The RC Circuit

The capacitor holds a charge. The resistor resists flow. Put them together and the story begins.

The Charge

At time zero, the capacitor is full. There is energy stored in its plates—positive charge here, negative charge there, separated by a gap that nothing can cross. The voltage across it is $V_0$. The energy sitting in that gap is $E = \frac{1}{2}CV_0^2$, waiting to be released.

The resistor sits in series, a narrow bridge between the two capacitor plates. Charges on the positive plate want to leave. They feel the electric field pulling them away from their like-charged neighbors. They want to flow through the resistor, neutralize the electrons on the other plate, and return to equilibrium.

The resistor lets them, but slowly. Every coulomb that flows through the resistor pays a price—heat, dissipation, the irreversible loss of organized charge into disorganized thermal motion. The resistor is the bottleneck. It controls the rate.

The Discharge

The circuit is closed. A switch throws. The story is written by differential equations.

The current at any moment is determined by the voltage across the resistor, which is the same as the voltage across the capacitor: $I(t) = \frac{V_C(t)}{R}$. But the capacitor voltage is falling because charge is leaving: $I(t) = -C\frac{dV_C}{dt}$. The minus sign matters—it says voltage decreases as charge flows out.

Equating these gives the governing equation: $RC\frac{dV_C}{dt} + V_C = 0$. The solution is inevitable:

$V_C(t) = V_0 e^{-t/RC}$

The voltage decays exponentially. The time constant $\tau = RC$ is the characteristic timescale. After one time constant, the voltage has dropped to $1/e \approx 36.8%$ of its initial value. After five time constants, it is below 1%. For practical purposes, the capacitor is empty.

What the Time Constant Means

$R$ in ohms, $C$ in farads, $\tau$ in seconds. The units combine naturally—resistance times capacitance is a time. Think of what each represents. A large resistor means a narrow bottleneck—charge flows slowly. A large capacitor means a big reservoir—there is more charge to lose. Both make the discharge slower. A large resistor and a small capacitor? The bottleneck is severe but the reservoir is tiny. The outcome depends on the product $RC$, not on either alone.

This is why time constants appear everywhere in electronics. Every circuit with resistance and capacitance has one. A filter that blocks high frequencies. A delay that turns on an LED half a second after a button press. A smoothing circuit that reduces ripple in a power supply. All the same physics, different numbers.

The Charge Story

Reverse the story: connect an uncharged capacitor in series with a resistor to a voltage source. The current starts high—instantly, the capacitor looks like a short circuit—and decays as the capacitor charges. $I(t) = \frac{V}{R}e^{-t/RC}$. The capacitor voltage rises as $V_C(t) = V(1 - e^{-t/RC})$, asymptotically approaching the source voltage.

Charge and discharge are mirror images. Both governed by the same time constant. Both exponential. Both inevitable.

The Energy

Here is the thing nobody tells you: exactly half the energy supplied by the source is stored in the capacitor during charging. The other half is dissipated as heat in the resistor. This is true regardless of resistance value. Even with zero resistance—where intuition suggests no dissipation should occur—half the energy is lost. (In the zero-resistance limit, the circuit oscillates and the energy loss mechanism changes, but the result holds for any nonzero $R$.)

The resistor does not merely slow things down. It destroys energy. Irreversibly. The capacitor can give it back, but only as electrical energy. The heat in the resistor is gone—scattered into the thermal motion of trillions of atoms.