NCERT Solution Class 10 Science Chapter 11 Electricity - Param Himalaya

NCERT Exercise Solution Class 10 Science Chapter 11 Electricity - Param Himalaya

Example 11.1 A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.

Solution:

Given 

I = 0.5 A , t = 10 min = 600 s.

We have Q = It = 0.5 A × 600 s = 300 C

Page Number 172 – Question with solution 1 to 3 :

1. What does an electric circuit mean?

Solution : It is a continuous and closed path through which electric current flows. It contains different components like • Cell or Battery Plug Key • Wires • Electric Components (like Ammeter,Voltmeter) • Bulb etc 

2. Define the unit of current.

Solution : 

The unit of electric current is ampere (A).

I  = $\frac{Q}{t}$

When Q = 1 C and t = 1 sec then

I = $\frac{1C}{1S}$

I = 1A

1 A is defined as the flow of 1 C of charge through a wire in 1 s.

3. Calculate the number of electrons constituting one coulomb of charge.

Solution : 

We know 

Q = ne 

n = $\frac{Q}{e}$

= $\frac{1}{1.6 \times 10^{-19}} = \frac{1 \times 10^{19}}{1.6}$ 

= $\frac{10 \times 10^{19}}{16}$

= $\frac{5}{8} \times 10^{19}$

= $0.625 \times 10^{19}$

= $6.25 \times 10^{-1} \times 10^{19}$

n = $6.25 \times 10^{18}$

Therefore, $n = 6.25 \times 10^{18}$ electrons constitute one coulomb of charge.

Example 11.2 How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?

Solution :

Given  , Q = 2 C , V = 12 volt

We know

W = VQ = 12 V × 2 C = 24 J.

W = 24 J

Page number 174 – Question 1-3 with  Solution

1. Name a device that helps to maintain a potential difference across a conductor.

Solution : 

Any source of electricity like battery, cell, power supply, etc. helps to maintain a potential difference across a conductor.

2. What is meant by saying that the potential difference between two points is 1 V?

Solution :

When W = 1 J , Q = 1 C then

V = $\frac{W}{Q}$

V= $\frac{1J}{1C}$

V = 1 volt

If 1 J of work is required to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Solution : 

The energy given to each coulomb of charge is equal to the amount of work which is done in moving it.

Now we know that,

V = $\frac{W}{Q}$

Where, Charge (Q)= 1 C and Potential difference = 6 V 

W = V × Q  = 6 × 1 

W= 6 Joule.

Example 11.3 (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω? 

Solution : (a) Given V = 220 V, R = 1200 Ω. 

We know V = IR

I = $\frac{V}{R}$

Then I = $\frac{220 V}{1200 Ω}$ = 0.18 A.

(b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω?

Solution: (b) Given V = 220 V, R = 100 Ω.

We know V = IR

 I = $\frac{220 V}{100 Ω}$ = 2.2 A. 

Example 11.4. The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?

Solution :

Given  V = 60 V,

current I = 4 A.

According to Ohm’s law,

V= IR

R = $\frac{V}{I}$

R = $\frac{60}{4}$

R= 15Ω .

When the potential difference is increased to 120 V 

Then V= 120 volt

R= 15Ω 

So according to ohm's law

V= IR

I = $\frac{V}{R}$

I = $\frac{120}{15}$

I = 8 A

Example 11.5 Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 11.2, predict the material of the wire.

Solution :

Given the resistance R = 26 Ω, 

the diameter d = 0.3 mm = 3 × $10^{-4}$ m,  radius r = d/2 = 1.5 × $10^{-4}$ m

the length l of the wire = 1 m. 

the resistivity of the given metallic wire is 

ρ = $\frac{RA}{l} = \frac{R\pi r^{2}}{l}$ 

ρ = $\frac{26 \times 3.14 \times (1.5 × 10^{-4})^{2}}{l}$

ρ = $183.69 \times 10^{-8}$Ω

ρ = $1.84 × 10^{-6}$ Ω m.

