ElectricalTransientsmediumFE ExamCoursework

Capacitor Voltage During RC Charging

An initially uncharged capacitor $C = 100\ \mu\text{F}$ charges through $R = 10\ \text{k}\Omega$ from a $10\ \text{V}$ source, switched on at $t = 0$ .

Find the capacitor voltage at $t = 2.0\ \text{s}$ .

Given

R	10 kohm	Series resistance
C	100 uF	Capacitance
V	10 V	Source voltage
t	2 s	Elapsed time

Hint 1

Find the time constant first, then note how many time constants have elapsed.

Hint 2

τ = RC = 1 s, so t = 2 s is exactly two time constants.

Hint 3

Charging rises toward the source: v = V(1 − e^(−t/τ)). At 2τ that is about 86.5%.

Worked solution — try the problem first

Time constant:

$\tau = RC = (10\times10^{3})(100\times10^{-6}) = 1.0\ \text{s}$

The unit prefixes conveniently cancel here: kΩ × μF gives milliseconds normally, but $10^4 \times 10^{-4} = 1$ s exactly.

Charging response for an initially uncharged capacitor:

$v_C(t) = V\left(1 - e^{-t/\tau}\right)$

$v_C(2.0) = 10\left(1 - e^{-2.0/1.0}\right) = 10\left(1 - e^{-2}\right)$

$= 10(1 - 0.1353) = 10(0.8647) = 8.647\ \text{V}$

$\boxed{v_C = 8.65\ \text{V}}$

The rule of thumb. After $1\tau$ the capacitor reaches 63.2%, after $2\tau$ 86.5%, after $3\tau$ 95%, and by $5\tau$ it is 99.3% charged — effectively steady state. This problem is exactly the $2\tau$ point.

Concepts:RC transientsTime constantExponential response

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