Physics

Physics

Problem

At a time equal to one time constant after a switch is closed in an ideal RC charging circuit, what percentage of its final charge has the capacitor accumulated?
Intuitive explanation

The key is recognizing that charge in an RC circuit grows exponentially, not linearly. At time t=τt = \tau (one time constant), the charge follows Q(t)=Qf(1−e−t/τ)Q(t) = Q_f(1 - e^{-t/\tau}), where QfQ_f is the final charge. Plug in t=τt = \tau and you'll see the exponential decay factor becomes e−1e^{-1} — a specific, memorable number that tells you exactly what fraction of the final charge has built up.

Intuitive explanation

The key is recognizing that charge in an RC circuit grows exponentially, not linearly. At time t=τt = \tau (one time constant), the charge follows Q(t)=Qf(1−e−t/τ)Q(t) = Q_f(1 - e^{-t/\tau}), where QfQ_f is the final charge. Plug in t=τt = \tau and you'll see the exponential decay factor becomes e−1e^{-1} — a specific, memorable number that tells you exactly what fraction of the final charge has built up.

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