Charging Of Capacitor And Inductor MCQ - Practice Questions with Answers

• Charging of a Capacitor:

When a capacitor with zero charges is connected to a battery of emf  V through connecting wires, total resistance including internal resistance of the battery and of the connecting wires be R then after a

time t let the charge on capacitor be q , current be i and $V_c=\frac{q}{C}$, charge deposited on the positive plate in time $d t$ is $d q=i d t$ so that

$
i=\frac{d q}{d t}
$

Using Kirchhoffs loop law, $\quad \frac{q}{C}+R i-V=0$
or,

$
R i=E-\frac{q}{C}
$

or, $\quad R \frac{d q}{d t}=\frac{V C-q}{C}$
or, $\quad \int_0^q \frac{d q}{V C-q}=\int_0^t \frac{1}{C R} d t$
or, $\quad-\ln \frac{V C-q}{V C}=\frac{t}{C R}$
or, $\quad 1-\frac{q}{V C}=e^{-t / C R}$
or, $\quad q=V C\left(1-e^{-t / C R}\right)$

Where RC is the time constant $(\tau)$ of the circuit.

$
\begin{aligned}
& \text { At } t=\tau=R C \\
& q=C V\left(1-\frac{1}{e}\right)=0.63 C V
\end{aligned}
$

- Discharging of capacitors:

If initially a capacitor has a charge $Q$ and is discharged through an external load. Let after a time $t$ the remaining charge in the capacitor be $q$ then Using Kirchhoff's loop law,

$
\frac{q}{C}-R i=0
$

Here $i=-\frac{d q}{d t}$ because the charge $q$ decreases as time passes.

Thus, $\quad R \frac{d q}{d t}=-\frac{q}{C}$
or, $\quad \frac{d q}{q}=-\frac{1}{C R} d t$
or, $\quad \int_Q^q \frac{d q}{q}=\int_0^t-\frac{1}{C R} d t$
or, $\quad \ln \frac{q}{Q}=-\frac{t}{C R}$
or, $\quad q=Q e^{-t / C R}$

• Note: At a steady-state capacitor connected to the DC battery acts as an open circuit. The capacitor does not allow a sudden change in voltage.

Charging and discharging of an inductor:

When an inductor is connected to a DC source of emf V through a resistance R the inductor charges to maximum current $\left(i_0=\frac{V}{R}\right)$ at steady state. If the inductor current is increased from zero at time $=0$ to i at time=t then-current i is given by

$
i=i_0\left(1-e^{\frac{-t}{\tau}}\right) \text { where } \tau=\frac{L}{R}
$

$\tau$ is the time constant of the circuit. Here the current is exponentially increasing.
- While discharging of inductor current decreases exponentially and is given by

$
i=i_0\left(e^{\frac{-t}{\tau}}\right)
$
 

• Note: At steady state, an inductor connected to the DC battery acts as a short circuit. The inductor does not allow a sudden change in current.