iii) Potential difference across each bulb = 4.5 V Wattage of each bulb = 1.5 W .’. Resistance of

Question. Two bulbs A and B are rated as 90W–120V and 60W–120V respectively. They are connected in parallel across a 120V source. Find the current in each bulb.
Which bulb will consume more energy?
Answer :  First Bulb: 90 W–120 V

so first bulb will consume more energy.

Question. Draw the nature of V–I graph for a nichrome wire. (V -Potential difference, I -Current)
A metallic wire of 625 mm length offers a 4 W resistance. If the resistivity of the metal is 4.8 X 10-7
ohm-metre, then calculate the area of cross-section of the wire.
Answer :  V–I graph for nichrome wire

Given: l = 625 mm = 0.625 m
R = 4 W
r = 4.8 X 10-7 Ohm-m
A = ?
R = P,l/A or A = P,l/R
A = 4.8 X 10-7 X 0.625/4
= 0.75 X 10-7 m2

Question. Derive the relation R = R₁ + R₂ + R₃ when three resistors R1, R2 and R3 are connected in series in an electric circuit.
Answer :: Three resistors R1, R2 and R3 are connected in series.
Therefore current in each resistor is same. Let current in the circuit is I.

Applied total potential = Sum of the potentials across
each resistor
V = V1+ V2+ V3
IReq = IR1+ IR2+ IR3
Req = R1+ R2+ R3

Question. (a) Nichrome wire of length L and radius R has resistance of 10 Q. How would the resistance of the wire change when:
(i) Only length of the wire is doubled?
(ii) Only diameter of the wire is doubled? Justify your answer.
(b) Why element of electrical heating devices aremade-up of alloys?
Answer : 
a. R = 10 W of length L, and area of cross-section A.
i. When only length is doubled Rl = 20 W
ii. When only diameter is doubled.

b. Alloys have high resistivity more than their constituent pure metals and do not oxide at high temperature easily.

Question. Resistivity of two elements A and B are = 1.62X10-8 Wm and 520#10-8 Wm respectively.
Out of these two, name the element that can be used to make:
a. filament of electric bulb.
b. wires for electrical transmission lines. Justify your answer in each case.
Answer :  Given: r A = 1.62X10-8 Ωm
r B = 520X10-8 Ωm
a. For filament of electric bulb resistivity must be higher i.e. 520X10-8 Ωm . So element B is used for filament.
b. For electric transmission resistivity must be lower which is of element A.

Question. State Ohm’s law. Calculate the resistance of a conductor, if the current flowing through it is 0.2 A when the applied potential difference is 0.8 V.
Answer :  If the physical conditions of a conductor are kept same then current is directly proportional to the potential difference across the ends of the conductor V ? I .
V = RI
I = 0.2 A
V = 0.8 Volt.
R = ?
R = V/I = 0.8/0.2 Ω
R = 4 Ω

Question. (a) Why are copper or aluminium wires generally used for electrical transmission and distribution purposes?
(b) Two wires, one of copper and other of manganin, have equal lengths and equal resistances. Which wire is thicker? Given that resistivity of copper is lower than that of manganin.
Answer : 
a. Copper or aluminium wires are used for transmission and distribution of electricity due to their low resistivity and high conductivity.
b. We know that R = P,l/A
r ∝ A
Thicker the wire, more the resistivity. The resistivity of manganin is more than copper. So manganin wire is thicker than copper.

Question. Two conducting wires of same material, equal length and equal diameter are connected in series. How does the heat produced by the combination of resistance change?
Answer :  Let the resistances of two wires are R each.
Heat produced by individual resistor 87

Question. A nichrome wire has a resistance of 10 W Find the resistance of another nichrome wire, whose length is three times and area of cross-section four times the first wire.
Answer :  We have R = 10 Ω
l1 = l
A1 = A
R = P,l/A = 10 Ω
or l/A = 10/P
For new wire,
L2 = 3l
A2 = 4A
R2 = P,3l/4A = 3/4 P,l/pΩ
R2 = 3/4 R = 3/4 X 10 = 15/2 Ω

Question. Define electric current and state its SI unit. With the help of Ohm’s law explain the meaning of 1 Ohm resistance.
Answer :  Electric current may be defined as the rate of flow of charge through a circuit. Its SI unit is Ampere.
According to Ohm’s law V = IR
R = V/I
If V = 1 Volt, I = 1A then R = 1 ohm.
A conductor has a resistance of 1 Ohm if a current of one amp flows through it when a potential difference of 1 volt is applied across it.

