RC Circuit & Time Constant

When a capacitor charges or discharges through a resistor, the governing equation is a first-order ODE derived from Kirchhoff's voltage law:

$V_R + V_C = V_0 \Rightarrow RC\frac{dV_C}{dt} + V_C = V_0$

This gives exponential behaviour characterised by the time constant $\tau = RC$ (in seconds when $R$ is in ohms and $C$ in farads).

• Charging ($V_C$ from 0 to $V_0$): voltage rises and current falls.
• Discharging ($V_C$ from $V_0$ to 0): both voltage and current decay exponentially.

Key milestones: after $1\tau$, the capacitor has completed ~63.2% of its charge/discharge. After $5\tau$, it is ~99.3% complete — considered "fully charged" in practice.

Charging:

$V_C(t) = V_0\!\left(1 - e^{-t/\tau}\right), \quad I(t) = \frac{V_0}{R}\,e^{-t/\tau}$

Discharging:

$V_C(t) = V_0\,e^{-t/\tau}, \quad I(t) = -\frac{V_0}{R}\,e^{-t/\tau}$

Energy stored in capacitor: $U = \frac{1}{2}CV_C^2$. Total energy dissipated in $R$ during charging $= \frac{1}{2}CV_0^2$ (exactly half of the energy supplied by the battery).

$R = 10\,\text{k}\Omega$, $C = 100\,\mu\text{F}$, $V_0 = 9\,\text{V}$.

$\tau = RC = 10{,}000 \times 100 \times 10^{-6} = 1\,\text{s}$

At $t = 1\,\text{s}$ (charging): $V_C = 9(1 - e^{-1}) = 9 \times 0.632 \approx 5.69\,\text{V}$

At $t = 2\,\text{s}$: $V_C = 9(1 - e^{-2}) \approx 7.78\,\text{V}$

Energy stored when fully charged: $U = \frac{1}{2}(100 \times 10^{-6})(9)^2 = 4.05\,\text{mJ}$

RC circuits set the timing in camera flashes, heartbeat pacemakers, and 555-timer ICs. Audio equalizers use RC filters to boost or cut frequency bands — the time constant determines the cutoff frequency $f_c = 1/(2\pi RC)$. Touch screens detect a finger's tiny capacitance change, altering the local $\tau$ of an RC grid.

1. An RC circuit has $R = 5\,\text{k}\Omega$ and $C = 200\,\mu\text{F}$. Find $\tau$ and the time for the voltage to reach 99% of $V_0$.

2. During charging, the battery supplies total energy $CV_0^2$. The capacitor stores $\frac{1}{2}CV_0^2$. Where does the other half go?