Question

An RC circuit consists of a resistance R = 5 MΩ and a capacitance C = 1.0 μF connected in series with a battery. In how much time will the potential diffrence across the capacitor become 8 times that across the resistor? (Given $lo{g}_e\left(3\right)=1.1$ )
(IIT-JEE 2005)

• A. 5.5 s
• B. 11 s
• C. 44 s
• D. 88 s

Answer 1

The correct option is B

11 s

If the circuit is like this,

If $V_c$ is the potential differnce across C, the power differences across R will be $V-V_C.$
We are asked when will be
$8(V-V_c)=V_c$
or, when is
$8V=9V_c$ ...(1)
We know, for a capacitor
$q_c=q_0(1-e^{-t/RC})$
$\frac{q_c}{C}=\frac{q_0}{C}(1-e^{t/RC})$
$V_c=V(1-e^{-t/RC}$
(since $\frac{q_0}{C}=V)$
$\therefore$ substitution in (1) gives,
$8V=9V(1-e^{-t/RC})$
$\frac{8}{9}=1-e^{-t/RC}$
$e^{-t/RC}=\frac{1}{9}$
$-t/RC=-lo{g}_{e}9$
$t/RC=2lo{g}_{e}3$
$t=2\times 1.1\times {10}^{-6}\times 5\times {10}^6$
= 11 s