Question. (a) What is the total resistance of n resistors each of resistance ‘R’ connected in: (i) series? (ii) parallel?
(b) Calculate the resultant resistance of 3 resistors 3 W, 4 W and 12 W connected in parallel.
Answer : 
a. In series combination

Question. (a) State Ohm’s law. Express it mathematically.
(b) Write symbols used in electric circuits to represent:
(i) variable resistance.
(ii) voltmeter.
(c) An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, what will be the power consumed?
Answer : 
a. Ohm’s Law states that if the physical conditions of a conductor are kept constant then current passing through a conductor is directly proportional to the potential difference across its ends.
P = 100/4 Ω
P = 25W

Question. (a) How is the direction of electric current related to the direction of flow of electrons in a wire?
(b) Calculate the current in a circuit, if 500 C of charge passes through it in 10 minutes.
Answer :  a. The direction of flow of electrons is opposite to the direction of conventional current.
b. Q = 500 C, t = 10 mts = 10 X 60 = 600 s 
I = Q/t = 500/600 A
I = 0.83 A

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. Potential difference b/w two points in an electric field is said to be 1 volt if the amount of work done in bringing a unit positive charge from one point to another point is 1 J.
b. Given: W = 100 J, Q = 20 C, V = ?
As V = W/Q ⇒ V = 100/20 JC-I
V = 5 JC-1
V = 5 Volt.

Question. State the formula co-relating the electric current flowing in a conductor and the voltage applied across it. Also, show this relationship by drawing a graph.
What would be the resistance of a conductor, if the current flowing through it is 0.35 ampere when the potential difference across it is 1.4 volt?
Answer : 
V = IR
i.e., V ? I
If we plot a graph b/w V and I , it is straight line.
Graph b/w V and I :

Given: I = 0.35 A
V = 1.4 Volt.
R = V/I = 1.4/0.35Ω
R = 4 Ω

Long Answer Type Questions :

Question. (a) In the circuit shown connect a nichrome wire of length “L” between points X and Y and note the ammeter reading.
(i) When this experiment is repeated by inserting another nichrome wire of the same thickness but twice the length (2L), what changes are observed in the ammeter reading?
(ii) State the changes that are observed in the ammeter reading if we double the area of cross-section without changing the length in the above experiment. Justify your answer in both the cases.
(b) “Potential difference between points A and B in an electric field is 1V”. Explain the statement.

Answer : 
a. (i) The resistance of two times long wire also becomes two times so current decreases in the circuit.
(ii) If area of the nichrome wire is doubled then its resistance decreases and hence current increases in the circuit.
b. Potential difference b/w A and B is 1 volt means that 1 J of work is to be done in moving a unit positive charge (+ 1C) from point A to B.

Question. (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 ohm?
(b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 ohm?
Answer : (a) Given:V = 220 V
R = 1200 ohm
As V = IR
220 = Ix1200 or I = 220/1200A
I = 0.18 A
(b) V = 220 V
R = 100 ohm
I = V/R = 220/100 A
I = 2.2 A

Question. The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw, if the potential difference is increased to 120 V?
Answer : 
V = 60 V, I = 4A

Question. Resistance of a metal wire of length 1 m is 26 W at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature?
Answer : 
We have l = 1m
R = 26 ohm
t = 20cC
r = 0.15 mm
r = ?
As

Question. Draw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 W resistor, a 10 W resistor and a 15 W resistor, an ammeter and a plug key; all connected in series. Also, connect a voltmeter to record the potential difference across the 15 W resistor and calculate:
a. the electric current passing through the above circuit and
b. potential difference across 5 W resistor when the key is closed.
Answer : 

b. P.D. across 5 W resistor V = IR
V = 0.33X5 = 1.65 Volt.
V = 1.65 Volt.

Question. Draw a labelled circuit diagram showing three resistors R1, R2 and R3 connected in series with a battery (E), a rheostat (Rh), a plug key (K) and an ammeter (A) using standard circuit symbols. Use this circuit to show that the same current flows through every part of the circuit. List two precautions you would observe while performing the experiment.
Answer :  By changing the position of ammeter, measure the current in it. Every time the magnitude of current is found same i.e., in series combination the current in every part of the circuit, remains same. 96
Precaution:
a. Connect the ammeter in series with battery.
b. All connection must be tight.
c. Current must not be greater than the range of ammeter.

Question. (a) Derive the formula for the calculation of work done when current flows through a resistor.
(b) One electric bulb is rated 40 W and 240 V and other 25 W and 240 V. Which bulb has higher resistance and how many times?
Answer :  a. Let R is the resistance of the resistor in which I current is passed for a time t by applying a p.d. ( V ) across the resistor.

Question. (a) What is an electric circuit?
(b) Calculate the number of electron that flow per second to constitute a current of one ampere.
Charge on an electron is 1.6#10-19 C.
(c) Draw an electric circuit for studying Ohm’s law.
Label the circuit component used to measure electric current and potential difference.
Answer : 
a. A continuous path in which current can flow when switch i6 plugged in.