The resistivity of the metal at 20°C is ρ = $1.84 × 10^{-6}$ Ω m From Table 11.2, we see that this is the resistivity of manganese.

Example 11.6: A wire of given material having length l and area of cross-section A has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length l/2 and area of cross-section 2A?

 Solution:

$R_{1} = \rho \frac{l}{A} = 4 \Omega$

$R_{2} = \rho \frac{l/2}{2A} = \frac{1}{4} \rho \frac{l}{A}$

$R_{2} = \frac{1}{4} R_{1}$

$R_{2} = \frac{1}{4} \times 4$

$R_{2} = 1\Omega$

The Resistance of the new wire is 1$ \Omega$

Page number 181 Intext Question :

1. On what factors does the resistance of a conductor depend?

Solution : $$R = \rho \frac{l}{A}$$

The resistance of a conductor depends upon the following factors :

1. Length of the conductor
2. Cross-Sectional area of the conductor
3. Material of the conductor
4. Temperature of the conductor

2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Solution : 

The current will flow more easily through thick wire. it is because the resistance of a conductor is inversely proportional to its area of cross-section. If thicker the wire, less is resistance and hence more easily the current flows.

$$R \propto \frac{1}{A}$$

3. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Solution : 

According to Ohm's law

V=IR

$I = \frac{V}{R}$

Now Potential difference is decreased to half

$\therefore$ New potential difference ${V}' = \frac{V}{2}$

Resistance remains constant 

So the new current 

${I}' = \frac{{V}'}{R} = \frac{\frac{V}{2}}{R}$ 

${I}'= \frac{1}{2}\frac{V}{R}$

${I}' = \frac{1}{2}I=\frac{I}{2}$

4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Solution : 

The resistivity of an alloy is higher than the pure metal. More ever , at high temperatures the alloys do not melt readily, Hence , the coils of heating appliances such as electric toasters and electric irons are made of any alloy rather than a pure metal.

5. Use the data in Table 11.2 to answer the following – (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?

Solution :

(a) Resistivity of iron = $10.0 \times 10^{-8} \Omega$

Resistivity of mercury = $94.0 \times 10^{-8} \Omega$

Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.

(b) It can be observed from Table 11.2 that the resistivity of silver is the lowest among the listed materials. Hence , it is the best conductor.

Example 11.7 An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery (Fig. 11.9). Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.

Solution : 

The resistance of electric lamp , $R_{1} = 20 \Omega$

The resistance of the conductor connected in series , $R_{2} = 4 \Omega$

Then the total resistance in the circuit 

$R = R_{1}+R_{2}$

$R_{s}$ = $20 \Omega+ 4 \Omega$ = $24 \Omega$

The total potential difference across the two terminals of the battery V = 6V.

Now by Ohm's law , the current through the circuit is given by 

$I =\frac{V}{R} = \frac{6V}{24 \Omega} = 0.25A$

Applying Ohm's law to the electric lamp and conductor separately we get potential difference across the electric lamp,

$V_{1} = 20 \Omega \times 0.25 \Omega = 5V$

and that across the conductor , $V_{2} = 4 \Omega \times 0.25 \Omega = 1V$

Suppose that we like to replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance must be such that a potential difference of 6V across the battery terminals will cause a current of 0.25 A in the circuit. The resistance R of this equivalent resistor would be.

$$R = \frac{V}{I} = \frac{6V}{0.25A} = 24 \Omega$$

This is the total resistance of the series circuit. It is equal to the sum of the two resistances.

Page 185 

1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Solution : 

Three cells of potential 2V , each connected in series therefore the potential difference of the battery will be 2V + 2V + 2V = 6V. The following circuit diagram shows three resistors of resistance 5Ω , 8Ω and 12Ω respectively connected in series and a battery of potential 6V and a plug key which is closed means the current is flowing in the circuit.

2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Solution : 

An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following fiqure.

The resistance are connected in series.

Ohm's law can be used to obtain the readings of ammeter and voltmeter. According to ohm's law.

