The capacitor 'C' is initially uncharged. Switch S_1 is closed fo… - Vidyadip

MCQ

The capacitor $'C'$ is initially uncharged. Switch $S_1$ is closed for a long time while $S_2$ remains open. Now at $t = 0$ , $S_2$ is closed while $S_ 1$ is opened. All the batteries are ideal and connecting wires are resistanceless. Find $INCORRECT$ statement

A
At time $t = 0$ (just after $S_2$ is closed), reading of ammeter is $\frac {E}{5R}$
✓
At time $t = 0$ (just after $S_2$ is closed), reading of ammeter is zero
C
Heat developed till time $t = 5\ RC\ ln\ 2$ in resistance $3R$ is $\frac {9}{40}CE^2$
D
After time $t > 0$ charge on the capacitor follows the equation $CE\ e^{-t/5RC}$

✓

Answer

Correct option: B.

At time $t = 0$ (just after $S_2$ is closed), reading of ammeter is zero

b
$\dot{\mathrm{i}}=\frac{\mathrm{E}}{5 \mathrm{R}}$

equation of charge of capacitor

$q = CE{e^{ - t/\tau }}$

$q = EC{e^{ - t/5RC}}$

at $\quad t = 5\,{\rm{RC}}\,\ell {\rm{n}}\,2$

$\mathrm{q}=\frac{\mathrm{EC}}{2}$

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