CBSE Class X

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.

Answer:

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

6 + 6 = 12 Ω

The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will

be

Therefore, the total resistance is 4 Ω.

Question 10:

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

Answer:

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

V= IR

R = V/I

where,

Supply voltage, V = 220 V

Current, I = 5 A

Equivalent resistance of the combination = R, given as

Therefore, four resistors of 176 Ω are required to draw the given amount of current.

Question 9:

A battery of 9 V is connected in series with resistors of 0.2 G, 0.3 G, 0.4 G, 0.5 G

and 12 G, respectively. How much current would flow through the 12 G resistor?

Answer:

There is no current division occurring in a series circuit. Current flow through the

component is the same, given by Ohm’s law as

V = IR

I = V/R

Where,

R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4 Ω, 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

Potential difference, V = 9 V

Therefore, the current that would flow through the 12 G resistor is 0.671 A.

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.

Answer:

Resistance (R) of a resistor is given by Ohm’s law as,

V= IR

R = V/I

Where,

Potential difference, V = 12 V

Current in the circuit, I = 2.5 mA = 2.5 × 10⁻³ A

Therefore, the resistance of the resistor is kΩ.

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.

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.

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 3.4 Ω.

Question 6:

A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ G 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:

Resistance (R) of a copper wire of length l and cross-section A is given by the

expression,

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 = 2 × 0.5 = 1 mm = 0.001 m

Therefore, resistance R’

Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω

Question 5:

How is a voltmeter connected in the circuit to 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.

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

Answer:

(c) Heat produced in the circuit is inversely proportional to the resistance R.

Let R_S and R_P be the equivalent resistances of the wires if connected in series

and parallel respectively. Hence, for same potential difference V, the ratio of

heat produced in the circuit is given by

Where,

Heat produced in the series circuit = H_S

Heat produced in the parallel circuit = H_P

Equivalent resistance, R_S = R + R = 2R

Equivalent resistance,

Therefore, the ratio of heat produced in series and parallel combinations is 1:4.

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

Answer:

(d)Energy consumed by an appliance is given by the expression,

Where,

Power rating, P = 100 W

Voltage, V = 220 V

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 2:

Which of the following terms does not represent electrical power in a circuit?

(a) I²R

(b) IR²

(c) VI

(d) V²/R

Answer:

(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

∴ P = VI

From equation (i), it can be written

P = (IR) × I

∴ P = I²R

From equation (ii), it can be written

Power P cannot be expressed as IR².

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) 1/25

(b) 1/5

(c) 5

(d) 25

Answer:

(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 = R/5

All the five parts are connected in parallel. Hence, equivalent resistance (R’) is

given as

Therefore, the ratio R/R’ is 25.

Question 2:

An electric motor takes 5 A from a 220 V line. Determine the power of the motor

and the energy consumed in 2 h.

Answer:

Power (P) is given by the expression

P = VI

Where,

Voltage, V = 220 V

Current, I = 5 A

P = 220 × 5 = 1100 W

Energy consumed by the motor = Pt

Where,

Time, t = 2 h = 2 × 60 × 60 = 7200 s

P = 1100 × 7200 = 7.92 × 10⁶ J

Therefore, power of the motor = 1100 W

Energy consumed by the motor = 7.92 × 10⁶ J

Question 1:

What determines the rate at which energy is delivered by a current?

Answer:

The rate of consumption of electric energy in an electric appliance is called electric

power. Hence, the rate at which energy is delivered by a current is the power of the

appliance.

Question 3:

An electric iron of resistance 20 G takes a current of 5 A. Calculate the heat

developed in 30 s.

Answer:

The amount of heat (H) produced is given by the joule’s law of heating as

H = VIt

Where,

Current, I = 5 A

Time, t = 30 s

Voltage, V = Current × Resistance = 5 × 20 = 100 V

H = 100 × 5 × 30 = 1.5 × 10⁴ J

Therefore, the amount of heat developed in the electric iron is .5 × 10⁴ J

Question 2:

Compute the heat generated while transferring 96000 coulomb of charge in one hour

through a potential difference of 50 V.

Answer:

The amount of heat (H) produced is given by the Joule’s law of heating as

H = VIt

Where,

Voltage, V = 50 V

Time, t = 1 h = 1 × 60 × 60 s

Amount of current,

Therefore, the heat generated is 4.8 × 10⁶ J

Question 1:

Why does the cord of an electric heater not glow while the heating element does?

Answer:

The heating element of an electric heater is a resistor. The amount of heat produced

by it is proportional to its resistance. The resistance of the element of an electric

heater is very high. As current flows through the heating element, it becomes too

hot and glows red. On the other hand, the resistance of the cord is low. It does not

become red when current flows through it.

Question 5:

What is (a) the highest, (b) the lowest total resistance that can be secured by

combinations of four coils of resistance 4 Ω , 8 Ω, 12 Ω, 24 Ω?

Answer:

There are four coils of resistances 4 Ω , 8 Ω, 12 Ω, and 24 Ω respectively.

(a) If these coils are connected in series, then the equivalent resistance will be the

highest, given by the sum 4 + 8 + 12 + 24 = 48 Ω

(b) If these coils are connected in parallel, then the equivalent resistance will be the

lowest, given by

Therefore, 2 Ω is the lowest total resistance.

Question 4:

How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total

resistance of (a) 4 Ω, (b) 1 Ω?

Answer:

There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.

(a) The following circuit diagram shows the connection of the three resistors.

Here, 6 Ω and 3 Ω resistors are connected in parallel.

Therefore, their equivalent resistance will be given by

This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.

Therefore, equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω

Hence, the total resistance of the circuit is 4 Ω

2. The following circuit diagram shows the connection of the three resistors.

All the resistors are connected in series. Therefore, their equivalent resistance will

be given as

Therefore, the total resistance of the circuit is 1 Ω

Question 3:

What are the advantages of connecting electrical devices in parallel with the battery

instead of connecting them in series?

Answer:

There is no division of voltage among the appliances when connected in parallel.

The potential difference across each appliance is equal to the supplied voltage.

The total effective resistance of the circuit can be reduced by connecting electrical

appliances in parallel.

Question 2:

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of

resistance 500 Ω are connected in parallel to a 220 V source. What is the

resistance of an electric iron connected to the same source that takes as much

current as all three appliances,and what is the current through it?

Answer:

Resistance of electric lamp, R₁ = 100 Ω

Resistance of toaster, R₂ = 50 Ω

Voltage of the source, V = 220 V

These are connected in parallel, as shown in the following figure.

Let R be the equivalent resistance of the circuit.

According to Ohm’s law,

V = IR

I = V/R

Where,

Current flowing through the circuit = I

7.04 A of current is drawn by all the three given appliances.

Therefore, current drawn by an electric iron connected to the same source of

potential 220 V = 7.04 A

Let R’  be the resistance of the electric iron. According to Ohm’s law,

V = IR’

Therefore, the resistance of the electric iron is 31. 25 Ω and the current flowing

through it is 7.04 A.