V=IR

Where Potential difference , V= 6V

Current flowing through the circuit/Resistor = I

Resistance of the circuit , R = 5+8 +12 = 25Ω

$I = \frac{V}{R} = \frac{6}{25} = 0.24A$

Potential difference across 12 $\Omega$ resistor = $V_{1}$

Current flowing through the 12 $\Omega$ resistor , I = 0.24A

Therefore, using Ohm's law , we obtain 

$V_{1}$ = IR = .24 $\times$ 12 = 2.88 V

Therefore, the reading of the ammeter will be 0.24A

The reading of the voltmeter will be 2.88V

Example 11.8 In the circuit diagram given in Fig. 11.10, suppose the resistors R1,R2 and R3 have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.

Solution : 

$R_{1} =5 \Omega$ , $R_{2} = 10 \Omega$ and $R_{3} = 30 \Omega$

Potential difference across the battery , V = 12 V

This is also the potential difference across each of the individual resistor , therefore , to calculate the current in the resistors , we use Ohm's law.

The Current $I_{1}$ , through $R_{1} = \frac{V}{R_{1}}$

$I_{1} = \frac{12V}{5\Omega}$ = $2.4 A$

The Current $I_{2}$ , through $R_{2} = \frac{V}{R_{2}}$

$I_{2} = \frac{12V}{10\Omega} = 1.2 A$

The Current $I_{3} , through R_{3} = \frac{V}{R_{3}}$

$I_{3} = \frac{12V}{30\Omega} = 0.4 A$

The total current in the circuit. 

$I = I_{1} + I_{2} + I_{3} = (2.4 + 1.2 + 0.4) A$

The total resistance $R_{p}$, is given by

$\frac{1}{R_{p}} = \frac{1}{5}+\frac{1}{10}+\frac{1}{30} = \frac{1}{3}$

thus , $R_{p} = 3 \Omega$

Example 11.9 If in Fig. 11.12, R1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω, R5 = 60 Ω, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.

Solution : 

Suppose we replace the parallel resistors $R_{1}$ and $R_{2}$ by an equivalent resistor or resistance , ${R}'$. Similarly we replace the parallel resistors $R_{3}$, $R_{4}$ and $R_{5}$ by an equivalent single resistor of resistance ${R}'' $ , then

$\frac{1}{{R}'} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{5}{40}$

$\frac{1}{{R}'} = \frac{1}{10} + \frac{1}{40} = \frac{5}{40}$

${R}'= 8 \Omega$

Similarly , $\frac{1}{{R}''} = \frac{1}{R_{3}} + \frac{1}{R_{4}} +\frac{1}{R_{5}}$

$\frac{1}{{R}''} = \frac{1}{30} + \frac{1}{20} +\frac{1}{60}$

$\frac{1}{{R}''} = \frac{6}{60} = \frac{1}{10}$

${R}'' = 10 \Omega$

Thus , the total resistance , $R = {R}' + {R}'' = 8 +10 = 18 \Omega$

So Total current $I = \frac{V}{R} = \frac{12V}{18\Omega} = 0.67 A$

Page 188 Intext Question with Solution : 

1.Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and $10^{6}$ Ω, (b) 1 Ω and $10^{3}$ Ω, and $10^{6}$Ω.

Solution :

(a) When $1\Omega$ and $10^{6} \Omega$ are connected in parallel.

Let R be the equivalent resistance 

$\therefore \frac{1}{R} = \frac{1}{1} + \frac{1}{10^{6}}$

$R= \frac{10^{6}}{1+10^{6}} \approx \frac{10^{6}}{10^{6}} = 1 \Omega$

Therefore , equivalant resistance = $1 \Omega$

(b) When $1 \Omega , 10^{3} \Omega , 10^{6} \Omega$ are connected in parallel.

