RC circuit — charging/discharging time constant and behavior

Question

How does a capacitor charge and discharge through a resistor in an RC circuit? What is the time constant and how do we use it to solve problems?

Solution — Step by Step

A capacitor $C$ in series with resistor $R$ , connected to a battery of emf $\varepsilon$ :

The charge on the capacitor grows exponentially:

q(t) = C\varepsilon\left(1 - e^{-t/RC}\right) = Q_0\left(1 - e^{-t/\tau}\right)

where $Q_0 = C\varepsilon$ is the final (maximum) charge and $\tau = RC$ is the time constant.

The current decreases exponentially:

i(t) = \frac{\varepsilon}{R} e^{-t/RC} = I_0 \, e^{-t/\tau}

At $t = 0$ : current is maximum ( $\varepsilon/R$ ), charge is zero. At $t = \infty$ : current is zero, charge is maximum ( $C\varepsilon$ ).

A fully charged capacitor ( $Q_0$ ) discharged through a resistor:

q(t) = Q_0 \, e^{-t/RC}

i(t) = \frac{Q_0}{RC} e^{-t/RC}

Both charge and current decay exponentially. After one time constant ( $t = \tau$ ), the charge drops to $Q_0/e \approx 0.37 Q_0$ — about 37% of the initial value.

\tau = RC

At $t = \tau$ : charging reaches 63% of max ( $1 - 1/e$ ); discharging drops to 37% ( $1/e$ )
At $t = 2\tau$ : charging reaches 86.5%
At $t = 3\tau$ : charging reaches 95%
At $t = 5\tau$ : practically fully charged (99.3%)

JEE loves asking: “After how many time constants does the capacitor reach 99% of its maximum charge?” Answer: $t = -\tau \ln(0.01) = \tau \ln(100) \approx 4.6\tau$ . For 99.3%, it is exactly $5\tau$ .

Units check: $\tau = RC$ has units of $\Omega \times F = \frac{V}{A} \times \frac{C}{V} = \frac{C}{A} = s$ (seconds).

During charging, the battery supplies energy $Q_0 \varepsilon = C\varepsilon^2$ . Of this:

Half ( $\frac{1}{2}C\varepsilon^2$ ) is stored in the capacitor
Half ( $\frac{1}{2}C\varepsilon^2$ ) is dissipated as heat in the resistor

This 50-50 split holds regardless of the value of $R$ .

flowchart TD
    A["RC Circuit Problem"] --> B{"Charging or discharging?"}
    B -->|"Charging"| C["q = Q0 times 1 minus exp of minus t/RC"]
    B -->|"Discharging"| D["q = Q0 times exp of minus t/RC"]
    C --> E["Current: i = emf/R times exp of minus t/RC"]
    D --> F["Current: i = Q0/RC times exp of minus t/RC"]
    A --> G["Time constant tau = RC"]
    G --> H["At t = tau: 63% charged or 37% remaining"]
    G --> I["At t = 5tau: practically complete"]

Why This Works

The exponential behaviour comes from solving the differential equation obtained by applying Kirchhoff’s voltage law:

\varepsilon = iR + \frac{q}{C}

Since $i = dq/dt$ , this becomes a first-order linear ODE:

\varepsilon = R\frac{dq}{dt} + \frac{q}{C}

The solution is the exponential growth/decay we see. The time constant $\tau = RC$ naturally appears as the characteristic timescale of the circuit — larger $R$ or $C$ means slower charging/discharging.

Alternative Method

For finding the time to reach a specific charge, rearrange the charging equation:

t = -RC \ln\left(1 - \frac{q}{Q_0}\right)

This is faster than trial-and-error with the original formula when the question gives a specific charge or voltage to reach.

Common Mistake

Students often assume that doubling $R$ halves the final charge stored. Wrong — the final charge $Q_0 = C\varepsilon$ does not depend on $R$ at all. Resistance only affects how fast the capacitor charges, not how much charge it finally stores. The time constant doubles (slower charging), but the same final charge is reached. JEE Advanced 2020 Paper 2 had an RC problem where this misconception was the primary distractor.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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