Charging of capacitor and inductor

The charging of capacitors and inductors plays a crucial role in understanding the behaviour of electrical circuits, particularly in energy storage and conversion. A capacitor stores energy in an electric field, while an inductor stores energy in a magnetic field. These components are essential in devices like cameras, where a capacitor helps store energy for the flash, or in power supplies, where inductors smooth out voltage variations. In real life, charging a capacitor is similar to filling a water tank—initially fast, but slowing down as it nears full capacity. Similarly, charging an inductor involves building up a magnetic field gradually, much like winding up a spring. These principles are fundamental to various applications, from smartphones to electric vehicles, showcasing their significance in modern technology.

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This Story also Contains

1. Charging of a Capacitor
2. Discharging of Capacitors
3. Charging and Discharging of an Inductor
4. Solved Examples Based on Charging of capacitor and inductor
5. Summary

Charging of a Capacitor

When a capacitor with zero charges is connected to a battery of emf V through connecting wires, the total resistance including internal resistance of the battery and of the connecting wires is R then after a time t let the charge on the capacitor be q, current be i and $V_c=\frac{q}{C}$,

charge deposited on the positive plate in time $dt$ is so that

$\begin{array}{rl}& dq=idt\\ & i=\frac{dq}{dt}\end{array}$
Using Kirchhoffs loop law, $\frac{q}{C}+Ri-V=0$
or,

$Ri=E-\frac{q}{C}$

or, $R\frac{dq}{dt}=\frac{VC-q}{C}$
or, ${\int }_0^q\frac{dq}{VC-q}={\int }_0^t\frac{1}{CR}dt$
or, $-\ln \frac{VC-q}{VC}=\frac{t}{CR}$
or, $1-\frac{q}{VC}=e^{-t/CR}$
or, $q=VC(1-e^{-t/CR})$

Where RC is the time constant $(\tau )$ of the circuit.
At $t=\tau =RC$
$q=CV(1-\frac{1}{e})=0.63CV$

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,

$\frac{q}{C}-Ri=0$
Here $i=-\frac{dq}{dt}$ because the charge $q$ decreases as time passes.
Thus, $R\frac{dq}{dt}=-\frac{q}{C}$
or, $\frac{dq}{q}=-\frac{1}{CR}dt$
or, ${\int }_Q^q\frac{dq}{q}={\int }_0^t-\frac{1}{CR}dt$
or, $\ln \frac{q}{Q}=-\frac{t}{CR}$
or, $q=Q{e}^{-t/CR}$

• Note: 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 $(i_0=\frac{V}{R})$ at steady state. If the inductor current is increased from zero at time $=0$ to i at times t then-current i is given by $i=i_0(1-e^{\frac{-t}{T}})$ where $\tau =\frac{L}{R}$

$\tau$ is the time constant of the circuit. Here the current is exponentially increasing.
While the 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.

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Solved Examples Based on Charging of capacitor and inductor

Example 1: Two identical capacitors A and B, charged to the same potential 5 V are connected in two different circuits as shown below at time $\mathrm{t}=0$. If the charge on capacitors A and B at time $\mathrm{t}=\mathrm{CR}$ is $Q_A$ and $Q_B$ respectively, then (Here e is the base of natural logarithm)

1) $Q_A=VC,Q_B=CV$
2) $Q_A=\frac{CV}{2},Q_B=\frac{VC}{e}$
3) $Q_A=\frac{VC}{e},Q_B=\frac{CV}{2}$
4) $Q_A=VC,Q_B=\frac{VC}{e}$

Solution:

Maximum charge on the capacitor $=5\mathrm{CV}$

(a) is reverse-biased and (b) is forward-biased

(a) (b)

So, $q=q_{max}\left[e^{-t/RC}\right]$

$Q_B=\frac{CV}{e}$

Hence the correct option is (4).

