Question

Solve the following example.
a. Heat energy is being produced in a resistance in a circuit at the rate of 100 W. The current of 3 A is flowing in the circuit. What must be the value of the resistance?
b. Two tungsten bulbs of wattage 100 W and 60 W power work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?
c. Who will spend more electrical energy? 500 W TV Set in 30 mins, or 600 W heater in 20 mins?
d. An electric iron of 1100 W is operated for 2 hrs daily. What will be the electrical consumption expenses for that in the month of April? (The electric company charges Rs 5 per unit of energy).

Answer 1

a. Given:
Power, P =100 W
Current, I = 3 A
Resistance, R = ?
We know, P = I²R
$\Rightarrow R=\frac{P}{I^2}=\frac{100}{9}\simeq 11\mathrm{\Omega }$

b. Power of first bulb, P₁ = 100 W
Power of second bulb, P₂ = 60 W
Now,
Resistance of first bulb, $R_1=\frac{V^2}{P_1}=\frac{220\times 220}{100}=484\mathrm{\Omega }$
Resistance of second bulb, $R_2=\frac{V^2}{P_2}=\frac{220\times 220}{60}=806.7\mathrm{\Omega }$
When the bulbs are connected in parallel, their equivalent resistance is
$R_{\mathrm{eq}}=\frac{R_1\times R_2}{R_1+R_2}=\frac{484\times 806.7}{484+806.7}=302.5\mathrm{\Omega }$
Current flowing in the main conductor is
$I=\frac{V}{R_{\mathrm{eq}}}=\frac{220}{302.5}=0.72\mathrm{A}$

c. We know,
$\mathrm{Electrical}\mathrm{energy}\left(E\right)=\mathrm{Power}\left(P\right)\times \mathrm{Time}\left(t\right)$
For TV set,
$E=500\times 30\times 60=900000\mathrm{J}$
For heater,
$E=600\times 20\times 60=720000\mathrm{J}$
Thus, TV set consumes more electrical energy.

d. Electric power required for working of iron, P = 1100 W
Duration for which the iron is operated daily = 2 h = $2\times 60\times 60=7200\mathrm{s}$
Electric energy consumed by iron in 7200 s is
$E=1100\times 7200=7920000\mathrm{J}$
Thus, total energy consumed in the month of April, $E\text{'}=E\times 30=7920000\times 30=237600000\mathrm{J}$
We know,
$$\begin{gathered} 1\mathrm{unit}=3.6\times {10}^6\mathrm{J} \\ \mathrm{or},1\mathrm{J}=\frac{1}{3.6\times {10}^6}\mathrm{unit} \\ \mathrm{Thus}, \\ 237600000\mathrm{J}=\frac{237600000}{3.6\times 1000000}=66\mathrm{units} \end{gathered}$$
Cost of 1 unit of energy = Rs 5
Thus, total electrical consumption expenses for the month of April = 66$\times$5 = Rs 330