# Electricity : NCERT Exercise Questions

Q.1       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
Sol.        (d) 25

Solution:-
The piece of wire having resistance equal to R is cut into five equal parts.
Therefore, resistance of each of the part will be equal to ${R \over 5}$
When all parts are connected in parallel, then equivalent resistance is = R' as given
Therefore, ${1 \over {R'}} = {5 \over R} + {5 \over R} + {5 \over R} + {5 \over R} + {5 \over R} = {{5 + 5 + 5 + 5 + 5} \over R} = {{25} \over R}$
Or, $R' = {R \over {25}}$
Thus, ${R \over {R'}} = {R \over {{R \over {25}}}} = R \times {{25} \over R} = 25$

Q.2       Which of the following terms does not represent electrical power in a circuit?
(a) I2R             (b) IR2              (c) VI             (d) V2/R
Sol.         (b) IR2

Solution:-
We know that Power (P) = VI
After substituting the value of V = IR in this we get
P = (IR) I = I x R x I = I2R, Thus P = I2R
Since $I = {V \over R}$, therefore, after substituting this value in above equation we get $P = {\left( {{V \over R}} \right)^2}$ $R = {{{V^2}} \over {{R^2}}} = {{{V^2}} \over R}$
Thus, P cannot be expressed as IR2

Q.3       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
Sol.        (d) 25 W

Solution:- Given,
Potential difference, V = 220V, Power, P = 100W
Therefore, power consumption at 100V =?
To solve this problem, first of all resistance of the bulb is to be calculated.
We know that $P = {{{V^2}} \over R}$
Therefore, $100W = {{{{\left( {220V} \right)}^2}} \over R}$ $\Rightarrow$ $R = {{220 \times 220} \over {100}}\Omega = 484\Omega$
Now, when bulb is operated at 110V
The power consumption $P = {{{V^2}} \over R} = {{{{\left( {110V} \right)}^2}} \over {484\Omega }} = {{12100{V^2}} \over {484\Omega }} = 25W$
Thus, bulb will consume power of 25W at 110V

Q.4       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
Sol.         (c) 1 : 4

Solution:
Let the potential difference = V,
Resistance of the wire = R
Resistance when the given wires connected in series = Rs
Resistance when the given wires connected in parallel = Rp
Heat produced when the given wires connected in series = Hs
Heat produced when the given wires connected in parallel = Hp
Thus, resistance Rs when the given two wires connected in series = R + R = 2R
Resistance ${R_P}$ when the given two wires connected in parallel
$= {1 \over R} + {1 \over R} = {2 \over R}$ or, ${R_P} = {R \over 2}$
We know that, $H = {I^2}Rt = {\left( {{V \over R}} \right)^2}R\,t, = {V \over R}t,\left( {\sin ce,I = {V \over R}} \right)$
Thus, ratio of heat produced in given two conditions
${H_S}:{H_P} = {{Vt} \over {{R_S}}}:{{Vt} \over {{R_P}}} = {{Vt} \over {{R_S}}} \times {{{R_P}} \over {Vt}} = {{{R_P}} \over {{R_S}}}$
After subsituting the value of ${{R_P}}$ and ${{R_S}}$ we get
${{{H_S}} \over {{H_P}}} = {{{R \over 2}} \over {2R}} = {R \over {2 \times 2R}} = {1 \over 4}$
Thus, ${H_S}:{H_P} = 1:4$

Q.5       How is a voltmeter connected in the circuit to measure the potential difference between two points?
Sol.          Voltmeter is connected into parallel to measure the potential difference between two points in a circuit. Q.6       A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Sol.

Given, Diameter of wire = 0.5 mm
Therefore, radius = ${{0.5mm} \over 2} = 0.25mm = {{0.25m} \over {1000}} = 0.00025m$
Resistivity, $\rho = 1.6 \times {10^{ - 8}}\Omega m$
Resistance (R) = $10\Omega$, therefore length (I) = ?
Resistance (R1) when diameter is doubled = ?
We know that, R = $\rho = {l \over A}$
Therefore, $l = {{RA} \over \rho } = {{R\pi {r^2}} \over \rho }$
$\Rightarrow$ $l = {{10\Omega \times 3.14 \times {{\left( {0.00025m} \right)}^2}} \over {1.6 \times {{10}^{ - 8}}\Omega m}}$
= ${{10\Omega \times 3.14 \times 0.00025m \times 0.00025m} \over {1.6 \times {{10}^{ - 8}}\Omega m}}$
${{10 \times 3.14 \times 0.0000000625 \times {{10}^8}} \over {1.6}}m$
=  ${{10 \times {{10}^8} \times 0.000000196250} \over {1.6}}m$
${{196.25} \over {1.6}}m = 122.656m = 122.7m$
When diameter of the wire is doubted, i.e. diameter = 0.5mm × 2 = 1 mm
Therefore, radius = ${{1mm} \over 2} = 0.5mm = {{0.5} \over {1000}}m = 0.0005m$
Therefore,  ${R_1} = \rho {l \over A} = \rho {1 \over {\pi {r^2}}}$
$1.6 \times {10^{ - 8}}\Omega m \times {{122.7m} \over {3.14 \times 0.0005m \times 0.0005m}}$
${{1.6 \times {{10}^{ - 8}} \times 122.7} \over {3.14 \times 0.000000025}}\Omega = {{1.6 \times 122.7} \over {3.14 \times 0.000000025 \times {{10}^8}}}\Omega$
${{196.32} \over {78.5}}\Omega = 2.5\Omega$
Thus, length of the wire =122.7 m
Resistance = $2.5\Omega$ (When diameter becomes double)

