NCERT Solutions for Class 10 Chapter 11 – Electricity

A battery consisting of one or more electric cells is one of the devices that help to maintain a potential difference across a conductor.

 

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

 

The unit of current is ampere. Ampere is defined by the flow of one coulomb of charge per second.

 

A continuous closed path made of electric components through which an electric current flows is known as an electric circuit. A simple circuit consists of the following components:

(a) Conductors

(b) Cell

(c) Switch

(d) Load

 

The value of the charge of an electron is 1.6 × 10-19 C.

According to charge quantization,

Q = nqe, where n is the number of electrons and qe is the charge of an electron.

Substituting the values in the above equation, the number of electrons in a coulomb of charge can be calculated as follows:

Therefore, the number of electrons constituting one coulomb of charge is 6. 25 × 1018.

In text 12.2 Page: 202

 

We know that the potential difference between two points is given by the equation,

V = W/Q, where,

W is the work done in moving the charge from one point to another

Q is the charge

From the above equation, we can find the energy given to each coulomb as follows:

W = V × Q

Substituting the values in the equation, we get

W = 6V × 1C = 6 J

Hence, 6 J of energy is given to each coulomb of charge passing through a 6 V of battery.

 

The resistance of the conductor depends on the following factors:

a. Temperature of the conductor

b. Cross-sectional area of the conductor

c. Length of the conductor

d. Nature of the material of the conductor

 

Resistance is given by the equation,

R = ρ l/A

where,

ρ is the resistivity of the material of the wire,

l is the length of the wire

A is the area of the cross-section of the wire.

From the equation, it is evident that the area of the cross-section of wire is inversely proportional to the resistance. Therefore, the thinner the wire, the more the resistance and vice versa. Hence, current flows more easily through a thick wire than a thin wire.

 

The change in the current flowing through the electrical component can be determined by Ohm’s Law.

According to Ohm’s Law, the current is given by

I = V/R

Now, the potential difference is reduced to half keeping the resistance constant,

Let the new voltage be V’ = V/2

Let the new resistance be R’ = R and the new amount of current be I’.

The change in the current can be determined using Ohm’s law as follows:

Therefore, the current flowing the electrical component is reduced by half.

 

The melting point of an alloy is much higher than a pure metal because of its high resistivity. At high temperatures, alloys do not melt readily. Therefore, alloys are used in heating appliances such as electric toasters and electric irons.

 

a. Iron is a better conductor than mercury because the resistivity of mercury is more than the resistivity of iron.

b. Among all the materials listed in the table, silver is the best conductor because the resistivity of silver is lowest among all, i.e., 1.60 × 10–8.

 

A battery of three cells of 2 V each equals to battery of potential 6 V. The circuit diagram below shows three resistors of resistance 12 Ω, 8 Ω and 5 Ω connected in series along with a battery of potential 6 V.

 

An ammeter should always be connected in series with resistors while the voltmeter should be connected in parallel to the resistor to measure the potential difference as shown in the figure below.

Using Ohm’s Law, we can obtain the reading of the ammeter and the voltmeter.

The total resistance of the circuit is 5 Ω + 8 Ω +12 Ω = 25 Ω.

We know that the potential difference of the circuit is 6 V, hence the current flowing through the circuit or the resistors can be calculated as follows:

I = V/R = 6/25 = 0.24A

Let the potential difference across the 12 Ω resistor be V1.

From the obtained current V1 can be calculated as follows:

V1 = 0.24A × 12 Ω = 2.88 V

Therefore, the ammeter reading will be 0.24 A and the voltmeter reading be 2.88 V.

 

(a) When 1 Ω and 106 are connected in parallel, the equivalent resistance is given by

Therefore, the equivalent resistance is 1 Ω.

(b) When 1 Ω, 103 Ω, and 106 Ω are connected in parallel, the equivalent resistance is given by

Therefore, the equivalent resistance is 0.999 Ω.

 

The electric lamp, the toaster and the water filter connected in parallel to a 220 V source can be shown as using a circuit diagram as follows:

The equivalent resistance of the resistors can be calculated as follows:

The resistance of the electric iron box is 31.25 Ω.

 

When the electrical devices are connected in parallel there is no division of voltage among the appliances. The potential difference across the devices is equal to supply voltage. Parallel connection of devices also reduces the effective resistance of the circuit.