Let R be the equivalent resistance 

$\frac{1}{R} = \frac{1}{1} + \frac{1}{10^{3}} + \frac{1}{10^{6}} = \frac{10^{6}+10^{3}+1}{10^{6}}$

$R = \frac{1000000}{1001001} = 0.999\Omega$

Therefore , equivalent resistance = $0.999 \Omega$

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?

Solution : 

Resistance of electric lamp .$R_{1} = 100 \Omega$

Resistance of toaster , $R_{2} = 50 \Omega$

Resistance of water filter , $R_{3} = 500 \Omega$

Potential difference of the Source , V = 220 V

These are connected in parallel , then

Let R be the equivalent resistance of the circuit.

$\frac{1}{R} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

$\frac{1}{R} = \frac{1}{100}+\frac{1}{50}+\frac{1}{500}$

$\frac{1}{R} = \frac{5+10+1}{500}$

$\frac{1}{R} = \frac{16}{500}$

$R = \frac{500}{16}$

According to Ohm's law,

V=IR

$I = \frac{V}{R}$

where  current flowing through the circuit = I

$I= \frac{220}{\frac{500}{16}}$

$I = \frac{220 \times 16}{500}$

I = 7.04 A

Let ${R}' = \frac{V}{I} = \frac{220}{7.04} = 31.25 \Omega$

Therefore , the resistance of the electric iron is $31.25 \Omega$ and the current flowing through it is 7.04 A

3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Solution : 

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.

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?

Solution : 

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

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

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

Therefore , their equivalent resistance will be given by 

$\frac{1}{{R}'} = \frac{1}{3}+\frac{1}{6}$

$\frac{1}{{R}'} = \frac{3}{6} =\frac{1}{2}$

${R}' = 2 \Omega$

This equivalent resistor of resistance $2 \Omega$ is connected  to $2 \Omega$ in series  therefore

Their equivalant resistance of the circuit = $2 \Omega + 2 \Omega = 4 \Omega$

hence the total resistance of the circuit is $4 \Omega$

(b) 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 

$\frac{1}{R} = \frac{1}{1}+\frac{1}{2}+\frac{1}{6}$

$\frac{1}{R} = \frac{1}{3}+\frac{1}{2}+\frac{1}{6}$

$\frac{1}{R} = \frac{2+3+1}{6} = \frac{6}{6} = 1$

R =$1 \Omega$

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? 

Solution : There are four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω respectively 

(a) if these coils are connected in series , then the equivalent resistance will be the highest , given by the 

R = 4+8+12+24 = 48Ω

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

$\frac{1}{R} = \frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24} = \frac{6+3+2+1}{24}$

$\frac{1}{R} = \frac{12}{24} = \frac{1}{2}$

$R = 2 \Omega$

Example 11.10 An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?

Solution : 

We know that the power input is

P = V I

Thus the current I = P/V

(a) When heating is at the maximum rate,

I = 840 W/220 V = 3.82 A;

and the resistance of the electric iron is

R = V/I = 220 V/3.82 A = 57.60 Ω.

(b) When heating is at the minimum rate,

I = 360 W/220 V = 1.64 A;

and the resistance of the electric iron is

R = V/I = 220 V/1.64 A = 134.15 Ω.

Example 11.11 100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor.

Solution

H = 100 J, R = 4 Ω, t = 1 s, V = ?

we have the current through the resistor as

I = √(H/Rt)

= √[100 J/(4 Ω × 1 s)]

= 5 A

Thus the potential difference across the resistor,

V = IR

V= 5 A × 4 Ω= 20 V.

Page no 190

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

Solution : 

The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminium is very law so it does not glow.

2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Solution : 

Given Charge , Q = 96000C

Time , t = 1 hr = 60 60 = 3600s

Potential difference , V = 50 volts

Now we know that H=Vit

So we have to calculate I first

I =$\frac{Q}{t}$

\therefore I = $\frac{96000}{3600} = \frac{80}{3}A$

then H=VIt = 50 $\times \frac{80}{3} \times 60 \times 60 $

H = $4.8 \times 10^{6}J$

3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Solution :

The amount of Heat (H) produced is given by the joule's law of heating as H=VIt

where Current , I = 5A

time , t = 30 s

Voltage , V = Current  $\times$  Resistance = 5 $\times$ 20 = 100 V

H = 100 $\times$ 5 $\times$ 30 = 1.5 \times $10^{4}J$

Example 11.12 An electric bulb is connected to a 220 V generator. The current is  0.50 A. What is the power of the bulb?

