Electricity — NCERT Solutions

Heat produced in time $t$ is
$$H=I^2Rt$$
Given $I=5\,\text{A}$, $R=20\,\Omega$, $t=30\,\text{s}$:
$$H=5^2\times20\times30=25\times20\times30=15000\,\text{J}$$
So the heat developed is 15000 J.

The rate at which energy is delivered by a current is determined by the power. The chapter gives:
$$P=VI$$
and also
$$P=I^2R=\frac{V^2}{R}$$
So power depends on the current and the potential difference across the circuit.

Power of the motor:
$$P=VI=220\times5=1100\,\text{W}$$
Energy consumed in 2 h:
$$E=Pt=1100\,\text{W}\times2\,\text{h}=2200\,\text{Wh}=2.2\,\text{kWh}$$
In joules:
$$2.2\times3.6\times10^6=7.92\times10^6\,\text{J}$$
So the motor's power is 1100 W and the energy consumed in 2 h is 2.2 kWh or $7.92\times10^6$ J.

Original wire has resistance $R$.

It is cut into 5 equal parts, so each part has resistance
$$R/5$$
If these 5 equal resistors are connected in parallel, equivalent resistance is
$$R' = \frac{R/5}{5} = \frac{R}{25}$$
Therefore,
$$\frac{R}{R'} = \frac{R}{R/25}=25$$
So the correct option is (d) 25.

Electrical power is given by
$$P=VI=I^2R=\frac{V^2}{R}$$
So $IR^2$ does not represent power.

For the same bulb, resistance is taken constant. Since $P=\frac{V^2}{R}$, halving the voltage makes power one-fourth:
$$P' = P\left(\frac{110}{220}\right)^2 = 100\times\frac{1}{4}=25\,\text{W}$$
So the correct option is 25 W.

Let each wire have resistance $R$.

• In series, total resistance = $2R$.
• In parallel, total resistance = $R/2$.

For the same potential difference, heat produced is
$$H=\frac{V^2t}{R}$$
So heat is inversely proportional to resistance.

Therefore,
$$H_{series}:H_{parallel}=\frac{1}{2R}:\frac{1}{R/2}=\frac{1}{2R}:\frac{2}{R}=1:4$$
So the ratio is 1:4.

A voltmeter is always connected in parallel across the points between which the potential difference is to be measured.

Given:

• $R=10\,\Omega$
• $\rho=1.6\times10^{-8}\,\Omega\,\text{m}$
• diameter $d=0.5\,\text{mm}=5\times10^{-4}\,\text{m}$

Area:
$$A=\frac{\pi d^2}{4}=\frac{\pi(5\times10^{-4})^2}{4}$$
$$=\frac{\pi\times25\times10^{-8}}{4}=1.9635\times10^{-7}\,\text{m}^2$$
Using $R=\rho l/A$,
$$l=\frac{RA}{\rho}= \frac{10\times1.9635\times10^{-7}}{1.6\times10^{-8}} \approx12.27\,\text{m}$$
So the length is about 12.3 m.

If diameter is doubled, area becomes 4 times. Since $R\propto 1/A$, resistance becomes one-fourth.
For 10 Ω, new resistance = 2.5 Ω.

From the table, the ratio $V/I$ is nearly constant:

• $1.6/0.5 = 3.2\,\Omega$
• $3.4/1.0 = 3.4\,\Omega$
• $6.7/2.0 = 3.35\,\Omega$
• $10.2/3.0 = 3.4\,\Omega$
• $13.2/4.0 = 3.3\,\Omega$

So the resistance is approximately 3.3 Ω. The $V$-$I$ graph is a straight line through the origin, showing Ohm’s law.

Use Ohm’s law:
$$R=\frac{V}{I}$$
Given $V=12\,\text{V}$ and $I=2.5\,\text{mA}=2.5\times10^{-3}\,\text{A}$,
$$R=\frac{12}{2.5\times10^{-3}}=4800\,\Omega$$
So the resistance is 4800 Ω.

