For the given input voltage waveform $V_{\text {in }}( t )$, the output voltage waveform $V _{ D }( t ),$ across the capacitor is correctly depicted by:
JEE MAIN 2020, Diffcult
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$V _{0}( t )= V _{\text {in }}\left(1- e ^{\frac{ t }{ RC }}\right)$
at $t =5 \mu s$
$V _{0}( t )=5\left(1- e ^{-\frac{5 \times 10^{-6}}{10^{3} \times 10 \times 10^{-9}}}\right)$
$=5\left(1- e ^{-0.5}\right)=2 V$
Now $V_{\text {in }}=0$ means discharging
$V _{0}( t )=2 e ^{-\frac{ t }{ RC }}=2 e ^{-0.5}$
$=1.21 V$
Now for next $5 \mu s$
$V _{0}( t )=5-3.79 e ^{-\frac{t}{ RC }}$
after $5 \mu s$ again
$V _{0}( t )=2.79 Volt \approx 3 V$
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