Two resistors of 200 , and 400 , are connected in series with a battery of 100 V. A bulb rated at 200 V, 100 W is connected across the 400 , resistance. The potential drop across the bulb is _______ V.

Step-by-step solution

The resistance of the bulb is calculated using its power rating: R_b = V^2 P = 200^2 100 = 400 , The bulb is connected in parallel with the 400 , resistor. The equivalent resistance of this parallel combination is: R_p = 400 400 400 + 400 = 200 , This parallel combination is in series with the 200 , resistor. The total equivalent resistance of the circuit is: R_ eq = 200 + R_p = 200 + 200 = 400 , The current drawn from the battery is: I = V R_ eq = 100 400 = 0.25 A The potential drop across the bulb is the same as