Question. Define power. State the difference between 1 watt and 1 watt hour. Establish the relationship between unit of electric energy and SI unit of energy. An electric heater rated 1000 W/220 V operates 2 hours daily.
Calculate the cost of energy to operate for 30 days at the rate of Rs 5.00 per kWh.
Answer :  Electric Power: P is defined as the rate at which electric energy is consumed in an electric circuit.
1 watt is the power consumed by an electric appliance that carries 1A current when a potential difference of 1 volt is applied across it. Whereas 1 Wh. is the unit of electric energy, which is consumed by an electric appliance of 1 W power in one hour.
Commercial unit of electric energy is kWh.
1 kWh = 1000Wx60x60 s
1 kWh = 3.6×106 J
Energy consumed by heater in one month
= 1000Wx2hx30 days = 60 kWh
= 60 units
Total cost @₹5/-kWh = 5×60 = ₹300

Question. A wire is 1.0 m long, 0.2 mm in diameter and has a resistance of 10 W . Calculate the resistivity of its material.
Answer :  L = 1.0 m
D = 0.2 mm

Question. In the circuit shown below, calculate:
a. total resistance in arm CE,
b. total current drawn from the battery, and
c. current in each arm, i.e., AB and CE of the circuit.

Answer : 
a. Total resistance in arm CE
= (2 + 4)W = 6 Ω
b. CE arm and AB arm in parallel.

c. Resistance of CE and AB arms is same = 6 Ω
Current in each arm = 3/6 = 1/2 A
I = 0.5 A

Question. (a) Why is the series arrangement not used for domestic circuits?
(b) Why is the tungsten used almost exclusively for filament of electric lamps?
(c) Why are the conductors of electric heating devices such as bread toasters and electric irons made of an alloy rather than a pure metal?
(d) Why are copper and aluminium wires usually employed for electricity transmission?
(e) Why does the cord of an electric heater not glow while the heating element does?
Answer : 
a. In the series arrangement if any device fails to work, circuit will be broken and other devices stop working. In series combination current remains same. Where as different devices requires different current.
b. Tungsten is used due to its high resistivity and high melting point.
c. Alloys have higher resistivity than their pure constituent and hence produces more heat.
d. Copper and aluminium are good conductors and have low values of resistivity so these are used for electric transmission.
e. Cords core made up of good conductor with low resistivity where as heating elements are made up of alloys of higher resistivity to produce more heat.

Question. (a) Name an instrument that measures electric current in a circuit. Define the unit of electric current.
(b) What do the following symbols mean in circuit diagrams?

(c) An electric circuit consisting of a 0.5 m long nichrome wire XY, 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:

What would be the values of V/I 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 : 
a. Ammeter: Ampere is the unit of current. If one coulomb charge flows in a circuit in 1 s then the current in the circuit will be 1 Ampere. (A)
b. (i) Variable resistor (ii) closed key

Question. (a) (i) What is meant by saying that the potential difference between two points is 1 volt?
(ii) How much energy is given to 5 coulomb of charge passing through a 12 V battery?
(b) Describe an activity with necessary electric circuit drawn to study the factors on which the resistance of a conducting wire depends.
Answer : 
a. (i) Potential difference b/w the two points is 1 volt mean if we bring a unit positive charge (+1C) from one point to other point then amount of work done is 1 J, W = VQ.
(ii) Given: Q =5 C, V =12 V, W =12 x 5 = 60 J
b. Activity to study the factors of affecting resistance of the conductor—Take a cell, an ammeter, manganin wires of different length and different area of cross- section and key. Connect
an ammeter, a manganin wire of (say) 10 cm long to a cell through a plug key.
(i). Now closed the key and note the current in the circuit with the help of ammeter.
(ii). Now replace the manganin wire with another manganin wire of twice the area of crosssection and again measure the current in the circuit with the help of ammeter.
(iii).Now repeat the activity by taking a copper wire of same length same area of cross-section, and note down the current in each case. You will find that current in the circuit in each
case is different. Which shows that resistance depends upon (1) nature of material (2) length of the conductor and (3) area of crosssection.

Question. (a) What do the following symbols represent in a circuit? Write the name and one function of each.

(b) Draw a schematic diagram of a circuit consisting of a battery of 12 V, three resistors of 5 W , 10 W and 20 W connected in parallel, an ammeter to measure the total current through the circuit, a voltmeter to measure the potential difference across the combination of resistors.
(c) State any one advantage of connecting electrical devices in parallel with the mains instead of connecting them in series in a household circuit.
Answer : 
a. (i) Variable resistor used to change the current in circuit.
(ii) Wires crossing each other which are not connected together. It is used when large number of connections are to be made with the help of wires out joining them.

c. (i) Let the devices are connected in series if one devices fails to work, the circuit will be broken and all devices stop working, which is not in the case of parallel combination. In parallel combination if one device fails others keep working.
(ii) In parallel combination the voltage will be same across each device.

