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

ANSWER: – 25
Which of the following
terms does not represent electrical power in a
(a) I2R
(b) IR 2
(c) VI
(d) V2 /R

ANSWER: – (b) IR 2
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

ANSWER 🙁 d) 25 W
Two conducting wires of the
same material and of equal
lengths and equal diameters are first connected in
then parallel in a circuit
across the same potential
difference. The ratio of heat produced in series and
combinations would be –
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

ANSWER:-c) 1:4
5) How is a voltmeter connected in the circuit to
measure the potential difference between two


ANSWER:-To measure the potential difference between
two points, a voltmeter should be
connected in parallel to the points.
A copper wire has diameter
0.5 mm and resistivity of 1.6 ×
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

ANSWER:-Area of cross-section of the wire, A =π (d/2) 2
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
We know that
The values of current I flowing in a given resistor for the corresponding
values of potential difference V across the resistor are given below
I(amperes ) 0.5 1.0 2.0 3.0
V (volts) 1.6 3.4 6.7 10.
Plot a graph between V and I
and calculate the resistance of that resistor.

ANSWER:-The plot between voltage and current is called
characteristic. The
voltage is
on x -axis and current is
plotted on y -axis. The values of the current for different values
of the voltage are shown in the given table.
V (volts) 1.6  
3.4   6.7    1
I(amperes ) 0.5    1.0    
2.0   3

The slope of the line gives the
value of resistance ( R ) as,
Slope = 1/R = BC/AC = 2/6.8
R = 6.8/2 = 3.4 Ω
Therefore, the resistance of the resistor is 3.4 Ω.
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.


Therefore, the resistance
the resistor is 4.8 kΩ
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?

I= V /R
Where, R is the equivalent resistance of resistances
Ω, 0.3 Ω, 0.4Ω, 0.5 Ω and 12 Ω. These are connected in
series. Hence,
sum of the resistances will
give the value of R.
R = 0.2 + 0.3 + 0.4 + 0.5
+ 12 =13.4
Potential difference, V =
9 V
I= 9/13.4 = 0.671 A
Therefore, the current
would flow through the 12
resistor is 0.671 A.
10) How many 176 Ω
(in parallel) are required to
carry 5 A on a 220 V line?

R = V /I
Supply voltage, V= 220 V
Current, I = 5 A
Equivalent resistance of
combination = R, given as
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.
ANSWER:-Two resistors in
Two 6 Ω resistors are connected
in parallel. Their equivalent resistance will be
The third 6 Ω resistor is in
series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω+ 3 Ω = 9 Ω.
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?

ANSWER:-Resistance R 1 of the bulb is
given by the expression,
Supply voltage, V = 220 V
Maximum allowable current, I =5 A
Rating of an electric bulb P=10watts
Because R =V 2/P
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 resistances, which may be use separately, in series, or in
parallel. What are the currents in the three? Cases?

ANSWER :-(i) Coils are used separately
According to Ohm’s law,
V = I 1R 1
I1 is the current flowing through the coil I1 = V/R 1 = 220/24 = 9.166 A
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series Total resistance, R 2 = 24 Ω +24
Ω = 48 Ω
According to Ohm’s law, V =I2 R2
I2 is the current flowing through the series circuit I2 = V/ R 2 = 220/48 = 4.58 A
Therefore, 4.58 a current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in Parallel Total resistance, R 3 is given as
According to Ohm’s law,
V = I 3R 3
I3 is the current flowing through the circuit I3 = V /R 3 =220/12 = 18.33 A
Therefore, 18.33 a current will
flow through the circuit when
coils are connected in parallel.
14) Compare the power used in the 2 Ω resistors in each of the
following circuits: (i) a 6 V
battery in series with 1
Ω and 2 Ω resistor, and (ii) a 4 V battery in parallel with 12 Ω
and 2 Ω

ANSWER :-(i) Potential difference, V = 6
1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent
resistance of the circuit, R = 1 + 2 = 3
According to Ohms
V = IR
I is the current through the circuit
I= 6/3 = 2 A
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 2 A. Power
is given by the expression,
P = ( I ) 2R = (2) 2 x 2 = 8 W
(ii) Potential difference, V = 4
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 resistors will be 4 V.
Power consumed by 2 Ω resistor is given by P = V 2/R = 42 /2 = 8 W
Therefore, the power used by 2Ω
resistor is 8 W.
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?

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, Power= Voltage x Current
Current = Power/Voltage = 60/220 A
Hence, current drawn from the
line = 100/220 + 60/220 = 0.727 A
16) Which uses more energy, a 250 W TV set in 1 hr,
or a 1200 W toaster in 10 minutes?

Power of the appliance =
Time = t
Energy consumed by a TV
of power 250 W in 1 h =
×3600 = 9 ×105 J
Energy consumed by a
of power 1200 W in 10
minutes = 1200
Energy consumed by a
of power 1200 W in 10
= 1200 ×600
= 7.2×10 5 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.
17) An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate
heat is developed in the

ANSWER:-P =I2 R 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.
18) Explain the following.
(a) Why is the 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 aluminum wires usually
employed for electricity
ANSWER :-(a) The melting point and of Tungsten is an
alloy which has very high melting point and very high resistivity so does not
burn easily at a high
(b) The conductors of electric heating devices such as
bread toasters and electric irons are made of alloy because resistivity of an
alloy is more
that of metals which produces large amount of heat.
(c) In series circuits voltage is
divided. Each component of a
series circuit receives a small
voltage so the amount of
current decreases and the
device becomes hot and does
not work properly. Hence, series
arrangement is not used
in domestic circuits.
(d) Resistance (R) of a wire is
inversely proportional to its
area of cross-section (A), i.e.
when area of cross section
increases the resistance
decreases or vice versa.
(e) Copper and aluminum are good conductors of electricity also they have low resistivity.
So they are usually used for
electricity transmission.