For the given input voltage waveform Vᵢₙ(t), the output voltage waveform V₀(t), across the capacitor is correctly depicted by [Figure: Vᵢₙ is a pulse train: +5 V from 0 to 5 μs, 0 from 5 to 10 μs, +5 V again from 10 to 15 μs. It is applied through 1 kΩ to a 10 nF capacitor, and V₀ is taken across t…

Question

For the given input voltage waveform Vᵢₙ(t), the output voltage waveform V₀(t), across the capacitor is correctly depicted by [Figure: Vᵢₙ is a pulse train: +5 V from 0 to 5 μs, 0 from 5 to 10 μs, +5 V again from 10 to 15 μs. It is applied through 1 kΩ to a 10 nF capacitor, and V₀ is taken across the capacitor. Options: (a) rises to about 2 V by 5 μs, sags slightly until 10 μs, then rises towards 3 V by 15 μs; (b) straight-line rise to 2 V and straight-line fall to 0, repeated; (c) rises to 2 V and stays flat; (d) rises to 2 V and decays back to 0 by 10 μs, then repeats.]

Accepted Answer

(1) Rises to about 2 V, sags, then climbs towards 3 V. τ=RC=10³×10×10⁻⁹=10 μs First pulse (0 to 5 μs): V=5(1-e^-0.5)=1.97 V≈2 V Gap (5 to 10 μs): V=1.97e^-0.5=1.19 V Second pulse: V=5-(5-1.19)e^-0.5=2.69 V, close to 3 V Only (a) shows the build-up over successive pulses. (Its sag during the gap is drawn smaller than the true drop.)

Step-by-step solution

Why the other options are wrong