KAHIBARO
Table of Contents
Charging and Discharging in RC Circuits
An RC circuit is a circuit that contains a resistor, with resistance $R$, and a capacitor, with capacitance $C$. These circuits are important because they show how voltages and currents can change with time, instead of staying constant. Unlike simple resistor circuits, an RC circuit does not usually respond instantly when a battery is connected or removed. The capacitor needs time to charge or discharge.
In this chapter, we focus on the time behavior of the circuit. The ideas of resistance, capacitance, current, voltage, and Kirchhoff's laws belong to nearby chapters, so here we use them directly to understand how RC circuits evolve.
The Basic Charging Circuit
Consider a battery, a resistor, and a capacitor connected in series with a switch. When the switch is closed, current begins to flow and charge builds up on the capacitor plates. As the capacitor charges, its voltage increases. This growing capacitor voltage opposes the battery, so the current gradually decreases.
Applying Kirchhoff's loop rule gives
$$
V - iR - \frac{q}{C} = 0
$$
where $q$ is the charge on the capacitor and $i = \dfrac{dq}{dt}$.
So the equation becomes
$$
V - R\frac{dq}{dt} - \frac{q}{C} = 0
$$
This differential equation describes the charging process.
Charge During Charging
The solution for the charge on the capacitor as a function of time is
$$
q(t) = CV\left(1 - e^{-t/RC}\right)
$$
At the beginning, when $t = 0$,
$$
q(0) = 0
$$
so the capacitor is uncharged. After a long time, as $t \to \infty$,
$$
q(t) \to CV
$$
which is the maximum charge the capacitor can hold in this circuit.
For a charging capacitor in a series RC circuit,
$$
q(t) = CV\left(1 - e^{-t/RC}\right)
$$
and the final charge is
$$
q_{\max} = CV
$$
Current During Charging
The current is the rate of change of charge:
$$
i(t) = \frac{dq}{dt}
$$
Differentiating the charge expression gives
$$
i(t) = \frac{V}{R} e^{-t/RC}
$$
So the current is largest at the moment the switch is closed:
$$
i(0) = \frac{V}{R}
$$
and then it decreases exponentially toward zero.
This makes physical sense. At first the capacitor is uncharged, so it behaves like a gap with no stored voltage across it. Later, as charge builds up, the capacitor voltage rises and reduces the current.
For charging,
$$
i(t) = \frac{V}{R} e^{-t/RC}
$$
The current starts at its maximum value and decreases to zero.
Voltage Across the Resistor and Capacitor
The resistor voltage is
$$
V_R(t) = i(t)R = V e^{-t/RC}
$$
The capacitor voltage is
$$
V_C(t) = \frac{q(t)}{C} = V\left(1 - e^{-t/RC}\right)
$$
At every moment, the loop rule gives
$$
V = V_R + V_C
$$
At the start, the resistor gets all the battery voltage and the capacitor gets none. After a long time, the capacitor gets all the battery voltage and the resistor gets none.
| Quantity | At $t=0$ | At $t \to \infty$ |
|---|---|---|
| $q$ | $0$ | $CV$ |
| $i$ | $\dfrac{V}{R}$ | $0$ |
| $V_R$ | $V$ | $0$ |
| $V_C$ | $0$ | $V$ |
The Time Constant
The quantity
$$
\tau = RC
$$
is called the time constant of the RC circuit. It sets the timescale for charging and discharging.
After one time constant, $t = \tau$,
$$
q(\tau) = CV\left(1 - e^{-1}\right) \approx 0.632\,CV
$$
So after one time constant, the capacitor has reached about $63.2\%$ of its final charge.
The current at that time is
$$
i(\tau) = \frac{V}{R}e^{-1} \approx 0.368\,\frac{V}{R}
$$
So the current has dropped to about $36.8\%$ of its initial value.
The RC time constant is
$$
\tau = RC
$$
For charging, after time $t=\tau$, the capacitor has reached about $63\%$ of its final charge.
The Discharging Circuit
Now suppose the capacitor is initially charged and the battery is removed. If the capacitor is connected across the resistor, it discharges through the resistor. Current flows for a while, and the stored charge decreases.
