RC and RL Circuit Transients for GATE: Time Constant, Initial and Final Values, Solved Examples

RC and RL transients for GATE: the one formula for first-order circuits, time constant from Rth, initial and final values, solved examples, traps and practice questions.

In the Thevenin and Norton post we squeezed whole circuits down to one source and one resistor. Now we add one more part, a capacitor or an inductor, and flip a switch. For a short while the circuit is in a hurry, and then it settles down. That short, busy stretch is called the transient, and it's one of the most dependable scoring areas in GATE Network Theory.

The good news: almost every first-order question falls to one formula and a three-step routine. The Thevenin skills you already have do half the work.

Why a capacitor or inductor makes the circuit wait

A resistor responds instantly. A capacitor and an inductor don't, because they store energy, and energy can't change in zero time.

A capacitor stores energy $\tfrac{1}{2}CV^2$ in its voltage, so its voltage can't jump. An inductor stores energy $\tfrac{1}{2}LI^2$ in its current, so its current can't jump.

That gives us the two rules that start every transient problem:

Here $0^-$ means just before the switch moves and $0^+$ just after. Everything else in the circuit, including the capacitor's current and the inductor's voltage, is free to jump.

What they look like after a long time

Once a DC circuit has settled, nothing changes any more, so

In plain words: at DC steady state, a capacitor behaves like an open circuit and an inductor behaves like a short circuit. Remember this and finding final values becomes simple circuit analysis.

Capacitor Inductor --------- Can't jump voltage $vC$ current $iL$ At $t = 0^+$, if uncharged behaves like a short behaves like an open At $t \to \infty$ (DC) open circuit short circuit Energy stored $\tfrac{1}{2}CV^2$ $\tfrac{1}{2}LI^2$

The one formula for every first-order circuit

Any voltage or current $x(t)$ in a circuit with one capacitor or one inductor and DC sources follows

where the time constant is

and $R{th}$ is the Thevenin resistance seen by the capacitor or inductor, with the independent sources switched off. This is exactly the $R{th}$ from the last post, just measured across different terminals.

So every problem becomes three small ones:

How fast is "fast"?

The time constant tells you everything about speed. After one $\tau$, a charging capacitor has covered 63.2% of the way to its final value. After five, it's at 99.3%, and engineers treat it as finished.

$t$ $\tau$ $2\tau$ $3\tau$ $4\tau$ $5\tau$ ------------------ Rising curve 63.2% 86.5% 95.0% 98.2% 99.3% Falling curve 36.8% 13.5% 5.0% 1.8% 0.7%

Keep the 63% and 37% in your head. They turn many GATE questions into mental arithmetic.

Worked example 1: charging a capacitor

In the left circuit of Fig. 1, the capacitor starts uncharged and the switch closes at $t = 0$.

Initial value. The capacitor was uncharged, so $vC(0^+) = vC(0^-) = 0$. Final value. Long after, the capacitor is an open circuit, no current flows, and it sits at the full supply: $vC(\infty) = 10$ V. Time constant. Switch off the source (short it) and look in from the capacitor: you see just the 2 kΩ. So $\tau = 2000 \times 5 \times 10^{-6} = 10$ ms.

At $t = 10$ ms the capacitor is at 6.32 V, and by 50 ms it's essentially at 10 V. Notice that the current $i = (10 - vC)/2000$ jumps straight to 5 mA at $t = 0^+$ and then dies away. The capacitor's current can jump. Only its voltage can't.

Worked example 2: an RL circuit that needs Thevenin

Now a GATE-style one. The switch closes at $t = 0$, and the inductor carries no current before that. Find $iL(t)$ and $vL(t)$.

Step 1: initial value. No current before the switch closes, so $iL(0^+) = 0$.

Step 2: final value. Long after, the inductor is a short circuit. It shorts out the 3 Ω resistor, so all the current goes through the 6 Ω and the inductor:

Step 3: time constant. Short the 12 V source and look in from the inductor. The 6 Ω and 3 Ω now sit in parallel:

Step 4: the answer.

Look at $vL(0^+) = 4$ V. At the instant the switch closes, the inductor refuses to carry current, so it behaves like an open circuit and takes the whole Thevenin voltage, $12 \times \tfrac{3}{9} = 4$ V. That's a lovely check: the Thevenin circuit you'd draw from the last post tells you both the starting voltage and the final current.

Common traps in GATE questions

These are the mistakes I see every year in class tests.

Using the wrong resistance for τ. It's the $R{th}$ seen from the capacitor or inductor, with sources switched off, not the resistance in series with the source. Letting the wrong quantity jump. Only $vC$ and $iL$ are continuous. Capacitor current and inductor voltage can, and usually do, jump at $t = 0$. Mixing up $0^-$ and $0^+$. Work out $vC$ or $iL$ in the circuit before the switch moves, then carry that value across to the new circuit. Assuming τ stays the same. If a switch opens and later closes, the circuit changes, and so does $R{th}$. Charging and discharging can have different time constants. Forgetting that the formula needs DC sources. With a sinusoidal source the final value isn't a constant. That's a different technique. Unit slips. kΩ times µF gives milliseconds. Write the powers of ten out once and you'll never get this wrong. Quitting too early. "After a long time" in GATE means after about five time constants, not one.

Practice questions

Give yourself about two minutes each.

A 10 kΩ resistor and a 10 µF capacitor are connected to a 5 V source at $t = 0$, with the capacitor uncharged. Find $\tau$ and $vC$ at $t = 0.2$ s. Find the time constant of the RL circuit in Fig. 1: 12 V, 2 Ω and 0.5 H. What current flows after a long time? A capacitor of 25 µF is charged to 20 V and then discharged through 4 kΩ. Find its voltage after 0.1 s and the energy it started with. In worked example 2, find $vL$ at $t = 0.5$ s. How long does a charging RC circuit take to reach 90% of its final value? A 10 V source feeds a 2 Ω resistor in series with a 2 Ω resistor and a 1 F capacitor in parallel. The capacitor starts uncharged. Find $vC(\infty)$ and $\tau$.

Answers

$\tau = 10^4 \times 10^{-5} = 0.1$ s, and $vC = 5\left(1 - e^{-2}\right) = 4.32$ V. $\tau = 0.5 / 2 = 0.25$ s, and the final current is $12 / 2 = 6$ A. $\tau = 4000 \times 25 \times 10^{-6} = 0.1$ s, so $vC = 20\,e^{-1} = 7.36$ V. The starting energy is $\tfrac{1}{2} \times 25 \times 10^{-6} \times 20^2 = 5$ mJ. $vL = 4\,e^{-1} = 1.47$ V. Solve $1 - e^{-t/\tau} = 0.9$, which gives $t = \tau \ln 10 \approx 2.3\,\tau$. The capacitor ends up across the lower 2 Ω, so $vC(\infty) = 10 \times \tfrac{2}{4} = 5$ V. With the source shorted, it sees $2 \parallel 2 = 1\ \Omega$, so $\tau = 1 \times 1 = 1$ s.

Question 6 is the one to watch. It's the same idea as worked example 2: the Thevenin resistance seen by the capacitor isn't the resistance next to the source. Next time we'll take this a step further and see what happens when a circuit has both an inductor and a capacitor, where the response can overshoot and ring. Tell me in the comments which question slowed you down.