 

Electric power is the rate of consumption of electrical energy by electric appliances. Hence, the rate at which energy is delivered by a current is the power of the appliance.

 

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

From the circuit above, it is understood that 3 Ω and 6 Ω are connected in parallel. Hence, their equivalent resistance is given by

The equivalent resistor 2 Ω is in series with the 2 Ω resistor. Now the equivalent resistance can be calculated as follows:

Req= 2 Ω +2 Ω = 4 Ω

Hence, the total resistance of the circuit is 4 Ω.

(b) The circuit diagram below, shows the connection of three resistors.

From the circuit, it is understood that all the resistors are connected in parallel. Therefore, their equivalent resistance can be calculated as follows:

The total resistance of the circuit is 1 Ω.

 

(a) If the four resistors are connected in series, their total resistance will be the sum of their individual resistances and it will be the highest. The total equivalent resistance of the resistors connected in series will be 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

(b) If the resistors are connected in parallel, then their equivalent resistances will be the lowest.

Their equivalent resistance connected in parallel is

Hence, the lowest total resistance is 2 Ω.

 

The heating element of an electric heater is made of an alloy which has a high resistance. When the current flows through the heating element, the heating element becomes too hot and glows red. The cord is usually made of copper or aluminum which has low resistance. Hence the cord doesn’t glow.

 

The heat generated can be computed by Joule’s law as follows:

H = VIt

where,

V is the voltage, V = 50 V

I is the current

t is the time in seconds, 1 hour = 3600 seconds

The amount of current can be calculated as follows:

 

The amount of heat generated can be calculated using the Joule’s law of heating, which is given by the equation

H = VIt

Substituting the values in the above equation, we get,

H = 100 × 5 × 30 = 1.5 × 104 J

The amount of heat developed by the electric iron in 30 s is 1.5 × 104 J.

 

The power of the motor can be calculated by the equation,

P = VI

Substituting the values in the above equation, we get

P = 220 V × 5 A = 1100 W

The energy consumed by the motor can be calculated using the equation,

E = P × T

Substituting the values in the above equation, we get

P = 1100 W × 7200 = 7.92 × 106 J

The power of the motor is 1100 W and the energy consumed by the motor in 2 hours is 7.92 × 106 J.

 

(d) 25

Explanation:

The resistance is cut into five equal parts, which means that the resistance of each part is R/5.

We know that each part is connected to each other in parallel, hence the equivalent resistance can be calculated as follows:

The ratio of R/R′ is 25.

 

 b) IR2

Explanation:

Electrical power is given by the expression P = VI. (1)

According to Ohm’s law,

V = IR

Substituting the value of V in (1), we get

P = (IR) × I

P = I2R

Similarly, from Ohm’s law,

I = V/R

Substituting the value of I in (1), we get

P = V × V/R = V2/R

From this, it is clear that the equation IR2 does not represent electrical power in a circuit.

 

To measure the voltage between any two points, the voltmeter should be connected in parallel between the two points.

25 W

Explanation:

The energy consumed by the appliance is given by the expression

P = VI = V2/R

The resistance of the light bulb can be calculated as follows:

R = V2/P

Substituting the values, we get

R = (220)2/100 = 484 Ω

Even if the supply voltage is reduced, the resistance remains the same. Hence, the power consumed can be calculated as follows:

P = V2/R

Substituting the value, we get

P = (110)2 V/484 Ω = 25 W

Therefore, the power consumed when the electric bulb operates at 110 V is 25 W.

 

Let Rs and Rp be the equivalent resistance of the wires when connected in series and parallel respectively.

For the same potential difference V, the ratio of the heat produced in the circuit is given by

Hence, the ratio of the heat produced is 1:4.

 

The resistance of the copper wire of length in meters and area of cross-section m2 is given by the formula

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

 

The plot between voltage and current is known as IV characteristic. The current is plotted in the y-axis while the voltage is plotted in the x-axis. The different values of current for different values of voltage are given in the table. The I V characteristics for the given resistor is shown below.

The slope of the line gives the value of resistance.

The slope can be calculated as follows:

Slope = 1/R = BC/AC = 2/6.8

To calculate R,

R = 6.8/2 = 3.4 Ω

The resistance of the resistor is 3.4 Ω.

 

The value of the resistor can be calculated using Ohm’s Law as follows:

 

 

In series connection, there is no division of current. The current flowing across all the resistors is the same.

To calculate the amount of current flowing across the resistors, we use Ohm’s law.

