RC and LR Circuits — Transient Behaviour

When you flip a switch in a circuit with a capacitor or an inductor, the current does not jump instantly to its steady-state value. It eases in (or out) over a characteristic timescale. That transient phase is what this chapter is about.

The equations are short, but JEE Main and Advanced love asking conceptual questions: what does the curve look like? What’s the energy budget? When does the inductor act like a wire versus an open switch? Get those right and you’ll bank 4–8 marks per attempt.

The Two Setups Side by Side

In an RC circuit (resistor + capacitor + battery), the capacitor charges from 0 to $\varepsilon$ over time. Initially the capacitor acts like a wire (uncharged plates, no voltage drop). Finally it acts like an open switch (fully charged, no current flows).

In an LR circuit (resistor + inductor + battery), the current grows from 0 to $\varepsilon/R$ . Initially the inductor acts like an open switch (resists current change). Finally it acts like a wire (steady current, no $dI/dt$ , no back EMF).

The behaviours are mirror images.

Key Terms & Definitions

Transient state — The time-dependent regime between switch-throw and steady-state.

Steady state — Long after switching ( $t \gg \tau$ ); all currents and voltages are constant.

Time constant ( $\tau$ ) — The characteristic timescale of the transient. For RC: $\tau = RC$ . For LR: $\tau = L/R$ . Units: seconds.

Half-life of decay — Time for current/charge to fall to half: $t_{1/2} = \tau \ln 2 \approx 0.693\tau$ .

Core Formulas

RC charging: $q(t) = q_0 (1 - e^{-t/\tau})$ , $i(t) = \dfrac{\varepsilon}{R} e^{-t/\tau}$ , $\tau = RC$ , $q_0 = C\varepsilon$ .

RC discharging: $q(t) = q_0 e^{-t/\tau}$ , $i(t) = -\dfrac{q_0}{RC} e^{-t/\tau}$ .

LR growth: $i(t) = i_0 (1 - e^{-t/\tau})$ , $\tau = L/R$ , $i_0 = \varepsilon/R$ .

LR decay: $i(t) = i_0 e^{-t/\tau}$ .

In all four, the exponential factor handles the timescale; the prefactor handles the steady-state value.

Methods & Concepts

Deriving the RC Charging Equation

Apply KVL around the charging loop:

\varepsilon = iR + \frac{q}{C} = R\frac{dq}{dt} + \frac{q}{C}

Separate variables:

\frac{dq}{C\varepsilon - q} = \frac{dt}{RC}

Integrate from $q=0$ at $t=0$ :

-\ln(C\varepsilon - q) + \ln(C\varepsilon) = \frac{t}{RC}

q(t) = C\varepsilon (1 - e^{-t/RC})

Differentiate to get current: $i(t) = (\varepsilon/R) e^{-t/RC}$ .

Deriving the LR Growth Equation

KVL: $\varepsilon = iR + L\, di/dt$ . Same form as RC. Solution:

i(t) = \frac{\varepsilon}{R}(1 - e^{-Rt/L})

The mathematics is identical to RC charging with the substitution $C \leftrightarrow 1/L$ (loosely).

Energy Bookkeeping

In RC charging:

Total energy supplied by battery: $W_{\text{batt}} = \varepsilon q_0 = \varepsilon \cdot C\varepsilon = C\varepsilon^2$ .
Energy stored in capacitor at end: $U_C = \tfrac{1}{2}C\varepsilon^2$ .
Energy dissipated in $R$ : $W_R = \tfrac{1}{2}C\varepsilon^2$ .

Exactly half the supplied energy is lost as heat — independent of $R$ . This is a famous result and a frequent JEE Main question.

In LR growth:

Energy from battery: large (depends on time).
Energy stored in $L$ at steady state: $\tfrac{1}{2}L i_0^2$ .
Rest dissipated in $R$ .

Solved Examples

Easy (CBSE) — Charge at half time-constant

Q. A 4 μF capacitor charges through a 5 kΩ resistor connected to a 12 V battery. Find the charge after one time constant.

Solution. $\tau = RC = 5000 \times 4 \times 10^{-6} = 0.02$ s. At $t = \tau$ , $q = q_0(1 - e^{-1}) = q_0 \times 0.632$ . $q_0 = C\varepsilon = 48~\mu$ C. So $q = 30.3~\mu$ C.

Medium (JEE Main) — LR circuit with switch

Q. A 0.5 H inductor and 10 Ω resistor are connected to a 20 V battery via a switch. After the switch is closed, find the time at which current reaches 1.5 A.

Solution. Steady-state $i_0 = 20/10 = 2$ A. Time constant $\tau = 0.05$ s.

