Electromagnetic Induction and Alternating Current 7 Question 15

(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: 15. (b)

Solution:

  1. 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)

$$ E _D=E _C $$



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