Two resistors of 200 Ω and 400 Ω are connected in series across a 100 V battery. A bulb rated 200 V, 100 W is connected in parallel with the 400 Ω resistor. Assuming the bulb resistance remains equal to its rated resistance, what is the potential difference across the bulb?
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Question

Two resistors of 200 Ω and 400 Ω are connected in series across a 100 V battery. A bulb rated 200 V, 100 W is connected in parallel with the 400 Ω resistor. Assuming the bulb resistance remains equal to its rated resistance, what is the potential difference across the bulb?

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G1 Physics · Accepted Answer

The resistance of the bulb at its rated operating values is R(bulb) = V² ÷ P = 200² ÷ 100 = 400 Ω. The bulb is connected in parallel with the existing 400 Ω resistor. Therefore their equivalent resistance is R(parallel) = 400×400 ÷ (400+400) = 200 Ω. This 200 Ω equivalent is in series with the other 200 Ω resistor. Hence the total resistance is R(total) = 200+200 = 400 Ω. The battery voltage of 100 V divides equally across the two equal 200 Ω series sections. Therefore the voltage across the pa…