Solution

P = VI

= 220 V × 0.50 A = 110 J/s

P = 110 W.

Example 11.13 An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h?

Solution

The total energy consumed by the refrigerator in 30 days would be

400 W × 8.0 hour/day × 30 days = 96000 W h = 96 kW h

Thus the cost of energy to operate the refrigerator for 30 days is

96 kW h × Rs 3.00 per kW h = Rs 288.00

Page number 192 

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

Solution : 

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.

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.

Solution : 

Power (P) is given by the expression , P =VI

where Voltage , V = 220 Volt 

Current , I = 5A

P = 220 $ \times 5$ = 1100 W

Energy consumed by the motor = Pt

where 

Time ,t = 2h = 2 $\times 60 \times$ 60 = 7200s

$\therefore P = 1100 \times 7200 = 7.92 \times 10^{6}J$

Exercise Solution : 

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

Solution : 

$\frac{1}{R^{'}} =\frac{1}{\frac{R}{5}} +\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}+\frac{1}{\frac{R}{5}}$

$\frac{1}{R^{'}} =\frac{5}{\frac{R}{5}}$

$\frac{1}{R^{'}} = \frac{25}{R}$

Then 

$\frac{R}{R^{'}} = 25$

Answer : (d) 25 

2. 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 :

 (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 (a) 100 W (b) 75 W (c) 50 W (d) 25 W

Solution : 

$P = \frac{V^{2}}{R} or R = \frac{V^{2}}{P}$ , 

Resistance of electric bulb will be same R

then

$\frac{V_{1}^{2}}{P_{1}} = \frac{V_{2}^{2}}{P_{2}}$

$P_{2} = \frac{V_{2}^{2} \times P_{1}}{V_{1}^{2}}$

$= \frac{110^{2} \times 100}{220^{2}}$

$P_{2} = 25 W$

Answer : (d) 25W

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

Solution :

When wires are connected in series 

$R_{eq} = R + R = 2R$

$\therefore$ Heat produced in combination ,

$H_{1} = \frac{V^{2}}{R_{eq}}t$

When wires are connected in parallel , 

$\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R}$

$\frac{1}{R_{eq}} = \frac{2}{R}$

$R_{eq} = \frac{R}{2}$

$\therefore$ Heat produced in combination ,

$H_{2} = \frac{V^{2}}{\frac{R}{2}}t$

$H_{2} = \frac{2V^{2}}{R}t$

Hence , $\frac{H_{1}}{H_{2}} = \frac{\frac{V^{2}}{R_{eq}}t}{\frac{2V^{2}}{R}t}$

$\frac{H_{1}}{H_{2}} = \frac{1}{4}$

Answer : Option (c) 1:4

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

Solution : 

To measure the potential difference between 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?

Solution : 

Given : Diameter = 0.5mm = 0.0005m 

Resistance , R = 10Ω

We know that 

$R = \rho\frac{l}{A}$

$l = \frac{RA}{\rho}$

$l = \frac{R \times \pi D^{2} }{\rho}$

$l = \frac{10 \times 3.14 \times(\frac{0.0005}{2})^{2}}{1.6 \times 10 ^{-8}}$

$l = \frac{10 \times 3.14 \times \frac{25}{4}\times 10^{-8}}{1.6 \times 10 ^{-8}}$

$l = \frac{10 \times 3.14 \times 25\times 10^{-8}}{4 \times 1.6 \times 10 ^{-8}}$

$l = \frac{785}{6.4} = 122.65$

if the diameter of the wire is doubled , new diameter = 2 $\times$ 0.0005 = 0.001 m