Example 2:

As shown in the figure, a battery of emf $ϵ$ is connected to an inductor $L$ and resistance $R$ in series. The switch is closed at $t=0$. The total charge that flows from the battery, between $t=0$ and $t=t_c$ ( ${\mathrm{t}}_{\mathrm{c}}$ is the time constant of the circuit ) is :

1) $\frac{ϵL}{R^2}(1-\frac{1}{e})$
2) $\frac{ϵL}{R^2}$
3) $\frac{ϵR}{e{L}^2}$
4) $\frac{ϵL}{e{R}^2}$

Solution:

As current at any time $t$ is given as

$\begin{array}{rl}& I=I_0(1-e^{\frac{-t}{T}})=\frac{ϵ}{R}(1-e^{\frac{-t}{T}})\\ & \text{ So }\mathrm{q}={\int }_0^{{\mathrm{T}}_{\mathrm{C}}}\text{ idt }\end{array}$
So integrating this will give a total charge

$\begin{array}{rl}& q=\frac{\epsilon }{\mathrm{R}}{[\mathrm{t}-\frac{{\mathrm{e}}^{-t/\tau }}{\frac{-1}{\tau }}]}_0^{\tau };=\frac{\epsilon }{\mathrm{R}}[\tau +\tau {\mathrm{e}}^{-1}-\tau ]\\ & q=\frac{\epsilon }{\mathrm{R}}\times \frac{1}{\mathrm{e}}\times \frac{\mathrm{L}}{\mathrm{R}};q=\frac{\epsilon \mathrm{L}}{{\mathrm{R}}^2\mathrm{e}}\end{array}$
Hence the correct option is (4).

Example 3 : The current (i) at time $t=0$ and $t=\mathrm{\infty }$ respectively for the given circuit is:

1) $\frac{5E}{18},\frac{10E}{33}$
2) $\frac{10E}{33},\frac{5E}{18}$
3) $\frac{5E}{18},\frac{18E}{33}$
4) $\frac{18E}{55},\frac{5E}{18}$

Solution:

At t = 0, the inductor is open

So the corresponding equivalent circuit is given below

$\begin{array}{rl}& {\mathrm{R}}_{\mathrm{eq}}=\frac{6\times 9}{6+9}=\frac{54}{15}\\ & I(\text{ at }t=0)=\frac{15E}{54}=\frac{5E}{18}\end{array}$
At $t={\mathrm{\infty }}_{\text{, }}$ For steady state inductor is replaced by plane wire

So the corresponding equivalent circuit is given below

We can reduce the above circuit to the below circuit.

$\begin{array}{rl}& {\mathrm{R}}_{\mathrm{eq}}=\left(\frac{1\times 4}{1+4}\right)+\left(\frac{5\times 5}{5+5}\right)=\frac{4}{5}+\frac{5}{2}=\frac{8+25}{10}=\frac{33}{10}\\ & \mathrm{I}=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{10\mathrm{E}}{33}\end{array}$

Hence the correct option is (1).

Example 4: Figure shows a circuit that contains four identical resistors with resistance $R=2.0\mathrm{\Omega }$, two identical inductors with inductance $L=2.0\mathrm{mH}$ and an ideal battery with emf $E=9\mathrm{V}$. The current 'i' just after the switch 's' is closed will be :

1) 9 A

2) 2.25 A

3) 3.0 A

4) 3.37 A

Solution:

Just when switch S is closed, the inductor will behave like an infinite resistance. Hence, the circuit will be like

Given: $\mathrm{V}=9\mathrm{V}$
From $\mathrm{V}=\mathrm{IR}$
$\mathrm{I}=\mathrm{V}/\mathrm{R}$
Req. $=2+2=4\mathrm{\Omega }$
$\mathrm{i}=9/4=2.25\mathrm{A}$

Hence the correct option is (2).

Example 5: In the following figure, the charge (in $\mu C$ ) on each condenser in the steady-state will be

1) 12

2) 6

3) 9

4) 3

Solution:

Charging Of Capacitors
$Q=Q_0(1-e^{\frac{-t}{R_c}})$

wherein
In the R-C circuit (Transient State).

$i=\frac{10}{(4+1)}=2\mathrm{amp}$

.The potential difference across 4 W resistance is $V=2\ast 4=8$ volt
In steady-state current flows through 4-ohm resistance only and it is Hence, the potential difference across each capacitor is 4 V

So charge on each capacitor $Q=3\ast 4=12\mu C$

Summary

The charging and discharging of capacitors and inductors follow exponential processes that are essential in various circuits. A capacitor charges by storing energy in an electric field and discharges through a gradual release of this energy, while an inductor stores energy in a magnetic field and behaves similarly in its charging and discharging process. The time constant, denoted as RC for capacitors and L/R for inductors, determines the rate of these processes. These principles are critical in many real-world applications, from power supplies to electronic devices.