Current Electricity

JEE Main level, single correct, medium

95 of 380

Question 95 of 380

Two resistors of 200 Ω200 \, \Omega and 400 Ω400 \, \Omega are connected in series with a battery of 100100 V. A bulb rated at 200200 V, 100100 W is connected across the 400 Ω400 \, \Omega resistance. The potential drop across the bulb is _______ V.

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Answer: (B) 5050

The resistance of the bulb is calculated using its power rating:

Rb=V2P=2002100=400 ΩR_b = \dfrac{V^2}{P} = \dfrac{200^2}{100} = 400 \, \Omega

The bulb is connected in parallel with the 400 Ω400 \, \Omega resistor. The equivalent resistance of this parallel combination is:

Rp=400×400400+400=200 ΩR_p = \dfrac{400 \times 400}{400 + 400} = 200 \, \Omega

This parallel combination is in series with the 200 Ω200 \, \Omega resistor. The total equivalent resistance of the circuit is:

Req=200+Rp=200+200=400 ΩR_{eq} = 200 + R_p = 200 + 200 = 400 \, \Omega

The current drawn from the battery is:

I=VReq=100400=0.25 AI = \dfrac{V}{R_{eq}} = \dfrac{100}{400} = 0.25 \text{ A}

The potential drop across the bulb is the same as the potential drop across the parallel combination:

Vp=I×Rp=0.25×200=50 VV_p = I \times R_p = 0.25 \times 200 = 50 \text{ V}

Answer: 5050

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