The switch in the circuit shifts from 1 to 2 | vedclass

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(B) During time $t_2$,the capacitor is discharging through the resistor $R$. The voltage across the capacitor is given by $V_C(t) = V_0 e^{-t/RC}$.Her

Vedclass · Accepted Answer

$2RC \ln 2$

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(B) During time $t_2$,the capacitor is discharging through the resistor $R$. The voltage across the capacitor is given by $V_C(t) = V_0 e^{-t/RC}$.Here,the capacitor discharges from $V_0 = 2V/3$ to $V = V/3$.So,$V/3 = (2V/3) e^{-t_2/RC} \implies 1/2 = e^{-t_2/RC} \implies e^{t_2/RC} = 2 \implies t_2 = RC \ln 2$.During time $t_1$,the capacitor is charging through the battery $V$. The voltage across the capacitor is given by $V_C(t) = V(1 - e^{-t/RC})$.Here,the capacitor charges from $V_{initial} = V/3$ to $V_{final} = 2V/3$.The charging equation is $V_C(t) = V - (V - V_{initial})e^{-t/RC}$.$2V/3 = V - (V - V/3)e^{-t_1/RC} \implies 2V/3 = V - (2V/3)e^{-t_1/RC} \implies (2V/3)e^{-t_1/RC} = V/3 \implies e^{-t_1/RC} = 1/2 \implies t_1 = RC \ln 2$.The total period $T = t_1 + t_2 = RC \ln 2 + RC \ln 2 = 2RC \ln 2$.

The switch in the circuit shifts from $1$ to $2$ when $V_C > 2V/3$ and goes back to $1$ from $2$ when $V_C < V/3$. The voltmeter reads the voltage as plotted. What is the period $T$ of the waveform in terms of $R$ and $C$?

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