Q.7       The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below – Plot a graph between V and I and calculate the resistance of that resistor.
Sol. Since slope of the graph will give the value of resistance, thus
Let consider two points A and B on the slope.
Draw two lines from B along X-axis and from A along Y-axis, which meets at point C
Now, BC = 10.2 V – 3.4 V = 6.8 V
AC = 3 – 1 = 2 ampere
Slope = ${1 \over R} = {{AC} \over {BC}} = {2 \over {6.8}} = {1 \over {3.4}}$
Thus, resistance, R = $3.4\Omega$

Q.8       When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Sol.

Given, Potential difference, V = 12V
Current ( I ) across the resistor = 2.5mA = 2.5 x 10 -3 = 0.0025 A
Resistance, R =?
We know that $R = {V \over I} = {{12V} \over {0.0025A}} = 4800\Omega$

Q.9       A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Sol.

Given, potential difference, V = 9V
Resistance of resistors which are connected in series = 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω respectively
Current through resistor having resistance equal to 12Ω =?
Total effective resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω
We know that current $\left( I \right) = {V \over R} = {{9V} \over {13.4\Omega }} = 0.671A$
Since, there is no division of electric current, in the circuit if resistors are connected in series, thus, resistance through the resistor having resistance equal to 12 Ω = 0.671 A

Q.10     How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Sol.

Given, resistance of each of the resistor = 176Ω
Electric current ( I ) = 5A
Potential difference (V) = 220V
Number of resistors connected in parallel =?
Let total x resistance are connected in parallel, and total effective resistance = R
Therefore, ${1 \over {{R_t}}} = x \times {1 \over {176\Omega }} = {x \over {176\Omega }}$
We know that, R = ${V \over I}$ Therefore, ${{176\Omega } \over x} = {{220V} \over {5A}}$
$\Rightarrow$ $x \times 220V = 176\Omega \times 5A$
$\Rightarrow$ $\Rightarrow x = {{176\Omega \times 5A} \over {220V}} = 4$
Thus, there are 4 resistors are to be connected.

Q.11     Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Sol.

Here we have four options to connect the three resistors in different ways.
(a) All the three resistors can be connected in series
(b) All the three can be connected in parallel
(c) Two of the three resistors can be connected in series and one in parallel and
(d) Two of the three resistors can be connected in parallel and one in series
Thus, effective resistance in the case
(a) When all the three resistors can be connected in series
Effective total resistance = 6 Ω + 6 Ω + 6 Ω = 18 Ω. This is not required
(b) All the three can be connected in parallel
Then, ${1 \over R} = {1 \over {6\Omega }} + {1 \over {6\Omega }} + {1 \over {6\Omega }} = {{1 + 1 + 1} \over 6}\Omega = {3 \over 6}\Omega = {1 \over 2}\Omega$
Thus, effective total resistance R = 2 Ω. This is also not required.

(c) Two of the three resistors can be connected in parallel and one in series
When two resistors are connected in parallel
Then, ${1 \over R} = {1 \over {6\Omega }} + {1 \over {6\Omega }} = {2 \over 6}\Omega = {1 \over 3}\Omega$
Therefore, R = 3Ω
And third one is connected in series, then total resistance = 3 Ω + 6 Ω = 9 Ω. This is required

(d) Two of the three resistors can be connected in parallel and one in series
When two resistors are connected in series, then total resistance = 6 Ω + 6 Ω = 12 Ω
And one resistor is connected in series with two in parallel
Then, ${1 \over R} = {1 \over {12\Omega }} + {1 \over {6\Omega }} = {{1 + 2} \over {12}}\Omega = {1 \over 4}\Omega$
Thus, R = 4 Ω. This is required.
Thus, when two resistors are connected in series and one in parallel then total effective resistance = 9 Ω
When two resistors are connected in parallel with one in series then total effective resistance = 4 Ω

Q.12     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?
Sol.

Given, potential difference (V) = 220 V
Power input (P) = 10W
Allowable electric current (I) = 5A
Number of lamps connected in parallel =?
To calculate this, first of all resistance of each of the lamp is to be calculated.
We know that, $P = {{{V^2}} \over R}$
Therefore, $10W = {{{{\left( {220V} \right)}^2}} \over R}$
$\Rightarrow$ $R = {{48400} \over {10W}} = 4840\Omega$
Let x bulb are to be connected in parallel to have the electric current (I) equal to 5A
Therefore, ${1 \over R} = x \times {1 \over {4840\Omega }} = {x \over {4840\Omega }}$
Therefore, effective resistance $R = {{4840} \over x}\Omega$
Now, we know that, $R = {V \over I}$
$\Rightarrow$${{4840} \over x}\Omega = {{220V} \over {5A}}$
$\Rightarrow$ $x = {{4840 \times 5} \over {220}} = 110$
Thus, total 110 bulbs are to be connected in parallel.

