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
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
2)
Which of the following terms does not represent electrical power in a
circuit?
Which of the following terms does not represent electrical power in a
circuit?
(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 –
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
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 –
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
ANSWER:-c) 1:4
5) How is a voltmeter connected in the circuit to
measure the potential difference between two
points?
measure the potential difference between two
points?
ANSWER:-To measure the potential difference between
two points, a voltmeter should be connected in parallel to the points.
two points, a voltmeter should be connected in parallel to the points.
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?
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?
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
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 −
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.
and calculate the resistance of that resistor.
ANSWER:-The plot between voltage and current is called
IV characteristic. The
voltage is plotted
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.
IV characteristic. The
voltage is plotted
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
3.4 6.7 1
I(amperes ) 0.5 1.0
2.0 3
2.0 3
The slope of the line gives the
value of resistance ( R ) as,
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.
unknown resistor, there is a current of 2.5 mA in the circuit. Find the value
of the resistance of the resistor.
ANSWER:-
Therefore, the resistance
of the resistor is 4.8 kΩ
of 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?
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?
ANSWER:-V = IR
I= V /R
Where, R is the equivalent resistance of resistances
0.2 Ω, 0.3 Ω, 0.4Ω, 0.5 Ω and 12 Ω. These are connected in
series. Hence, the
sum of the resistances will give the value of R.
0.2 Ω, 0.3 Ω, 0.4Ω, 0.5 Ω and 12 Ω. 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
+ 12 =13.4 Ω
Potential difference, V =
9 V
9 V
I= 9/13.4 = 0.671 A
Therefore, the current
that would flow through the 12
Ω
resistor is 0.671 A.
that would flow through the 12
Ω
resistor is 0.671 A.
10) How many 176 Ω
resistors (in parallel) are required to
carry 5 A on a 220 V line?
resistors (in parallel) are required to
carry 5 A on a 220 V line?
ANSWER:-V = IR
R = V /I
Where,
Supply voltage, V= 220 V
Current, I = 5 A
Equivalent resistance of
the combination = R, given as
the 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.
of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
ANSWER:-Two resistors in
parallel
parallel
Two 6 Ω resistors are connected
in parallel. Their equivalent resistance will be
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.
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?
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,
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?
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,
According to Ohm’s law,
V = I 1R 1
Where,
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
Ω = 48 Ω
According to Ohm’s law, V =I2 R2
Where,
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
Where,
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.
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
resistors.
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
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
resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s
law,
law,
V = IR
Where,
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.
current in series circuits.
Hence, current flowing through the 2 Ω resistor is 2 A. Power
is given by the expression,
is given by the expression,
P = ( I ) 2R = (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 resistors will be 4 V.
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.
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?
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.
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?
or a 1200 W toaster in 10 minutes?
ANSWER:-H= Pt
Where,
Power of the appliance =
P
P
Time = t
Energy consumed by a TV
set of power 250 W in 1 h =
250×3600 = 9 ×105 J
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
toaster of power 1200 W in 10
minutes = 1200 ×600
Energy consumed by a
toaster of power 1200 W in 10
minutes
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.
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.
at which
heat is developed in the heater.
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?
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?
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?
domestic circuits?
(d) How does the resistance of a wire vary with its
area of cross-section?
area of cross-section?
(e) Why are copper and aluminum wires usually
employed for electricity transmission?
employed for electricity transmission?
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 temperature.
alloy which has very high melting point and very high resistivity so does not
burn easily at a high temperature.
(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 which produces large amount of heat.
bread toasters and electric irons are made of alloy because resistivity of an
alloy is more than
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.
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.
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.
So they are usually used for
electricity transmission.