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

Answer:

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) I^{2}R (b) IR^{2} (c) VI (d) V^{2}/R

Answer:

Correct option :- __(b) IR ^{2}__

As Power P = VI = I^{2}R = (V^{2})/R

So the term IR^{2} 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:

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 = V^{2}/R or R =V^{2}/P

Therefore R = (220)^{2}/100 =484 Ω

The resistance remains unchanged.

Second Case:-

Potential difference V = 110V,

Resistance = 484 Ω

=> P= (V^{2})/(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

Answer:

Correct option __(c) 1:4__

Potential difference = V

Resistance of wire = R

Resistance when connected in series = R_{s}

Resistance when connected in parallel = R_{p}

Resistance R_{p} = (1/R) + (1/R) = (2/R)

Heat produced H = I^{2}Rt = (V/R)^{ 2} Rt (Using I = (V/R))

= (V/R) t

Therefore ratio of heat produced within in series (H_{S}) to the ratio of heat produced when in parallel (H_{P}) is given as:-

H_{S}: H_{P} = (V_{t} /R_{s}): (V_{t}/R_{P}) = (R_{p}/R_{S})

H_{S}: H_{P}_{ =} (R/2)/ (2R) = (1/4)

Therefore H_{S}: H_{P }= 1:4

**Question 5.**

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

Answer:

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?

Answer:

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.

Answer:

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 = (V_{2}-V_{1})/ (I_{2}-I_{1}); Resistance is given by the slope of the graph.

Therefore using values, V_{2} = 3.4V, V_{1} = 1.6V, I_{2} = 1A and I_{1} =0.5A

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

R =3.6 Ω

Similarly, R = (V_{3}-V_{2})/ (I_{3}-I_{2});

= (6.7 – 3.4) / (2 -1) = 3.3 Ω

Also R = (V_{4}-V_{3})/ (I_{4}-I_{3});

= (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,

R_{av} = (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.

Answer:

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?

Answer:

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?

Answer:

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.

Answer:

- 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?

Answer:

Given,

Current I =5A

Power = 10 W

Potential difference V = 220V

Using Power P = (V^{2})/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/R_{1}) + (1/R_{2}) + … + 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?

Answer:

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.

Answer:

- 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= I^{2}R

P_{1}= (2)^{2} x 2

=4 x 2

P_{1}=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 P_{2}= (V^{2})/R

P_{2} = (4)^{2}/2 = (16/2)

P_{2} = 8W

Therefore 8W power is used in 2Ω resistor

Comparing the 2 powers (P_{2}/ P_{1}) = (8W)/ (8W)

Or (P_{2}/ P_{1}) =1

Or P_{2} = P_{2}

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?

Answer:

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 I_{1} = (P/V)

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

- For 60W lamp,

Power P = 60W

Voltage V = 220V

Using P =VI

Or Current I_{2} = (P/V)

= (60W/220V)

= (3/11)

Therefore total current I = I_{1} + I_{2}

= (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?

Answer:

- 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.

Answer:

Given:-

Current I = 15A

Resistance R = 8 Ω

Time t = 2 hours

Using, Power P = I^{2} 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?

Answer:

- 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.

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