Q.13      A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Sol.

Given, Potential difference (V) = 220 V
Resistance of each coil = 24 Ω
The current in given three case, i.e. when used separately, when used in parallel, when used in series =?
Case – 1 –
When used separately, then resistance, R = 24 Ω and V = 220V
We know that electric current, $I = {V \over R} = {{220V} \over {24\Omega }} = 9.16A$
Case – 2 – When the two resistors are connected in series.
Total effective resistance = 24 Ω + 24Ω = 48 Ω
Therefore, electric current, $I = {V \over R} = {{220V} \over {48\Omega }} = 4.58A$
Case – 3 – When the two resistors are connected in parallel,
Then ${1 \over R} = {1 \over {24\Omega }} + {1 \over {24\Omega }} = {{1 + 1} \over {24\Omega }} = {2 \over {24\Omega }} = {1 \over {12}}\Omega$
Therefore, total effective resistance, $R = 12\Omega$
Thus, electric current, $I = {V \over R} = {{220V} \over {12\Omega }} = 18.33A$
Thus, electric through the circuit
(a) When coil is used separately = 9.16A
(b) When coils are used in series = 4.58 A
(c) When coils are used in parallel = 18.33 A

Q.14     Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Sol.

Case – 1 –
Potential difference = 6 V
Resistance of resistors = 1 Ω and 2 Ω
Power used through resistors of 2 Ω =?
Since, resistors are connected in series, thus, total effective resitance, R = 1 Ω + 2 Ω = 3 Ω
We know that, $I = {V \over R} = {{6V} \over {3\Omega }} = 2A$
Since, current remains same when resistors are connected in series.
Therefore, current through the resistors or $2\Omega = 2A$
Thus, Power $\left( P \right) = {I^2} \times R = {\left( {2A} \right)^2} \times 2\Omega = 8W$
Case – 2 –
Potential difference, V = 4V
Resistance of resistors connected in parallel = 12 Ω and 2 Ω
Power used by resistors having resistance = 2 Ω
Since, voltage across the circuit remains same if resistors are connected in parallel.
Thus, Power (P) used by resistance of $2\Omega = {{{V^2}} \over R} = {{{{\left( {4V} \right)}^2}} \over {2\Omega }} = {{16} \over 2}W = 8W$
Thus, power used by resistance of 2 Ω in both the case = 8 W

Q.15    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?
Sol.

Since, both the lamps are connected in parallel, thus, potential difference will be equal
Thus, potential difference = 220 V
Power of one lamp, P1 = 100W
Power of second lamp, P2 = 60W
We know that, Power (P) = VI, or I = ${P \over V}$
Thus, total current through the circuit, I = ${{{P_1}} \over V} + {{{P_2}} \over V}$
$\Rightarrow$ I = ${{100W} \over {220V}} + {{60W} \over {220V}} = {{100 + 60} \over {220}}A = {{160} \over {220}}A = 0.727A$ answer

Q.16    Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Sol.

Given, power of TV (P) = 250W,
time (t) = 1 hr = 60 x 60 s = 3600 s
Thus, energy used by it =?
We know that energy used by appliance = Power x time
Thus, energy used by TV = 250 W x 3600 s = 900000 J = 9 x 105 J
Power of toaster = 1200W
Time (t) = 10 minute = 60 x 10 = 600 s
Thus, energy used by toaster = P x t = 1200W x 600 s = 720000 J = 7.2 x 105 J
Thus, given TV set will use more energy than toaster.

Q.17     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.
Sol.

Given, Resistance, R = 8 Ω
Electric current ( I ) = 15A
Time (t) = 2 h = 2 x 60 x 60 s = 7200 s
Rate at which heat is developed in heater =?
We know that rate of heat produced = I2R = (15A)2 x 8 Ω = 225 x 8 J/s = 1800 J/s Answer

Q.18     Explain the following.
(a) Why is tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires, usually employed for electricity transmission?
Sol.

(a) The melting point of tungsten is very high, i.e. 33800C, which enables it not to melt at high temperature and to retain most of the heat. The heating of tungsten makes it glow. This is the cause that tungsten is used almost exclusively for filament of electric lamps.

(b) To produce more heat, the high melting point of conductors is necessary. The alloys of metal have higher melting points than pure metals. Thus, to retain more heat alloy is used rather than pure metal in electrical heating devices, such as bread-toaster, electric iron, etc.

(c) There is loss of voltage in the series arrangement in the circuits because of add on effect of resistances. So, series arrangement is not used for domestic circuits.

(d) Resistance of a wire is indirectly proportional to the area of cross section. Resistance increases with decrease in area of cross section and vice versa.

(e) The resistivity of copper and aluminium wires are lower than that of iron but more than that of silver. These wires are cheaper than silver, that’s why copper and aluminium wires usually employed for electricity transmission.