The circuit shown has been connected for a long time. The voltage across the capacitor is ............. $V$
KVPY 2013, Medium
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(d)
After a long time capacitor is charged and it acts like an open circuit. So, equivalent circuit is
So, current in circuit is
$I=\frac{6}{(1+2) \times 10^3}=2 \times 10^{-3} \,A$
Potential drop across $2 \,ks \Omega$ resistor is
$V_{A B}=I \cdot R_{A B}$
$=2 \times 10^3 \times 2 \times 10^3=4 \,V$
Clearly, potential drop across capacitor is also $4 \,V$.
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