# Electricity

## NCERT In Text Solution

### Part 2

Question 13: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

1. Voltage remains same in all the appliances.
2. Lower total effective resistance

Question 14: Why does the cord of an electric heater not glow while the heating element does?

Answer: The resistance of heating element is very high compared to the cord of an electric heater. Since, heat produced is directly proportional to the resistance, thus element of electric heater glows because of production of more heat and cord does not.

Question 15: Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 106 Ω, (b) 1 Ω and 103 Ω, and 106 Ω.

Answer: (a) 1 Ω and 106 Ω

We know; 1/R = (1)/(R_1) + (1)/(R_2)

Or, 1/R = 1/1 + (1)/(10^6)

Or, 1/R = (10^6 + 1)/(10^6)

Or, R = (10^6)/(10^6) [approx]

Or, R = 1 Ω

Answer: (b) 1 Ω and 103 Ω, and 106 Ω

We know, 1/R = (1)/(R_1) + (1)/(R_2) + (1)/(R_3)

Or, 1/R = 1/1 + (1)/(10^3) + (1)/(10^6)

Or, 1/R = (10^6 + 10^3 + 1)/(10^6)

Or, R = (10^6)/(10^6) [approx]

Or, R = 1 Ω

Question 16: 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.

Answer: Given: R1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel

Potential difference = 220 V

Sum of resistance can be calculated as follows:

1/R=(1)/(100Ω)+(1)/(50Ω)+(1)/(500Ω)

=(5+10+1)/(500)Ω=(16)/(500)Ω

Therefore, R=(500)/(16)Ω=31.25Ω

Electric current (I)=V/R

Or, I=(220V)/(31.25Ω)=7.04A

Since the electric iron takes same amount of current as three appliances, thus electric current through it = 7.04 A

The electric current = 7.04 A and potential difference = 220 V

Thus, Resistance of electric iron = Total resistance of three appliances = 31.25 Ω

Thus, electric current through the electric iron = 7.04 A

Resistance of electric iron = 31.25 Ω

Question 17: 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?

Answer: (a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series

Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R1

(1)/(R_1)=(1)/(6Ω)+(1)/(3Ω)

=(1+2)/(6)Ω=3/6Ω=1/2Ω

Thus, R_1=2Ω

Now, total effective resistance in the circuit = R1 + 2 Ω = 2 Ω + 2 Ω = 4 Ω

Hence, when resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series, then the total effective resistance in the circuit = 4 Ω

(b) When all the three resistance is connected in parallel then

1/R=(1)/(2Ω)+(1)/(3Ω)+(1)/(6Ω)

=(3+2+1)/(6)Ω

=6/6Ω=1Ω

Thus, R=1Ω

When all the three resistance will be connected in parallel, then the total effective resistance in the circuit = 1 Ω

Question 18: 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?

Answer: When all the resistors are connected in parallel

1/R=(1)/(4Ω)+(1)/(8Ω)+(1)/(12Ω)+(1)/(24Ω)

=(6+3+2+1)/(24)Ω

=(12)/(24)Ω=1/2Ω

Thus, R=2Ω

Thus, R = 2 Ω

(b) When all the resistors are connected in series

Thus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω

Thus, highest resistance = 48 Ω

Lowest resistance = 2 Ω

Question 19: Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer: Given, Potential difference (V) = 50V,

Charge (Q) = 96000 coulomb

Time, t = 1 hour = 60 xx 60 s = 3600 s

Heat produced =?

We know that electric current (I) = Q/t

= 96000 ÷ 3600 s

H=50Vxx(96000)/(3600)Axx3600

=50xx96000 J

=4800000 J=4.8xx10^6 text(Joule)

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

Answer: Given, Electric current (I) = 5A, Resistance (R) = 20 Ω, Time (t) = 30s

Therefore, Heat produced (H) =?

Since, V = I xx R = 5 A xx 20 Ω = 100 V

We know, H = V I t

Or, H = 100 V xx 5 A xx 30 s

= 15000 J = 1.5 xx 10^4  J

Question 21: What determines the rate at which energy is delivered by a current?

Answer: Since electric power is the rate of consumption of electric energy in any electrical appliance. Hence, rate at which energy is delivered by a current is the power of electric appliance.

Question 22: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer: Given, Electric current (I) = 5A,

Potential difference (V) = 220V,

Time (t) = 2h = 2 xx 60 xx 60 s = 7200 s

Power (P) =?

Energy consumed =?

We know, P = V I = 220 V xx 5 A = 1100 W

Energy consumed = P xx t

= 1100 W xx 7200 s

= 7920000 J = 7.92 xx 10^6  J