Question

An capacitance C and a resistance R are connected in series to a battery of voltage V and negligible internal resistance through a switch. The switch is closed at $t=0$. The charge that passes through the battery in one time constant is
[e is base of natural logarithms]

• A. $eVC$
• B. $\frac{VC}{e}$
• C. $\frac{VC}{e^2}$
• D. $CV{e}^2$

Answer 1

Answer: (B)

$\frac{VC}{e}$

$I_0=\frac{V}{R}$

and $I=I_0(1-e^{-\frac{t}{RC}})$

$Q={\int }_0^{T}Idt$

$=I_0{\int }_0^T(1-e^{\frac{-t}{RC}})dt$

$=I_0\left[\right(T-0)+RC(e^{-t\frac{1}{RC}}\left){\int }_0^T\right]$

$=I_0\left(RC\right(1+(e^{-1}-1))$

$=\frac{I_{0}RC}{e}$

$Q=\frac{VC}{e}$