Question

It is given that when switch S is open, charge on one of the capacitors is $q^{\prime }=\frac{c\epsilon }{2}$ and another one is zero. At t = 0 switch S is closed, then

• A. Current i through R is $i=\frac{\epsilon }{R}{e}^{-\frac{2t}{RC}}$
• B. Charge on capacitor $C_2$ is $\frac{\epsilon C}{2}(2-e^{-\frac{2t}{RC}})$
• C. Heat generated across R in long time is $\frac{{\epsilon }^{2}C}{16}$
• D. Charge on capacitor $C_1$ is $\frac{\epsilon C}{4}(1-e^{-\frac{2t}{RC}})$

Answer 1

Answer: (C), (D)

The correct options are
C Heat generated across R in long time is $\frac{{\epsilon }^{2}C}{16}$
D Charge on capacitor $C_1$ is $\frac{\epsilon C}{4}(1-e^{-\frac{2t}{RC}})$


By $K.V.L:$
$\epsilon =iR+\frac{q}{C}+\frac{q+q^{\prime }}{C}$
As $\frac{q^{\prime }}{c}=\frac{\epsilon }{2}$
$\frac{\epsilon }{2}-\frac{2q}{c}=iR$

As $i=\frac{dq}{dt}$
$\frac{\epsilon C-4q}{2CR}=\frac{dq}{dt}\Rightarrow {\int }_0^q\frac{dq}{\epsilon C-4q}={\int }_0^t\frac{dt}{2RC}$
$q=\frac{\epsilon C}{4}(1-e^{-\frac{2t}{RC}})$
$\therefore i=\frac{dq}{dt}=\frac{\epsilon }{2R}{e}^{-\frac{2t}{RC}}$
So, heat generated after a long duration, $H={\int }_0^{\infty }{i}^{2}Rdt=\frac{{\epsilon }^{2}C}{16}$

$q^{\prime }+q=\frac{\epsilon C}{4}(3-e^{-\frac{2t}{RC}})$