Question

An uncharged capacitor of capacitance $4\mu F$, a battery of emf 12 volt and a resistor of 2.5 $M\Omega$ are connected
in series. The time after which $v_c$?

$3v_R$ =0.693

• A. 6.93 sec
• B. 13.86 sec
• C. 20.52 sec
• D. none of these

Answer 1

The correct option is B 13.86 sec

$\begin{array}{c}C=4\mu F\\ V=12V\\ R=2.5\times {10}^6\Omega \\ \Rightarrow q=qss\left(1-e^{\frac{-t}{Rc}}\right)\\ \Rightarrow q=4\times {10}^{-6}\times 12\left(1-e^{\frac{-t}{Rc}}\right)\\ \because V_c=3V_R\\ \Rightarrow \frac{q}{c}=3iR\\ \Rightarrow \frac{4\times {10}^{-6}\times 12}{4\times {10}^{-6}}\left(1-e^{\frac{-t}{Rc}}\right)=3.\frac{qss}{Rc}\left(1-e^{\frac{-t}{Rc}}\right).R\\ \Rightarrow 12\left(1-e^{\frac{-t}{10}}\right)=3.\frac{4\times {10}^{-6}\times 12}{10}\left(1-e^{\frac{-t}{10}}\right).R\\ \Rightarrow 12\left(1-e^{\frac{-t}{10}}\right)=3\times 12\Rightarrow 12\left(1-e^{\frac{-t}{10}}\right)\\ \Rightarrow 12-12e^{\frac{-t}{10}}=36e^{{}^{\frac{-t}{10}}}\\ \Rightarrow 12=48e^{{}^{\frac{-t}{10}}}\\ \Rightarrow e^{{}^{\frac{-t}{10}}}=\frac{1}{4}\\ \Rightarrow e^{{}^{\frac{-t}{10}}}=4\\ \Rightarrow t=20\left(n_2\right)\\ =13.86\sec \end{array}$

$\therefore$ Option $B$ is correct.