JEE Mains Physics · Current Electricity

Capacitors and Inductors in DC Circuits

In a DC circuit a capacitor ends up blocking current and an inductor ends up as a plain wire; just after a switch closes it is the other way round, and in between both change exponentially with time constant RC or L/R.

All Current Electricity notes JEE Mains Physics strategy

Why this matters

Twenty-nine PYQs, seventeen of them multiple choice, and six from 2026; twenty-four come with a figure. Fourteen ask for a capacitor's charge, voltage or energy in steady state. Eight ask for the current just after a switch closes or long after, mostly with inductors. Seven use the time constant of an RC or LR circuit, including one that compares a capacitor's discharge with radioactive decay.

Concept 1 of 3: Capacitors in steady state

Once a capacitor is fully charged, no more charge flows onto it, so its branch carries no current. Any resistor in series with it then has no current and no voltage across it. Solve the rest of the circuit as if that branch were missing, and read the capacitor's voltage as the potential difference between the two points it is joined to.

Definition

In steady DC, the capacitor branch carries no current. Remove it to solve the rest of the circuit.
A resistor in series with the capacitor drops no voltage, so it does not affect $V_C$ .
$V_C$ = the potential difference between the capacitor's two nodes. Then $Q = CV_C$ and $U = \tfrac{1}{2}CV_C^{2}$ .
A capacitor joining the midpoints of two dividers: $V_C$ is the difference of the two divider outputs.
Capacitors in series in one branch carry the same charge and share $V_C$ in the ratio of $1/C$ .

Steady state

I_{\text{C branch}} = 0, \qquad Q = CV_C, \qquad U = \tfrac{1}{2}CV_C^{2}

Worked example

A 12 V battery drives a 2 Ω and a 4 Ω resistor in series. A 5 μF capacitor in series with a 10 Ω resistor is connected across the 4 Ω resistor. Find the capacitor's charge and energy in steady state.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q98Moderate

Example 1 · Current Electricity · Capacitors and Inductors in DC Circuits

At steady state the charge on the capacitor, as shown in the circuit below, is

\_\_\_

\mu C

The series resistor drops nothing

With no current in the capacitor branch, a resistor in that branch has no voltage across it. Subtracting an IR drop there gives a wrong capacitor voltage.

Leave the capacitor branch out of the equivalent resistance

In steady state a capacitor branch is an open circuit. Including its resistor in the equivalent resistance changes the current everywhere else.

The capacitor voltage is a difference of potentials

Find the potential at each plate's node from the rest of the circuit and subtract. Do not assume the capacitor has the battery's full emf.

Drill 13 more on capacitors in steady state

Concept 2 of 3: Just after switching and long after

An inductor resists any sudden change in its current, and a capacitor resists any sudden change in its voltage. So just after a switch closes, an inductor that carried no current still carries none, as if its branch were cut, and an uncharged capacitor still has no voltage, as if it were a plain wire. Long after, nothing is changing: the inductor is just a wire and the capacitor blocks current.

Definition

Just after closing ( $t = 0^{+}$ ): an inductor with no current is an open circuit; an uncharged capacitor is a short circuit.
Long after ( $t \to \infty$ ): an ideal inductor is a plain wire (or its own resistance if it has one); a capacitor is an open circuit.
The current in an inductor and the voltage on a capacitor never jump; use their values just before the switch moved.
Just after closing, an ideal inductor in series with R has the whole emf across it; the current then grows.

The two moments

t = 0^{+}: \ L \to \text{open},\ C \to \text{wire}; \qquad t \to \infty: \ L \to \text{wire},\ C \to \text{open}

Worked example

A 10 V battery feeds a 5 Ω resistor in series with two parallel branches: a plain 5 Ω, and an ideal inductor in series with another 5 Ω. Find the battery current just after the switch closes and long after.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q119Moderate

Example 2 · Current Electricity · Capacitors and Inductors in DC Circuits

Three identical resistors with resistance

R = 12\Omega

and two identical inductors with self-inductance

L

= 5mH

are connected to an ideal battery with emf of

12\text{ }V

as shown in figure. The current through the battery long after the switch has been closed will be____ A.