All resistors are in series, so total resistance is
$$R=0.2+0.3+0.4+0.5+12=13.4\,\Omega$$
Current in the series circuit:
$$I=\frac{V}{R}=\frac{9}{13.4}\approx0.67\,\text{A}$$
Since current is the same through every resistor in series, current through the 12 Ω resistor is 0.67 A.

The computed value is 0.67 A; if the printed answer key or option set differs, the chapter's laws give this result.

To carry 5 A on a 220 V line, the equivalent resistance needed is
$$R=\frac{V}{I}=\frac{220}{5}=44\,\Omega$$
For $n$ identical $176\,\Omega$ resistors in parallel,
$$R_p=\frac{176}{n}$$
Set this equal to 44 Ω:
$$\frac{176}{n}=44 \Rightarrow n=4$$
So 4 resistors are required.

Let each resistor be 6 Ω.

(i) 9 Ω

Connect two 6 Ω resistors in parallel:
$$\frac{1}{R} = \frac{1}{6}+\frac{1}{6}=\frac{2}{6} \Rightarrow R=3\,\Omega$$
Then connect this 3 Ω combination in series with the third 6 Ω resistor:
$$R_{total}=3+6=9\,\Omega$$

(ii) 4 Ω

Connect two 6 Ω resistors in series:
$$R=6+6=12\,\Omega$$
Then connect this 12 Ω combination in parallel with the third 6 Ω resistor:
$$\frac{1}{R_{total}}=\frac{1}{12}+\frac{1}{6}=\frac{1}{12}+\frac{2}{12}=\frac{3}{12} \Rightarrow R_{total}=4\,\Omega$$

Each lamp is rated 10 W at 220 V. Current drawn by one lamp:
$$I=\frac{P}{V}=\frac{10}{220}\,\text{A}$$
Maximum current allowed = 5 A.
Number of lamps:
$$N=\frac{5}{10/220}=\frac{5\times220}{10}=110$$
So 110 lamps can be connected in parallel.

Given each coil has resistance 24 Ω and supply voltage is 220 V.

Separate use

For one coil:
$$I=\frac{220}{24}=9.17\,\text{A}$$

Series use

Total resistance:
$$R=24+24=48\,\Omega$$
Current:
$$I=\frac{220}{48}=4.58\,\text{A}$$

Parallel use

Equivalent resistance:
$$\frac{1}{R} = \frac{1}{24}+\frac{1}{24}=\frac{2}{24} \Rightarrow R=12\,\Omega$$
Current:
$$I=\frac{220}{12}=18.33\,\text{A}$$
So the currents are 9.17 A, 4.58 A, and 18.33 A respectively.

(i) 6 V battery in series with 1 Ω and 2 Ω

Total resistance:
$$R=1+2=3\,\Omega$$
Current:
$$I=\frac{6}{3}=2\,\text{A}$$
Power in 2 Ω resistor:
$$P=I^2R=2^2\times2=8\,\text{W}$$

(ii) 4 V battery in parallel with 12 Ω and 2 Ω

The 2 Ω resistor gets full 4 V.
$$P=\frac{V^2}{R}=\frac{4^2}{2}=8\,\text{W}$$
So the power used in the 2 Ω resistor is 8 W in both cases.

For bulbs in parallel, current drawn from the line is the sum of individual currents.

For 100 W bulb:
$$I_1=\frac{P}{V}=\frac{100}{220}=0.455\,\text{A}$$
For 60 W bulb:
$$I_2=\frac{60}{220}=0.273\,\text{A}$$
Total current:
$$I=0.455+0.273=0.728\,\text{A}$$
So the current drawn is about 0.73 A.

Compute energy used by each:

• 250 W TV in 1 h

$$E=250\times1=250\,\text{Wh}=0.25\,\text{kWh}$$

• 1200 W toaster in 10 min = 1/6 h

$$E=1200\times\frac{1}{6}=200\,\text{Wh}=0.2\,\text{kWh}$$

Since 0.25 kWh > 0.2 kWh, the 250 W TV set in 1 hour uses more energy.

Rate at which heat is developed = power:
$$P=I^2R=5^2\times44=25\times44=1100\,\text{W}$$
So the heater develops heat at a rate of 1100 W.