Capacitor voltage after one time constant in a series RC circuit
BlinkNBuild practice problem · GATE standard, authored and verified in-house
In a series $RC$ circuit with $R = 1\ \text{k}\Omega$ and $C = 1\ \mu\text{F}$, the capacitor is initially uncharged. At $t = 0$ a switch connects the circuit to a $10\ \text{V}$ DC source.
The capacitor voltage at $t = 1\ \text{ms}$, in volts, is ________.
Show the step-by-step derivation
Answer
6.32 V (accept 6.30 to 6.34) V (accept 6.30 to 6.34)
Step-by-step derivation
- Compute the time constant: $$\tau = RC = (1 \times 10^{3})(1 \times 10^{-6}) = 1 \times 10^{-3}\ \text{s} = 1\ \text{ms}.$$
- Write the charging response. With zero initial charge and a step to $V_s = 10$ V, the capacitor voltage is $$v_C(t) = V_s\left(1 - e^{-t/\tau}\right).$$
- The instant asked for is $t = 1\ \text{ms}$, which is exactly one time constant, so $t/\tau = 1$: $$v_C = 10\left(1 - e^{-1}\right).$$
- With $e^{-1} = 0.36788$: $$v_C = 10\,(1 - 0.36788) = 10 \times 0.63212 = \mathbf{6.32\ V}.$$
- Worth memorising. After one time constant a charging exponential has reached $63.2\%$ of its final value, and a discharging one has fallen to $36.8\%$. After $5\tau$ the response is within $0.7\%$ of final, which is the usual engineering definition of "settled".
The trap this question is built around
Answering $3.68$ V by using the discharge form $V_s e^{-t/\tau}$. The capacitor starts at zero and rises toward the source, so the correct shape is $1 - e^{-t/\tau}$. Always ask which way the exponential is going.