For the given circuit, in the steady state, $\left| V _{ B }- V _{ D }\right|=.......V.$
JEE MAIN 2023, Diffcult
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In steady state, capacitor behaves as an open circuit. Circuit is
$i_{A B}=\frac{6}{3}=2 A$
$i_{A D}=\frac{6}{12}=0.5 A$
$V_B+2 \times 2-10 \times 0.5=V_D$
$V_B-V_D=1\;volt$
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