Question. An electric iron has a rating of 750 W; 200 V. Calculate:
a. the current required.
b. the resistance of its heating element,
c. energy consumed by the iron in 2 hours.
Answer :  Rating of iron is 750 W–200V.
P = 750W, V = 200 Volt.
a. P = VI or I = P/V
I = 750/200 = 3.75 A
b. Resistance = V/I = 200/3.75 = 53.3 Volt.
c. Energy consumed in 2 hr = Pxt
E = 750X2Wh = 1500Wh

Question. (a) Define potential difference between two points in a conductor.
(b) Name the instrument used to measure the potential difference in a circuit. How is it connected?
(c) A current of 2 A passes through a circuit for 1 minute. If potential difference between the terminals of the circuit is 3 V, what is the work done in transferring the charges?
Answer : 
a. Electric potential is the amount of work done in bringing a unit positive charge from one point to another.
b. Voltmeter. It is connected in parallel in the circuit.
c. I = 2A, t = 1 min. = 60 s, V = 3 V
W = VQ = V^Ith
W = 3X2X60 J
W = 360 J

Question. When two resistors of resistances R1 and R2 are connected in parallel, the net resistance is 3 W . When connected in series, its value is 16 W . Calculate the values of R1 and R2.
Answer :  R1 and R2 are in parallel combination.
1/RP = 1/R1 + 1/R2
R1 X R2 /R1 + R2 = 3
When R1 and R2 are in series combination.
Rs = R1+ R2 = 16 W
R1+ R2 = 16 …(2)
Solving eq. (1) and eq. (2)
R12 – 16R2 + 48 = 0
(R1- 4)(R1- 12) = 0
R1 = 4 Ω, 12 W
R2 = 12 Ω or 4 Ω

Question. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A? 
Answer. Resistance R₁ of the bulb is given by the expression,

Where,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb  P₁ = 10 W
R₁ = (220)² / 10 = 4840 Ω
According to Ohm’s law,
V = I R
Where,
R is the total resistance of the circuit for x number of electric bulbs
R = V / I = 220 / 5 =44 Ω
Resistance of each electric bulb, R₁ = 4840 Ω

Therefore, 110 electric bulbs are connected in parallel.

Question. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer. Supply voltage, V = 220 V
Resistance of one coil, R = 24 Ω
(i) Coils are used separately
According to Ohm’s law,

Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series 
Total resistance, R₂ = 24 Ω + 24 Ω = 48 Ω
According to Ohm’s law,
V= I₂R₂
Where, 
I₂ is the current flowing through the series circuit

Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel

Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

Question. Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1 G and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer. (i) Potential difference, V = 6 V
1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law,
V = IR
Where,
I is the current through the circuit
I = 6 / 3 = 2A
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor is . Power is given by the expression,
P = (I)² R = (2)² X 2 = 8 W
(ii) Potential difference, V = 4 V
12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistor will be 4 V. 
Power consumed by 2 Ω resistor is given by 
P = V² / R = 4² / 2 = 8W
Therefore, the power used by 2 Ω resistor is 8 W.

Question. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer. Both the bulbs are connected in parallel. Therefore, potential difference across each of  them will be 220 V, because no division of voltage occurs in a parallel circuit. 
Current drawn by the bulb of rating 100 W is given by,

Question. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer. Energy consumed by an electrical appliance is given by the expression,
H = Pt
Where,
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = 250 × 3600 = 9 × 10⁵ J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 × 600 = 7.2× 10⁵ J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.

Question. An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Answer. Rate of heat produced by a device is given by the expression for power as
P = I²R
Where,
Resistance of the electric heater, R = 8 Ω
Current drawn, I = 15 A
P = (15 )² X 8  = 1800 J / s
Therefore, heat is produced by the heater at the rate of 1800 J/s.

Question. An electric lamp of 100 G, a toaster of resistance 50 G, and a water filter of resistance 500 G are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it? 
Answer. Resistance of electric lamp, R₁ = 100 Ω
Resistance of toaster, R₂ = 50 Ω
Resistance of water filter, R₃ = 500 Ω
Voltage of the source, V = 220 V
These are connected in parallel, as shown in the following figure.

Let R be the equivalent resistance of the circuit.

According to Ohm’s law,
V = IR
I = V / R
Where,
Current flowing through the circuit = I

7.04 A of current is drawn by all the three given appliances.
Therefore, current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A
Let be the resistance of the electric iron. According to Ohm’s law, 

Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.

Question. A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
Answer :  Given: I = 0.5 A
t = 10X60 s
I = Q/t or Q = It
Q = 0.5X10x60 C
Q = 300 C