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:
[tex]V(t)=V_0(1-e^{-\frac{t}{\tau}})[/tex]
where
[tex]V_0[/tex] is the final potential difference on the capacitor, which is equal to the potential difference of the battery
[tex]\tau[/tex] 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
[tex]V(t)=0.70V_0[/tex]
Substituting into the equation and solving for t, we find:
[tex]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[/tex]
Therefore, the capacitor reaches 70% of its final potential difference after 1.21 time constants.
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