Question

If the condenser shown in the circuit is charged to $5 \,V$ and left in the circuit, in $12 \,s$ the charge on the condenser will become:

• A. $ \frac{10}{e}$ coulomb
• B. $ \frac{e^{2}}{10}$ coulomb
• C. $ \frac{10}{e^{2}}$ coulomb
• D. $ \frac{e}{10}$ coulomb

Answer 1

Answer: (A)

Exponential decay of charge takes place.
In a C-R circuit, discharging takes place.
The $q-t$ equation is $q=q_{0} e^{-t / R C}$
where $R$ is resistance, $C$ is capacitance
and $q_{0}=C V_{0}$
Hence, $q=C V_{0} e^{-t / R C}$
Given, $V_{0}=5\, V, C=2 \,F, t=12 \,s, R=6 \,\Omega$
$ \therefore q=2 \times 5 e^{-12 / 6 \times 2}=\frac{10}{e}$ coulomb