1.5 = 2(1 - e^{-t/0.05}) \Rightarrow e^{-t/0.05} = 0.25 \Rightarrow t = 0.05 \ln 4 \approx 0.0693 \text{ s}

Hard (JEE Advanced) — Energy half-rule

Q. A capacitor $C$ is charged through resistor $R$ to a final voltage $V$ from an EMF source $V$ . Show that exactly half the energy supplied is dissipated as heat — independent of $R$ .

Solution. Charge delivered $= CV$ . Energy supplied by battery $= V \cdot CV = CV^2$ . Energy stored at end $= \tfrac{1}{2}CV^2$ . Heat dissipated = supplied − stored = $\tfrac{1}{2}CV^2$ . Ratio is $\tfrac{1}{2}$ , regardless of $R$ . ( $R$ only sets how long the dissipation takes.)

Exam-Specific Tips

CBSE 3-mark: “Derive the equation for charging of a capacitor through a resistor.” Memorise the integration steps.

JEE Main: Curve shape questions — sketch $i(t)$ for charging vs discharging. Charging current decays from peak; charge grows from 0.

JEE Advanced: Multi-loop networks with one capacitor or inductor. Reduce the rest of the circuit to a Thevenin equivalent first, then apply $\tau = R_{\text{th}}C$ or $L/R_{\text{th}}$ .

NEET: Time constant identification and steady-state current questions.

Common Mistakes to Avoid

1. Wrong $\tau$ for LR. $\tau = L/R$ , not $LR$ . Units check: H/Ω = (V·s/A)/(V/A) = s. ✓

2. Treating capacitor as wire at $t = \infty$ . At steady state, capacitor blocks DC — so it’s an open switch.

3. Treating inductor as open at $t = \infty$ . At steady state, inductor passes DC freely — so it’s a wire.

4. Forgetting the $i_0$ factor. Current never exceeds $\varepsilon/R$ ; the exponential just controls how fast you approach it.

5. Sign error in discharge. During discharge, current flows opposite to charging direction. Some books write the formula with positive $i$ , others with negative — pick a convention and stick with it.

Practice Questions

Q1. $\tau$ of an RC circuit with $R = 100$ kΩ, $C = 10~\mu$ F.

$\tau = 10^5 \times 10^{-5} = 1$ s.

Q2. When does charge reach half its final value in RC charging?

$0.5 = 1 - e^{-t/\tau} \Rightarrow t = \tau \ln 2 \approx 0.693\tau$ .

Q3. What’s the current immediately after closing the switch in an LR circuit?

Zero. Inductor opposes any sudden change.

Q4. What’s the current immediately after closing the switch in an RC circuit?

$\varepsilon/R$ — the capacitor acts like a short circuit at $t=0$ .

Q5. Energy stored in a 2 H inductor carrying 3 A.

$U = \tfrac{1}{2}(2)(9) = 9$ J.

Q6. During RC discharge through a resistor, where does the energy go?

It is dissipated as heat in the resistor. Total heat = total energy initially stored = $\tfrac{1}{2}CV^2$ .

Q7. Why is exactly half the battery’s energy lost in charging a capacitor through any resistor?

The 1/2 comes from integrating $i^2 R$ over the exponential charging current. Algebraically, $\int_0^\infty (\varepsilon/R)^2 R e^{-2t/RC} dt = \tfrac{1}{2}C\varepsilon^2$ — the $R$ cancels. The lost fraction is universal.

Q8. What does $\tau = 0$ correspond to physically?

Either $R = 0$ (perfect wire) or $C = 0$ / $L = 0$ (no reactive element). Transient is instantaneous — the circuit jumps to steady state with no delay.

FAQs

Why don’t transients last forever mathematically?

The exponential decay never exactly reaches zero — it’s asymptotic. After $5\tau$ , the current is at $e^{-5} \approx 0.7\%$ of its initial value, which is below most measurement thresholds. Practically, we say the transient is “over” by then.

Can RC and LR circuits oscillate?

No — both have only one energy-storage element. Oscillations need two (LC or RLC). RC and LR show pure exponential approach without overshoot.

What if $R = 0$ in an LR circuit?

The current would grow linearly forever — there’s nothing to limit it. In practice, every real inductor has some series resistance.

Does the time constant depend on EMF?

No. $\tau$ is purely a property of the components. EMF affects only the amplitude (steady-state value), not the timescale.

Why are RC circuits used in flash bulbs and pacemakers?

The slow charge-discharge cycle of an RC pair gives a controlled energy release. Flash bulbs charge a capacitor over seconds and dump it through the bulb in milliseconds — high peak power, low average power.