Let new resistance be $R^{`}$

$R^{'} = \rho \frac{l}{A}$

$R^{'} = \frac{1.6 \times 10^{-8} \times 122.65}{\pi(\frac{0.0001}{2})^{2}}$

$R^{'} = \frac{1.6 \times 10^{-8} \times 122.65 \times 4}{3.14 \times 10^{-6}}$

$R^{'} = \frac{784.96}{3.14} \times 10^{-2}$

$R^{'} = 250 \times 10^{-2}$

$R^{'} = 2.5 \Omega$

Therefore , the length of the wire is 122.65 m and the new resistance is 2.5 Ω

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.

V(Volt)	1.6	3.4	6.7	10.2	13.2
I (Amperes)	0.5	1.0	2.0	3.0	4.0

Solution : 

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 fiqure.

The slope of the line gives the value of resistance (R) as,

$Slope = \frac{1}{R} = \frac{BC}{AC} = \frac{2}{6.8}$

$R = \frac{6.8}{2} = 3.4 \Omega$

Therefore , the resistance of the resistor is $3.4 \Omega$

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.

Solution : Resistance (R) of a resistor is given by Ohm's law as , V=IR

$R =\frac{V}{I}$

Where , Potential difference , V = 12 Volt

Current in the circuit , I = 2.5 mA = $2.5 \times 10^{-3}$ A

$R = \frac{12}{2.5 \times 10^{-2}} = 4.8 \times 10^{3} \Omega$

$R= 4.8k\Omega$

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?

Solution : 

Current flow through the component is the same , given by Ohm's las as

V = I/R

I = V/R

where ,

R is the equivalent resistance

then R = 0.2 Ω+0.3 Ω+0.4 Ω +0.5 Ω +12 Ω

R = 13.4 Ω

Potential difference , V = 9 Volt

so , I = 9/13.4

I = 0.671 A

Therefore , the current 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?

Solution : For x number of resistor of resistance 176  Ω , the equivalent resistance of the resistors connected in parallel is given by Ohm's las as V = IR

R = V/I

Where ,

Supply voltage , V = 220 volt 

Current , I = 5A

Equivalent resistance of the combination = R , given as

$\frac{1}{R} = x \times \frac{1}{176}$

$R = \frac{176}{x}$

From OHM'S LAW

$\frac{V}{I} = \frac{176}{x}$

$x = \frac{176 \times I}{V} = \frac{176 \times 5}{220}$

x = 4

Therefore , four resistors of 176 $\Omega$ are required to draw the given amount of current.

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.

Solution : 

(a) When they are connected in parallel Two 6Ω resistors are connected in parallel.

Resistance $\frac{1}{R} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

$R = 3 \Omega$ 

If 3rd resistor of 6Ω and 3Ω are connected in series, it becomes $R^{'} = R + 6\Omega = 3 \Omega + 6 \Omega = 9 \Omega$

(b) Two  6 \Omega resistors are in series . Their equivalent resistance will be the 

R = 6 \Omega + 6 \Omega  = 12 \Omega 

The third 6 \Omega  resistor is in parallel with 12 \Omega . Hence , equivalent resistance will be 

$\frac{1}{R^{'}} = \frac{1}{R} + \frac{1}{6}$

$\frac{1}{R^{'}} = \frac{1}{12} + \frac{1}{6}$

$\frac{1}{R^{'}} = \frac{3}{12}$

$\frac{1}{R^{'}} = \frac{1}{4}$

$R^{'} = 4 \Omega$

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?

Solution : 

Resistance $R_{1}$ of the bulb is given by the expression,

Supply Voltage , V = 220 Volt

Maximum allowable current , I = 5A

Rating of an electric bulb P = 10 walts 

Because $R_{1} = \frac{V^{2}}{P}$

then ,

$R_{1} = \frac{(220)^{2}}{10} = 4840 \Omega$

According to Ohm's Law 

V=IR

Let R is the total resistance of the circuit for x number of electric bulbs 

$R = \frac{V}{I}$

$R = \frac{220}{5}$

$R = 44 \Omega$

Resistance of each electric bulb , R_{1} = 4840 \Omega

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} +..... upto x times$

$\frac{1}{R} = \frac{1}{R_{1}}\times x$

$x= \frac{R_{1}}{R} = \frac{4840}{44} = 110$

$\therefore$ Number of electric bulbs connected in parallel are 110

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?

