Problem — which graph for an RC circuit

Problem

Which graph for an RC circuit? A resistor $R$ and an initially uncharged capacitor $C$ are connected to a battery $\mathcal{E}$ at instant $t = 0$. Among the following graphs, which one correctly represents the current $i(t)$ in the circuit?

A: constant current. B: exponential decay. C: exponential growth.

(a) Identify the correct answer. (b) What does the excluded graph represent? (c) How does the correct graph change if $R$ doubles? And if $C$ doubles?

Solution

(a) Answer: graph B (exponential decay). At $t = 0$ the capacitor is uncharged and behaves like a short circuit: the current starts at its maximum $i_0 = \mathcal{E}/R$. As $C$ charges, its voltage grows and opposes the battery, so the current decreases. At steady state the capacitor is charged ($V_C = \mathcal{E}$) and the current vanishes. The trend is $i(t) = \frac{\mathcal{E}}{R}\,e^{-t/RC}$ i.e. an exponential decay from $\mathcal{E}/R$ to $0$: exactly graph B.

(b) The excluded graphs.

• A (constant): would be the current in a purely resistive circuit (without a capacitor), or the steady-state case of an inductor — not a charging RC circuit.
• C (exponential growth): instead describes the charge $q(t)$ or the voltage $V_C(t) = \mathcal{E}(1 - e^{-t/RC})$ on the capacitor, which starts at zero and saturates. It is the quantity complementary to the current, not the current itself.

(c) Effect of $R$ and $C$. The time constant is $\tau = RC$.

• If $R$ doubles: $\tau$ doubles (slower decay) and the initial current $i_0 = \mathcal{E}/R$ halves: the curve starts lower and falls more gently.
• If $C$ doubles: $\tau$ doubles (slower decay) but the initial current $i_0 = \mathcal{E}/R$ does not change: the curve starts from the same point but falls more slowly.

$\ev{\text{graph B: } i(t) = (\mathcal{E}/R)\,e^{-t/RC}}$

Links

Topics: Circuiti elettrici Concepts: Circuito RC Skills: Lettura dei grafici