Electromagnetic Induction And Alternating Current Ques 87
- In process 1 , the energy stored in the capacitor $E _C$ and heat dissipated across resistance $E _D$ are related by
(a) $E _C=E _D \ln 2$
(b) $E _C=E _D$
(c) $E _C=2 E _D$
(d) $E _C=\frac{1}{2} E _D$
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Answer:
Correct Answer: 87.(b)
Solution:
- 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)
$$ E _D=E _C $$
BETA