But first, let us find out the equivalent resistance as follows:

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

Now, using Ohm’s law,

The current flowing across the 12 Ω is 0.671 A.

 

Let us consider the number of resistors required as ‘x.’

The equivalent resistance of the parallel combination of resistor R is given by

The number of resistors required is 4.

 

If we connect all the three resistors in series, their equivalent resistor would 6 Ω + 6 Ω + 6 Ω =18 Ω, which is not the desired value. Similarly, if we connect all the three resistors in parallel, their equivalent resistor would be

which is again not the desired value.

We can obtain the desired value by connecting any two of the resistors in either series or parallel.

Case (i)

If two resistors are connected in parallel, then their equivalent resistance is

The third resistor is in series, hence the equivalent resistance is calculated as follows:

R = 6 Ω + 3 Ω = 9 Ω

Case (ii)

When two resistors are connected in series, their equivalent resistance is given by

R = 6 Ω + 6 Ω = 12 Ω

The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is calculated as follows:

 

The resistance of the bulb can be calculated using the expression

P1 = V2/R1

R1 = V2/P1

Substituting the values, we get

Hence, 110 lamps can be connected in parallel.

 

Case (i) When coils are used separately

Using Ohm’s law, we can find the current flowing through each coil as follows:

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

Case (ii) When coils connected in series

The total resistance in the series circuit is 24 Ω + 24 Ω = 48 Ω

The current flowing through the series circuit is calculated as follows:

Therefore, a current of 4.58 A flows through the series circuit.

Case (iii) When coils connected in parallel

When the coils are connected in parallel, the equivalent resistance is calculated as follows:

The current in the parallel circuit is 18.33 A.

 

(i) The potential difference is 6 V and the resistors 1 Ω and 2 Ω are connected in series, hence their equivalent resistance is given by 1 Ω + 2 Ω = 3 Ω. The current in the circuit can be calculated using the Ohm’s law as follows:

Therefore, the power consumed by the 2 Ω is 8 W.

(ii) When 12 Ω and 2 Ω resistors are connected in parallel, the voltage across the resistors remains the same. Knowing that the voltage across 2 Ω resistor is 4 V, we can calculate the power consumed by the resistor as follows:

The power consumed by the 2 Ω resistor is 8 W.

 

Since both the bulbs are connected in parallel, the voltage across each of them will be the same.

Current drawn by the bulb of rating 100 W can be calculated as follows:

P = V × I

I = P/V

Substituting the values in the equation, we get

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

Similarly, the current drawn by the bulb of rating 60 W can be calculated as follows:

I = 60 W/220 V = 60/220 A

Therefore, the current drawn from the line is

 

The energy consumed by electrical appliances is given by the equation

H = Pt, where P is the power of the appliance and t is the time

Using this formula, the energy consumed by a TV of power ration 250 W, can be calculated as follows:

H = 250 W × 3600 seconds = 9 × 105 J

Similarly, the energy consumed by a toaster of power rating 1200 W is

H = 1200 W × 600 s = 7.2 × 105 J

From the calculations, it can be said that the energy consumed by the TV is greater than the toaster.

 

The rate at which the heat develops in the heater can be calculated using the following formula

P = I2 R

Substituting the values in the equation, we get

P = (15A) 2 × 8 Ω = 1800 watt

The electric heater produces heat at the rate of 1800 watt

 

a. The resistivity and melting point of tungsten is very high. Due to this property, it doesn’t burn readily when heated. Electric lamps operate at high temperature. Hence, tungsten is a choice of metal for the filament of electric lamps.

b. The conductors of electric heating devices are alloys because of their high resistivity. Alloys have higher resistivity than pure metals. Due to its high resistivity, a large amount of heat is produced when current passes through it.

c. The series arrangement is not used for domestic circuits due to the following reasons:

• The overall voltage gets distributed in a series circuit. As a result, electric appliances may not get the rated power for their operation.

• All the connected appliances cannot be operated independently. If one device is defective, then the entire circuit will not function.

• The total resistance becomes large, and as a result, the current is reduced.

d. Resistance is inversely proportional to the area of cross section. When the area of cross section increases the resistance decreases and vice versa.

e. Copper and aluminium are good conductors of electricity and have low resistivity, because of which they are usually employed for electricity transmission. Due to low resistivity, the power losses in the form of heat are also significantly less when electricity is transmitted through them.