Question 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)
(b)
(c) 5
(d) 25
(d) Resistance of a
piece of wire is proportional to its length. A piece of wire has a
resistance R.
The wire is cut into five equal parts.
Therefore,
resistance of each part =
All the five parts are connected in parallel. Hence, equivalent
resistance (R’) is given as
Therefore,
the ratio
is 25.
Question 2:
Which of the following terms does not represent electrical power in a
circuit?
(a) I^{2}R
(b) IR^{2}
(c) VI
(d)
(a) I^{2}R
(b) IR^{2}
(c) VI
(d)
(b) Electrical
power is given by the expression, P
= VI …
(i)
According
to Ohm’s law, V
= IR …
(ii)
Where,
V
= Potential difference
I
= Current
R
= Resistance
From equation (i), it can be written
P
= (IR) ×
I
From equation (ii), it can be written
Power
P cannot
be expressed as IR^{2}.
Question 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
(d)Energy
consumed by an appliance is given by the expression,
Where,
Power
rating, P =
100 W
Voltage,
V = 220 V
Resistance,
R =
The resistance of the bulb remains constant if the supply voltage is
reduced to 110 V. If the bulb is operated on 110 V, then the energy
consumed by it is given by the expression for power as
Therefore, the power consumed will be 25 W.
Question 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
(c) The Joule heating
is given by, H = i^{2}Rt
Let, R be the
resistance of the two wires.
The equivalent
resistance of the series connection is R_{S}
= R + R = 2R
If V is the applied
potential difference, then it is the voltage across the equivalent
resistance.
The heat dissipated in
time t is,
The equivalent
resistance of the parallel connection is R_{P}
=
V is the applied
potential difference across this R_{P}.
The heat dissipated in
time t is,
So, the ratio of heat
produced is,
Note: H
R also H
i^{2} and H
t. In this question, t is same for both the circuit. But
the current through the equivalent resistance of both the circuit is
different. We could have solved the question directly using H
R if in case the current was also same. As we know the voltage
and resistance of the circuits, we have calculated i in terms
of voltage and resistance and used in the equation H = i^{2}Rt
to find the ratio.
Question 5:
How is a
voltmeter connected in the circuit to measure the potential
difference between two points?
To measure
the potential difference between two points, a voltmeter should be
connected in parallel to the points.
Question 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?
Resistance
(R) of a
copper wire of length l
and crosssection A
is given by the expression,
Where,
Resistivity
of copper,
Area
of crosssection of the wire, A
=
Diameter= 0.5 mm = 0.0005 m
Resistance,
R = 10 Ω
Hence,
length of the wire,
If
the diameter of the wire is doubled, new diameter
Therefore,
resistance
Therefore,
the length of the wire is 122.7 m and the new resistance is
Question 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 −
I (amperes
)

0.5

1.0

2.0

3.0

4.0

V
(volts)

1.6

3.4

6.7

10.2

13.2

Plot
a graph between V
and I and
calculate the resistance of that resistor.
The
plot between voltage and current is called IV
characteristic. The voltage is plotted on xaxis
and current is plotted on yaxis.
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

10.2

13.2

I (amperes
)

0.5

1.0

2.0

3.0

4.0

The
IV
characteristic of the given resistor is plotted in the following
figure.
The
slope of the line gives the value of resistance (R)
as,
Therefore,
the resistance of the resistor is.
Question 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.
Resistance
(R) of a
resistor is given by Ohm’s law as,
Where,
Potential
difference, V
= 12 V
Current
in the circuit, I
= 2.5 mA =
Therefore,
the resistance of the resistor is
.
Question 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?
There is
no current division occurring in a series circuit. Current flow
through the component is the same, given by Ohm’s law as
Where,
R
is the equivalent resistance of resistances.
These are connected in series. Hence, the sum of the resistances will
give the value of R.
R
= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
Potential
difference, V
= 9 V
Therefore,
the current that would flow through the 12 Ω
resistor is 0.671 A.
Question 10:
How
many 176 Ω
resistors (in parallel) are required to carry 5 A on a 220 V line?
For
x number
of resistors of resistance 176 Ω, the equivalent resistance of
the resistors connected in parallel is given by Ohm’s law as
Where,
Supply
voltage, V
= 220 V
Current,
I = 5 A
Equivalent
resistance of the combination = R,given as
From Ohm’s
law,
Therefore,
four resistors of 176 Ω
are required to draw the given amount of current.
Question 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.
If
we connect the resistors in series, then the equivalent resistance
will be the sum of the resistors, i.e., 6 Ω
+ 6 Ω + 6 Ω = 18 Ω, which is not desired. If we
connect the resistors in parallel, then the equivalent resistance
will be
Hence, we should either connect the two resistors
in series or parallel.
(i) Two
resistors in parallel
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.
(ii) Two
resistors in series
Two
6 Ω
resistors are in series. Their equivalent resistance will be the sum
The
third 6 Ω
resistor is in parallel with 12 Ω.
Hence, equivalent resistance will be
Therefore,
the total resistance is.
Question 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?
Resistance
R_{1}
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
According
to Ohm’s law,
V
= I R
Where,
R
is the total resistance of the circuit for x
number of electric bulbs
Resistance
of each electric bulb,
Ω
Therefore,
110 electric bulbs are connected in parallel.
Question 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 used separately, in series, or in parallel.
What are the currents in the three cases?
Supply
voltage, V
= 220 V
Resistance
of one coil, R
=
(i) Coils
are used separately
According
to Ohm’s law,
Where,
is
the current flowing through the coil
Therefore, 9.16 A current will flow through the coil when used
separately.
(ii) Coils
are connected in series
Total
resistance,
According
to Ohm’s law,
Where,
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
Total
resistance,
is given as
According
to Ohm’s law,
Where,
is the current flowing through the circuit
Therefore, 18.33 A current will flow through the circuit when coils
are connected in parallel.
Question 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.
(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
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,
(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
Therefore,
the power used by 2 Ω
resistor is 8 W.
Question 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?
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,
Similarly,
current drawn by the bulb of rating 100 W is given by,
Question 16:
Which uses
more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10
minutes?
Energy
consumed by an electrical appliance is given by the expression,
Where,
Power
of the appliance = P
Time
= t
Energy
consumed by a TV set of power 250 W in 1 h = 250 ×
3600 = 9 ×
10^{5} J
Energy
consumed by a toaster of power 1200 W in 10 minutes = 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.
Question 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.
Rate of
heat produced by a device is given by the expression for power as
Where,
Resistance
of the electric heater, R
= 8 Ω
Current
drawn, I =
15 A
Therefore,
heat is produced by the heater at the rate of 1800 J/s.
Question 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
breadtoasters 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
crosssection?
(e) Why are copper and aluminium wires usually employed for
electricity transmission?
(a) The melting point and resistivity of tungsten are very high. It
does not burn readily at a high temperature. The electric lamps glow
at very high temperatures. Hence, tungsten is mainly used as heating
element of electric bulbs.
(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 than that of metals. It produces large amount of heat.
(c) There is voltage division in series circuits. Each component of a
series circuit receives a small voltage for a large supply voltage.
As a result, the amount of current decreases and the device becomes
hot. Hence, series arrangement is not used in domestic circuits.
(d) Resistance (R)
of a wire is inversely proportional to its area of crosssection (A),
i.e.,
(e) Copper and aluminium wires have low resistivity. They are good
conductors of electricity. Hence, they are usually employed for
electricity transmission.