Current Electricity

JEE Main level, single correct, medium

204 of 380

Question 204 of 380

Under steady state condition the potential difference across the capacitor in the circuit is _______ V.

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Answer: (A) 0.50.5

In steady state, the capacitor acts as an open circuit, meaning no current flows through the middle branch.

The circuit simplifies to a single loop containing the 2 V2 \text{ V} battery, the 6Ω6 \Omega resistor, and the 2Ω2 \Omega resistor.

The total resistance of this loop is Req=2+6=8ΩR_{\text{eq}} = 2 + 6 = 8 \Omega.

The current flowing through the loop is I=28=0.25 AI = \dfrac{2}{8} = 0.25 \text{ A}.

The potential difference across the parallel branches is equal to the voltage drop across the 2Ω2 \Omega resistor:
V=I×2=0.25×2=0.5 VV = I \times 2 = 0.25 \times 2 = 0.5 \text{ V}

Since there is no current in the middle branch, the voltage drop across the 4Ω4 \Omega resistor is zero. Thus, the entire potential difference of 0.5 V0.5 \text{ V} appears across the capacitor.

Answer: 0.50.5

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