Question

A charged capacitor of capacitance C and having charge Q is to be connected with another uncharged capasitor of capasitance C' as shown till the steady state is reached , find the value of C' for heat liberated through the wires to be minimum.

• A. zero
• B. C
• C. C /2
• D. 2C

Answer 1

The correct option is A zero

$u_i=\frac{Q^2}{2c}$---------------$\left(1\right)$

$c\left(x-0\right)+\left(x-0\right)c^{\prime }=Q$

$\Rightarrow x\left(c+c^{\prime }\right)=Q$

$\Rightarrow x=\frac{Qc}{c+c^{\prime }}$

charge on $c$ $=xc$

$=\frac{Qc}{c+c^{\prime }}$

$\&$ charge on $c^{\prime }=\frac{Q{c}^{\prime }}{c+c^{\prime }}$

$\therefore u_f=\frac{Q^2{c}^2}{{\left(c+c^{\prime }\right)}^2\cdot 2c}+\frac{Q^{2}c{}^{\prime 2}}{{\left(c+c^{\prime }\right)}^2\cdot 2c^{\prime }}$

$=\frac{Q^2}{2{\left(c+c^{\prime }\right)}^2}\left(c+c^{\prime }\right)=\frac{Q^2}{2\left(c+c^{\prime }\right)}$

$heat=u_i-u_y=\frac{Q^2{c}^{\prime }}{2\left(c+c^{\prime }\right)}$

$\therefore$ for heat to be min $c^{\prime }\to 0$

Hence,

option $\left(A\right)$ is correct answer.