Class 10 - Physics - Electricity

Question1.

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

Correct option: - (d) 25

The piece of wire of resistance R is cut into 5 equal parts.

Therefore, resistance of each part =(R/5)

As when all these parts are connected in parallel then the equivalent resistance R’ is given as,

R’ = (5/R) + (5/R) + (5/R) + (5/R) + (5/R) = (5+5+5+5+5)/(R)

= (25/R)

=>R’ = (R/25)

Therefore, (R/R’) = (R/R/25)

=R X (25/R)

=25.

Question2.

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

(a) I2R (b) IR2 (c) VI (d) V2/R

Correct option :- (b) IR2

As Power P = VI = I2R = (V2)/R

So the term 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

Given;

P = 100W and

Potential difference V = 220V

In First Case:-

Power P = VI   (equation 1)

Where I = current through the resistance where voltage is applied.

Also V =IR   (equation 2)

=>Using (equation 2) in (equation 1)

P = V2/R or R =V2/P

Therefore R = (220)2/100 =484 Ω

The resistance remains unchanged.

Second Case:-

Potential difference V = 110V,

Resistance = 484 Ω

=> P= (V2)/(R)

P = ((110)2/484) = 25W.

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

Correct option (c) 1:4

Potential difference = V

Resistance of wire = R

Resistance when connected in series = Rs

Resistance when connected in parallel = Rp

Resistance Rp = (1/R) + (1/R) = (2/R)

Heat produced H = I2Rt = (V/R) 2 Rt   (Using I = (V/R))

= (V/R) t

Therefore ratio of heat produced within in series (HS) to the ratio of heat produced when in parallel (HP) is given as:-

HS: HP = (Vt /Rs): (Vt/RP) = (Rp/RS)

HS: HP = (R/2)/ (2R) = (1/4)

Therefore HS: HP = 1:4

Question 5.

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

In order to measure potential difference between two points, a voltmeter should be connected in parallel.

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?

Given:-

Diameter d = 0.5 mm = 0.0005 m

Resistance, R = 10 Ω

Area of cross-section of the wire, A =π (d/2)2

Also,

R = ρ (l/A) where l= length of the wire, A = area of the wire, ρ =resistivity.

=> l = (RA/ ρ)

= (10 x 3.14 x (0.0005/2)2) / (1.6 x 10-8)

= (10 x 3.14 x 25)/ (4 x 1.6)

R=122.72m.

Therefore, length of the wire = 122.72m

If the diameter of the wire is doubled, new diameter=2×0.5=1mm=0.001m

New resistance = Rʹ

R’ = ρ (l/A)

= (1.6 x 10-8) x (122.72)/ ((3.14 x ½ x10-3)2)

=250.2 x 10-2 Ω

Therefore, the new resistance R’ = 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 –

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 voltage is to be plotted on x-axis and current on y-axis.

The graph will look like:-

Using Ohm’s law:-

R = (V/I) where R = resistance, I = current and V = potential difference

So, R = (V2-V1)/ (I2-I1); Resistance is given by the slope of the graph.

Therefore using values, V2 = 3.4V, V1 = 1.6V, I2 = 1A and I1 =0.5A

So, R = (3.4-1.6)/ (1-0.5)

R =3.6 Ω

Similarly, R = (V3-V2)/ (I3-I2);

= (6.7 – 3.4) / (2 -1) = 3.3 Ω

Also R = (V4-V3)/ (I4-I3);

= (10.2 -6.7)/ (3-2) = 3.5 Ω

R= (V₅ - V₄)/ (I₅ - I₄) = (13.2 - 10.2)/ (4-3) = 3 Ω

As R is variable, therefore taking average of the resistances,

Rav = (3.6 + 3.3 + 3.5 + 3)/4 = (13.4/4) = 3.35Ω

This shows, resistance = 3.35Ω

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.

Given:-

Potential difference V = 12V

Electric current I= 2.5mA = (2.5/1000) A

=0.0025A

According to Ohm’s law:-

V = IR

Or R= (V/I)

R = (12/0.0025)

Resistance R = 4800 Ω or R =4.8 kΩ

Question 9.

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, and 0.5 Ω and 12 Ω, respectively.

How much current would flow through the 12 Ω resistors?

Given,

Potential difference V = 9V,

Resistors are connected in series.

So R = 0.2 Ω+ 0.3 Ω+ 0.4 Ω+0.5 Ω+12 Ω

=13.4Ω

Using Ohm’s law:-

R = (V/I)

13.4 = (9/I)

Therefore current I = 0.67A

Question 10.

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

Given,

Current I = 5A

Potential difference V = 220V

Resistance of each resistor = 176 Ω

Let number of resistors connected in parallel =’n’

Total effective resistance = R

According to Ohm’s law;

V = IR

Therefore, R = (V/I)

= (220/5) = 44 Ω

Since resistors are connected in parallel,

(1/R) = (n/176)

n= (176/44)

n = 4 resistors.

Therefore 4 resistors are connected in parallel.

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.

• In order to get a resistance of 9 Ω from three resistors each of 6Ω.