Which element is the wire, and when

At the first instant the capacitor is the wire and the inductor is the gap; long after it is the reverse. Swapping them gives the other moment's answer, which is usually an option.

A real inductor keeps its resistance

Long after switching, an inductor with its own resistance acts as that resistor, not as a plain wire.

Nothing jumps

An inductor's current and a capacitor's voltage just after a switch equal their values just before it. Use that, not the steady-state rule, for the first instant.

Drill 7 more on just after switching and long after

Concept 3 of 3: Charging, discharging and the time constant

Between the first instant and the steady state, the change is exponential. In each time constant the remaining gap shrinks by the same factor e. For a capacitor the time constant is RC; for an inductor it is L/R. Energy goes as the square of charge or current, so it decays twice as fast.

Definition

RC charging: $q = Q_0\left(1 - e^{-t/RC}\right)$ , and the same for $V_C$ . Discharging: $q = q_0e^{-t/RC}$ .
Time to fall to $1/n$ of the start: $t = \tau\ln n$ . Charge halves at $\tau\ln 2$ ; energy halves at $\tfrac{1}{2}\tau\ln 2$ .
The field between the plates is $V/d$ , so it falls exactly like the charge.
LR growth: $i = \dfrac{E}{R}\left(1 - e^{-t/\tau}\right)$ with $\tau = L/R$ ; the inductor's voltage is $E - iR$ . Decay: $i = i_0e^{-t/\tau}$ .
After one time constant a charging capacitor or growing current is at $1 - 1/e \approx 63\%$ of its final value; a decaying one is at $1/e \approx 37\%$ .
A square-wave input across RC gives alternate exponential rises and falls at the capacitor.

Exponential change

q = q_0e^{-t/RC}, \qquad i = \frac{E}{R}\left(1 - e^{-tR/L}\right), \qquad t_{1/n} = \tau\ln n

Worked example

A 2 μF capacitor discharges through a 5 kΩ resistor. Find the time constant, the time for the charge to fall to a quarter, and the time for the energy to fall to a quarter.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q23Moderate

Example 3 · Current Electricity · Capacitors and Inductors in DC Circuits

The electric field between the two parallel plates of a capacitor of

1.5\mu F

capacitance drops to one third of its initial value in

6.6\mu s

when the plates are connected by a thin wire. The resistance of this wire is ................

\Omega

. (Given,

\log3 = 1.1

)

Energy decays twice as fast

Energy goes as q², so it falls as e^(−2t/RC). The energy halves in half the time the charge takes to halve.

Growth uses 1 − e^(−t/τ)

For charging or growing current, set 1 − e^(−t/τ) equal to the fraction asked, not e^(−t/τ). The two give different times except at one half.

ln, not log₁₀

The time is τ ln n with the natural logarithm. A value printed as log 3 = 1.1 is really ln 3: log₁₀ 3 is only about 0.48.

Drill 6 more on charging, discharging and the time constant

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Capacitors in steady state
Steady state
$I_{\text{C branch}} = 0, \qquad Q = CV_C, \qquad U = \tfrac{1}{2}CV_C^{2}$
Just after switching and long after
The two moments
$t = 0^{+}: \ L \to \text{open},\ C \to \text{wire}; \qquad t \to \infty: \ L \to \text{wire},\ C \to \text{open}$
Charging, discharging and the time constant
Exponential change
$q = q_0e^{-t/RC}, \qquad i = \frac{E}{R}\left(1 - e^{-tR/L}\right), \qquad t_{1/n} = \tau\ln n$

Watch out for (9)

The series resistor drops nothing
→ Capacitors in steady state
Leave the capacitor branch out of the equivalent resistance
→ Capacitors in steady state
The capacitor voltage is a difference of potentials
→ Capacitors in steady state
Which element is the wire, and when
→ Just after switching and long after
A real inductor keeps its resistance
→ Just after switching and long after
Nothing jumps
→ Just after switching and long after
Energy decays twice as fast
→ Charging, discharging and the time constant
Growth uses 1 − e^(−t/τ)
→ Charging, discharging and the time constant
ln, not log₁₀
→ Charging, discharging and the time constant

Test yourself on Current Electricity

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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