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.