Connecting 2 resistors in parallel and then connecting the third resistor in series.

i.e. (1/R) = (1/6) + (1/6)

= (1+1)/ (6) = (2/6)

= (1/3)

Or R = 3 Ω;

Therefore total resistance R = 3 Ω + 6Ω = 9 Ω

• In order to get resistance of 4 Ω from 3 resistors of 6 Ω each, connect all the resistors in parallel and then connect 2 such parallel in series.

Therefore (1/R) = (1/6) + (1/6) + (1/6)

= (1+1+1)/ (6)

= (3/6)

= (1/2)

R= 2 Ω;

Therefore R = 2 Ω + 2 Ω = 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?

Given,

Current I =5A

Power = 10 W

Potential difference V = 220V

Using Power P = (V2)/R

Therefore, 10 = (220)2/R

=> R = (220 x 220)/10

Therefore resistance of 1 bulb R = 4840 Ω.

By Ohm’s law, V = IR

Let Resistance of the circuit = R

Number of bulbs =z

Therefore, R = (V/I)

= (220/5) = 44 Ω

Total resistance R = 44 Ω

Resistance of each bulb = 4840 Ω.

When connected in parallel, (1/R) = (1/R1) + (1/R2) + … + z times

Therefore, (1/44) = (1/4840) xz

z= (4840)/ (44)

z=110

Therefore 110bulbs are connected in parallel in this circuit.

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?

Case 1: When each coil is used separately,

Given;

Potential difference V = 220V

Resistance =24 Ω. (one coil only)

Using Ohm’s law; V=IR

=> I = (V/R) = (220)/ (24)

I =9.2A.

Case 2: When the two coils are connected in series

Combined resistance when each of two coils having resistance 24 Ω is connected in series will be 24 Ω + 24 Ω = 48 Ω.

Resistance R = 48 Ω

Therefore I = (V/R)

= (220)/ (48)

I = 4.58Amp

Case 3: When the two coils are connected in parallel

Combined resistance when resistances are connected in parallel;

(1/R) = (1/24) + (1/24)

= (1+1)/ (24)

= (2/24)

(1/R) = (1/12)

Therefore R = 12 Ω (when two coils are in parallel)

Using I = (V/R)

= (220)/ (12)

=18.34Amp

Question 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.

• First Case:-

Potential difference V = 6V

Resistance R = 1 Ω + 2 Ω (resistors are in series)

=> R =3 Ω

Using Ohm’s law: R = (V/I)

So, 3 = (6/I)

I = (6/3) = 2A

This shows in a series circuit same current flows throughout the circuit. So current through 2 Ω is also 2 Amp.

So, P= I2R

P1= (2)2 x 2

=4 x 2

P1=8W

Therefore 8W power is used in 2Ω resistor.

• Second Case:-

In this case 4V battery is connected in parallel combination of 12 Ω and 2 Ω resistors.

Potential difference V = 4V (across the 2 Ω resistor)

Resistance R = 2 Ω

Power P2= (V2)/R

P2 = (4)2/2 = (16/2)

P2 = 8W

Therefore 8W power is used in 2Ω resistor

Comparing the 2 powers (P2/ P1) = (8W)/ (8W)

Or (P2/ P1) =1

Or P2 = P2

The 2Ω resistor uses equal power in both the circuits.

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?

As both the bulbs are connected in parallel, therefore the potential difference across each of them is = 220V.

• For 100W lamp,

Power = 100W

Voltage V = 220V

Using P =VI

Or Current I1 = (P/V)

= (100/220) = (5/11) A.

• For 60W lamp,

Power P = 60W

Voltage V = 220V

Using P =VI

Or Current I2 = (P/V)

= (60W/220V)

= (3/11)

Therefore total current I = I1 + I2

= (5/11) + (3/11) = (8/11)

I = 0.727A

Therefore total current is 0.727A.

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 250W in 1 hr;

Given:-

Power P =250W

Time t = 1 hr

Energy consumed, E = (P x t)

Therefore E = 250W x 1h = 250Wh

• Energy consumed by 1200W in 10 min;

Given:-

Power P = 1200W

Time t = 10 min = (10/60) h = (1/6) h

Energy consumed E = P x t

=1200W x (1/6) h = 200Wh

Therefore 250W TV set uses more power in 1h than a 1200W toaster in 10min.

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.

Given:-

Current I = 15A

Resistance R = 8 Ω

Time t = 2 hours

Using, Power P = I2 X R

= (15)2 x 8

P= 225 x 8

=1800W or 1800J/s

Therefore the rate at which heat is developed in the heater is 1800J/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 are copper and aluminium wires usually employed for electricity     transmission?

• Tungsten is used almost exclusively for filament of electric lamps as they have very high melting point and high resistivity.
• They don’t burn easily at high temperatures.
• The conductors of electric heating devices, such as bread- toasters and electric irons, made of an alloy rather than a pure metal because resistivity of the alloy is more than the resistivity of the metals which produces large amount of heat.
• There is voltage division in series circuit as each component receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly.
• Resistance of wire is inversely proportional to it area of cross-section.

(R ∝ l/A), this shows when the area of cross-section increases then the resistance decreases and vice-versa.

As copper and aluminium wires have low resistivity. They are good conductors of electricity therefore they are usually used in electrical transmission.