Using Kirchhoff's loop rule,
$$
iR + \frac{q}{C} = 0
$$
With $i = \dfrac{dq}{dt}$, this gives
$$
R\frac{dq}{dt} + \frac{q}{C} = 0
$$
The solution is
$$
q(t) = q_0 e^{-t/RC}
$$
where $q_0$ is the initial charge on the capacitor.
Current and Voltage During Discharging
The current during discharge is
$$
i(t) = \frac{dq}{dt} = -\frac{q_0}{RC}e^{-t/RC}
$$
The negative sign shows that the current direction is opposite to the direction used during charging. If we are only interested in magnitude, we can write
$$
|i(t)| = \frac{q_0}{RC}e^{-t/RC}
$$
The capacitor voltage is
$$
V_C(t) = \frac{q(t)}{C} = \frac{q_0}{C}e^{-t/RC}
$$
If the initial capacitor voltage is $V_0 = \dfrac{q_0}{C}$, then
$$
V_C(t) = V_0 e^{-t/RC}
$$
and the current magnitude becomes
$$
|i(t)| = \frac{V_0}{R} e^{-t/RC}
$$
For discharging,
$$
q(t) = q_0 e^{-t/RC}
$$
$$
V_C(t) = V_0 e^{-t/RC}
$$
$$
|i(t)| = \frac{V_0}{R} e^{-t/RC}
$$
All of these decrease exponentially with time.
Exponential Behavior
The main mathematical feature of RC circuits is exponential change. Charging does not happen at a constant rate, and discharging does not happen at a constant rate either. The rate is fastest at the beginning and slows down with time.
This is different from motion with constant acceleration or current in a simple resistor circuit. In RC circuits, the changing capacitor voltage affects the current, and that feedback creates the exponential form.
A Simple Numerical Example
Suppose
$$
R = 2.0 \times 10^3\ \Omega
$$
and
$$
C = 5.0 \times 10^{-6}\ \text{F}
$$
Then the time constant is
$$
\tau = RC = (2.0 \times 10^3)(5.0 \times 10^{-6}) = 1.0 \times 10^{-2}\ \text{s}
$$
So
$$
\tau = 0.010\ \text{s}
$$
If the battery voltage is $12\ \text{V}$, then during charging the maximum capacitor charge is
$$
q_{\max} = CV = (5.0 \times 10^{-6})(12) = 6.0 \times 10^{-5}\ \text{C}
$$
After one time constant,
$$
q(\tau) \approx 0.632\,q_{\max} \approx 3.79 \times 10^{-5}\ \text{C}
$$
and the current is
$$
i(\tau) = \frac{12}{2000}e^{-1} \approx 2.2 \times 10^{-3}\ \text{A}
$$
Graph Shapes
In a charging process, the charge and capacitor voltage rise quickly at first and then level off. The current and resistor voltage start high and then fall toward zero.
In a discharging process, charge, current magnitude, and capacitor voltage all fall exponentially toward zero.
Physical Interpretation
A capacitor stores energy and charge, but it cannot change its voltage instantly in an ideal RC circuit. A resistor limits how fast charge can move. Together, these two effects create a delayed response.
If $R$ is larger, current is more strongly limited, so charging and discharging take longer. If $C$ is larger, more charge is needed to produce a given voltage, so charging and discharging also take longer. That is why the product $RC$ controls the timescale.
Summary Formula Table
| Process | Charge | Capacitor voltage | Current |
|---|---|---|---|
| Charging | $q(t)=CV\left(1-e^{-t/RC}\right)$ | $V_C(t)=V\left(1-e^{-t/RC}\right)$ | $i(t)=\dfrac{V}{R}e^{-t/RC}$ |
| Discharging | $q(t)=q_0 e^{-t/RC}$ | $V_C(t)=V_0 e^{-t/RC}$ | $i(t)=-\dfrac{q_0}{RC}e^{-t/RC}$ |
Key idea of RC circuits:
A resistor and capacitor together produce time-dependent behavior described by exponential functions, with timescale
$$
\tau = RC
$$
Final Remarks
RC circuits are among the simplest circuits whose behavior changes with time. They are used in timing devices, signal filters, camera flashes, and many electronic systems. The central idea is always the same, the capacitor stores charge, the resistor controls the rate, and the circuit changes exponentially with the time constant $RC$.