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
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 idea behind this question
Every first-order circuit moves exponentially from its initial value to its final value with time constant $\tau = RC$: $v(t) = V_f + (V_i - V_f)e^{-t/\tau}$. After one time constant it has covered 63.2% of the way; after five it is within 1% of the end, which is why $5\tau$ is used as "fully charged".
Try a variation
In the same circuit, what is the capacitor voltage at $t = 2$ ms?
Show the answer
Answer: $8.65$ V
$t = 2\tau$, so $10(1 - e^{-2}) = 8.65$ V.
Other mistakes to avoid
- Calculating $\tau$ with the wrong units - kilohms times microfarads is milliseconds.
- Using the capacitor current formula when the question asks for voltage, or the other way round.