First slide

Electrical power

Question

A charged capacitor of capacitance C and having charge Q is to be connected with another uncharged capacitor of capacitance $C_{1}$ through a resistor R as shown in figure. The value of $C_{1}$ for heat liberated in R to be minimum is :

Easy

A

$\frac{C}{2}$
B

C
C

infinite
D

zero

Solution

Common potential = $\frac{Q}{C + C_{1}}$
Liberated heat
$H = \frac{Q^{2}}{2 C} - \frac{1}{2} (C + C_{1}) {(\frac{Q}{C + C_{1}})}^{2} = \frac{Q^{2}}{2 C} - \frac{Q^{2}}{2 (C + C_{1})} = \frac{Q^{2} C_{1}}{2 C (C + C_{1})}$
which is minimum at $C_{1}$ = 0 (It means open circuit).

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