Solution : 

Supply Voltage  , V = 220 volt 

Resistance of one coil , R = 24 $\Omega$

(i) Coils are used  separately 

According to Ohm's law 

$V = I_{1}R_{1}$

Where 

$I_{1}$ is the current flowing through the coil

$I_{1} = \frac{V}{R_{1}} = \frac{220}{24} = 9.166 A$

Therefore , 9.16 A current will flow through the coil when used separately

(ii) Coils are connected in series

Total resistance , $R_{2} = 24 \Omega + 24 \Omega = 48 \Omega$

According to Ohm's law , $V = I_{2}R_{2}$

Where 

$I_{2}$ is the current flowing through the series circuit.

$I_{2} = \frac{V}{R_{2}}n= \frac{220}{48} = 4.58 A$

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,

$\frac{1}{R_{3}} = \frac{1}{24 \Omega} + \frac{1}{24 \Omega}$

$\frac{1}{R_{3}} = \frac{2}{24} = \frac{1}{12}$

$R_{3} = 12 \Omega$

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.

Solution : 

(i) Potential difference , V = 6 Volt 

1 Ω and 2 Ω resistors are connected in series . so 

$R = R_{1} + R_{2}$

$R =  1+ 2 = 3Ω$

According to Ohm's Law 

$V = \frac{I}{R}$

Where , 

I is the current through the circuit

I = \frac{6}{3} = 2A

Same current will flow through each component of the circuit hence current flowing through the 2Ω  resistor is 2A. Power is given by the expression.

$P = I^{2}R = 2^{2} \times 2 = 8 W$

(ii) Potential Difference , V = 4 V

12 $\Omega$ and $2 \Omega$ resistors are connected in parallel. The voltage across each components of a parallel circuit remains the same.

Hence , the voltage across 2 \Omega resistor will be 4 V.

$P = \frac{V^{2}}{R} = \frac{4^{2}}{2} = 8 W$

Therefore , the power used by $2 \Omega$ resistor is 8W.

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?

Solution : both the lamp are connected in parallel , Therefore , Potential difference across each of them will be 220 V , 

Current drawn by the bulb of rating 100W is given by  Power = Voltage \times Current

$Current = \frac{Power}{Voltage}$ 

$I_{1}= \frac{100}{220}$

$I_{2}= \frac{60}{220}$

Then Total current drawn from the line $I = I_{1} + I_{2}$

$I= \frac{100}{220} +\frac{60}{220}$

$I = \frac{100+60}{220}$

$I = \frac{160}{220}$

I = 0.727 A

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

Solution : 

Energy consumed by an electrical appliance is given by the expression , H = Pt 

Where , Power of the appliance = P

Time = t

Energy consumed by a TV set of power 250 W in 1 h = $250 \times 3600$ = $9 \times 10^{5}J$

Energy consumed by a Toaster of power 1200 W in 10 minutes = $1200 \times 600 = 7.2 \times 10^{5}J$

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.

Solution : Rate of heat produced by a device is given by the expression for power as $P = I^{2}R$ where ,

Resistance of the electric heater , R =  8 Ω  

Current drawn , I = 15 A

$P = 15^{2} \times 8 = 1800 \frac{j}{s}$

Therefore , heat is produced by the heater at the rate of $1800 \frac{j}{s}$

18. Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

Solution : 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.

(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?

Solution :  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.

(c) Why is the series arrangement not used for domestic circuits?

Solution : 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.

(d) How does the resistance of a wire vary with its area of cross-section?

Solution : 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.

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Solution : Copper and aluminium are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.