NCERT Solutions for class 10 Science Chapter 12:Electricity

NCERT Solutions for class 10 Science Chapter 12 Exercise Question

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

Here each part is

For equivalent resistance :

= + + + + = =
= 25

Answer: d) 25

Question 2

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

(a) I²R

(b) IR²

(c) VI

(d) V²/R

Answer

Answer: b) IR²

Explanation:

We know P = VI. ............(i)

According to Ohm’s law,

V = IR

Puting Value of V in eq (i)

P = (IR) × I

P = I²R

Again Ohm’s law,

I = ..............(ii)

From eq (i) and (ii)

P = V × = V²/R

So IR² does not represent electrical power

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

Answer: 25 W

We know P = VI = V²/R

Again

R = V²/P

R = (220)²/100 = 484 Ω

Again for V = 110

P = V²/R

P = (110⁾²/484 Ω = 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

Answer

Let R_s and R_p be the equivalent resistance of the wires when connected in series and parallel respectively.

So
R_s = R + R =2R
= + =
R_p =
We know for V constant
H_s = or H_s ∝ Similarly
H_p = or H_p ∝ So
= = =

Question 5

How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer

the voltmeter should be connected parallel in circuit to measure the potential difference

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?

Answer

Here D = 0.5mm = 5 x 10 ^-4 m
r = = 2.5 x 10 ^-4 m
ρ =1.6 × 10^–8 Ω m
R = 10Ω
We know R = ρ
l = = =
= 122.7m
Again R' = ρ
R' = ρ ............[r = ]
If D' =2D R = ρ
R' = ρ =R = ρ = 2.5 Ω

The length of the wire is 122.72 m and the new resistance is 2.5 Ω.

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 –

Plot a graph between V and I and calculate the resistance of that resistor.

Answer

The slope graph give resistance.

Slope = =

R = 6.8/2 = 3.4 Ω

The resistance is 3.4 Ω.

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

We know V= IR

R = V/I

R = = 4.8 x 10³ Ω

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?

Answer

For R_s = R₁ + R₂ + R₃ +.........R_n

So R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

Now, using Ohm’s law,

V =IR

I = V/R = 9 /31.4 = 0.67 A

The current flowing across the 12 Ω is 0.671 A.

Question 10

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

Answer

Let number of resistors is n

We know
= n x =
R_p=

Now, using Ohm’s law,

V= IR
=
n = = 4

The number of resistors is 4.

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

When two resistance connect in parallel and third resistance with series to get total resistance 9 Ω

So

= + = + = =
Now R_p connected in series with third resistance
R_e = R_p + 6
= 3 + 6 = 9 Ω

ii)

When two resistance connect in series and third resistance with parallel to get total resistance 4 Ω

R_s= R₁ + R₂
= 6 + 6 =12
Now R_s connected in parallel with third resistance
= + = + = =
R_e = 4 Ω

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?

Answer

Given

P = 10 W and V = 220 V

Let the total number of blub is 'n'

Current I = 5 A

We know

P =
R = = = 4840 Ω

We also know

V = IR

R == = 44 Ω
So 'n' number of resistance each 4840 Ω connect in parrallel have equivalent resistance 44 Ω
= + + +...........+ 'n'times
= x n
=
n = = 110

So, 110 lamps can be 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 resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer

Case (i) When coils are used separately

Given
V = 220 V and R = 24 Ω

Using Ohm’s law

I = = =9.18 A

when they resistor used separately,9.166 A of current flows through each resistor

Case (ii) When coils connected in series

The total resistance = 24 Ω + 24 Ω = 48 Ω

I = = = 4.58 A

So, 4.58 A flows through the series circuit.

Case (iii) When coils connected in parallel

Total resistance = + = +
= =
= =12 Ω
I = = =18.33 A

18.33 A current flow in parallel circuit .

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.

Answer

(i) Here R₁ =1 Ω
R₂ = 2 Ω
V = 6V

Total resistance R_e= R₁ + R₂

= 1+ 2= 3 Ω

Using ohm's law

I = = = 2 A
Now Power
P =I²R
= 2²2= 4 x 2 = 8 W

The power consumed by the 2 Ω is 8 W.

(ii) In Parallel combination Voltage remains same So

P = = = = 8 A

The power consumed by the 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?

Answer

Here bulbs are connected in parallel So the voltage remains same.

For P = 100 W

P = V × I

I₁ = P/V

I = 100 W/220 V = 100/220 = 0.455A

For P = 60 W

I₂ = 60 W/220 V = 60/220 =0.273A

So net current I = I₁ + I₂ = 0.455 + 0.273 =0.728 A

Question 16

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer

Given W = 250 W and time = 1hr = 3600 seconds
We know

H = Pt

H = 250 × 3600
= 9 × 10⁵ J

Similarly For 1200 W and time =10 min = 600 seconds

H = 1200 W × 600 s = 7.2 × 10⁵ J

Energy consumed by the TV is more than the toaster.

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.

Answer

We know

P = I² R

R = 8 Ω , I = 15 A
time = 2 hour

P = (15A) ² × 8 Ω = 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 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 copper and aluminium wires are usually employed for electricity transmission?

a)

The resistivity and melting point of tungsten is very high.So, it doesn’t burn readily when heated. Electric lamps need very high temperature. That why tungsten used as filament of electric lamps.

b)

conductors of electric heating devices are made up of alloys because alloys don't burns readily at high temperature unlike metals and alloys have a greater resistivity compare to pure metal.

c)

The volatge will be divided across all component and each componenet of a series recives a samall voltage and if one electrical appliances stops working then all appliances will stop working. Hence series arrangement are not used in domestic circuits.

d)

R ∝
As We see resistance is inversely proportional to the area of cross section. When the area of cross section increases the resistance decreases and vice versa

e)

Due to low resistivity and good conductors of electricity Copper and aluminium are usually employed for electricity transmission.