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QJEEM23Phfb3f: JEE Main 2023 Physics - Class 12 JEEM23Phfb3f Solution: For the given circuit, in the steady state, ________ V.

For the given circuit, in the steady state, ∣VB−VD∣=\left|\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{D}}\right|= ________ V.

JEE Main 2023 (Online) 31st January Evening Shift Physics - Capacitor Question 42 English

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In steady state, capacitor behaves as an open circuit. Circuit is :

JEE Main 2023 (Online) 31st January Evening Shift Physics - Capacitor Question 42 English Explanation

⇒iAB=63=2 A\Rightarrow i_{A B}=\frac{6}{3}=2 \mathrm{~A}

iAD=612=0.5 Ai_{A D}=\frac{6}{12}=0.5 \mathrm{~A}

⇒VB+2×2−10×0.5=VD\Rightarrow V_{B}+2 \times 2-10 \times 0.5=V_{D}

⇒VB−VD=1\Rightarrow V_{B}-V_{D}=1 volt

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