Question

Four identical bulbs each rated 100 watt, 220 volts are connected across a battery as shown. The total electric power consumed by the bulb (in Watts) is ____.

Answer 1

$\frac{7500}{121}$

Answer 2

1. Calculate the resistance of each bulb: The power rating of a bulb is given by $P = \frac{V^2}{R}$. From this, the resistance $R$ can be calculated as $R = \frac{V_{rated}^2}{P_{rated}}$. Given $P_{rated} = 100$ W and $V_{rated} = 220$ V, the resistance of each bulb is: $R = \frac{(220 \text{ V})^2}{100 \text{ W}} = \frac{48400}{100} \Omega = 484 \Omega$.

2. Determine the equivalent resistance of the circuit: The circuit diagram shows one bulb connected in series with a parallel combination of three identical bulbs. The equivalent resistance of the three identical bulbs connected in parallel is: $R_{parallel} = \frac{R}{3} = \frac{484 \Omega}{3}$. The total equivalent resistance of the circuit is the sum of the resistance of the first bulb and the equivalent resistance of the parallel combination: $R_{total} = R + R_{parallel} = 484 \Omega + \frac{484 \Omega}{3} = 484 \left(1 + \frac{1}{3}\right) = 484 \times \frac{4}{3} = \frac{1936}{3} \Omega$.

3. Calculate the total current drawn from the battery: The battery voltage is $V_{battery} = 200$ V. Using Ohm's law ($I = \frac{V}{R}$), the total current flowing from the battery is: $I = \frac{V_{battery}}{R_{total}} = \frac{200 \text{ V}}{\frac{1936}{3} \Omega} = \frac{200 \times 3}{1936} \text{ A} = \frac{600}{1936} \text{ A}$. Simplifying the fraction: $I = \frac{600}{1936} = \frac{75}{242}$ A.

4. Calculate the total power consumed by the bulbs: The total electric power consumed by the circuit is given by $P_{total} = V_{battery} \times I$. $P_{total} = 200 \text{ V} \times \frac{75}{242} \text{ A} = \frac{15000}{242} \text{ W}$. Simplifying the fraction: $P_{total} = \frac{7500}{121}$ W.