Question

Initially the capacitor is uncharged , find the charge on capacitor as a function of time, if switch is closed at $t=0$.

• A. $q=\frac{\epsilon C}{2}(1-e^{\frac{-2t}{3RC}})$
• B. $q=\frac{\epsilon C}{2}(1-e^{\frac{-t}{RC}})$
• C. $q=\frac{\epsilon C}{2}(1-e^{\frac{-t}{2RC}})$
• D. $q=\frac{3\epsilon C}{2}[1-e^{\left[\frac{-2t}{RC}\right]}]$

Answer 1

The correct option is A $q=\frac{\epsilon C}{2}(1-e^{\frac{-2t}{3RC}})$

Applying KVL in loop ABCDA
$\epsilon -iR-(i-i_1)R=0$
$\epsilon -2iR+i_{1}R=0.......\left(1\right)$


Applying KVL in loop ABCEFDA
$E-iR-i_{1}R-\frac{q}{C}=0.....\left(2\right)$

From equation (1) and (2):
$EC-3i_{1}RC=2q$

$EC-2q=3\left(\frac{dq}{dt}\right)RC$

${\int }_0^q\frac{dq}{EC-2q}={\int }_0^t\frac{dt}{3RC}\Rightarrow -\frac{1}{2}\ln \frac{\epsilon C-2q}{\epsilon C}=\frac{t}{3RC}$

$q=\frac{\epsilon C}{2}(1-e^{\frac{-2t}{3RC}})$

Hence, option (a) is the correct answer.