Electromagnetic Induction And Alternating Current Ques 87

Answer:

Correct Answer: 87.(b)

Solution:

87. When switch is closed for a very long time capacitor will get fully charged and charge on capacitor will be $q=C V$

Energy stored in capacitor, $E _C=\frac{1}{2} C V^{2}$

Work done by a battery, $W=V q=V C V=C V^{2}$

Energy dissipated across resistance

$E _D=($ work done by battery $)-($ energy stored $)$

$E _D=C V^{2}-\frac{1}{2} C V^{2}=\frac{1}{2} C V^{2}$

From Eqs. (i) and (ii)

$$ E _D=E _C $$