In the RC circuit shown,the switch is closed | vedclass

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(B) The potential across the resistor is given by $V_R = \varepsilon e^{-t/RC}$.The potential across the capacitor is given by $V_C = \varepsilon (1 -

Vedclass · Accepted Answer

$145$

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(B) The potential across the resistor is given by $V_R = \varepsilon e^{-t/RC}$.The potential across the capacitor is given by $V_C = \varepsilon (1 - e^{-t/RC})$.From the graph,at $t = 100 \, ms$,the potentials $V_R$ and $V_C$ are equal,i.e.,$V_R = V_C$.Substituting the expressions: $\varepsilon e^{-t/RC} = \varepsilon (1 - e^{-t/RC})$.Dividing by $\varepsilon$: $e^{-t/RC} = 1 - e^{-t/RC}$.Rearranging gives $2e^{-t/RC} = 1$,or $e^{-t/RC} = 1/2$.Taking the natural logarithm on both sides: $-t/RC = \ln(1/2) = -\ln(2)$.Thus,$t/RC = \ln(2)$.Given $t = 100 \, ms$ and $\ln(2) \approx 0.693$,we have $RC = t / \ln(2) = 100 / 0.693 \approx 144.3 \, ms$.Rounding to the nearest option,the time constant is approximately $145 \, ms$.

In the $RC$ circuit shown,the switch is closed at $t = 0$. Graphs showing the variation of potential $(V_R)$ across the resistor and potential $(V_C)$ across the capacitor are given. The time constant of the circuit is approximately equal to.....$ms$

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