Question

In an RC circuit, how many time constants must elapse if an initially uncharged capacitor is to

Answer 1

The capacitor reaches 70% of its final potential difference after 1.21 time constants

Explanation:

In an RC circuit, when an initially uncharged capacitor is connected to a battery, charge starts to flow through the circuit and accumulates on the capacitor.

As a result, the potential difference across the capacitor also increases, until the capacitor becomes fully charged.

The potential difference across the capacitor follows the equation:

$V(t)=V_0(1-e^{-\frac{t}{\tau}})$

where

$V_0$ is the final potential difference on the capacitor, which is equal to the potential difference of the battery

$\tau$ is the time constant of the circuit

In this problem, we are asked after what time the capacitor reaches 70% of its final potential difference, therefore the time t at which

$V(t)=0.70V_0$

Substituting into the equation and solving for t, we find:

$0.70V_0 = V_0(1-e^{-\frac{t}{\tau}})\\0.70=1-e^{-\frac{t}{\tau}}\\e^{-\frac{t}{\tau}}=0.30\\\frac{t}{\tau}=-ln(0.30)=1.21\\t=1.21\tau$

Therefore, the capacitor reaches 70% of its final potential difference after 1.21 time constants.

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