An 11 V pulse of 10 micro second duration is applied to the circuit in the given figure. Assuming that the capacitor is completely discharged prior to applying the pulse, the peak value of the capacitor voltage is [Figure description: A pulse source (11 V, 10 µs) in series with a 1 kΩ resistor feeding a node; from the node a 10 kΩ resistor and an 11 µF capacitor are connected in parallel to the bottom rail.]
- A. 11 V
- B. 5.5 V
- C. 6.32 V
- D. 10 V
Answer: C. 6.32 V
Explanation: The Thevenin equivalent seen by C has Vth = 11×(10/11) = 10 V and Rth = 1k∥10k = 10/11 kΩ. With C = 11 µF, time constant τ = 10 ms. After 10 ms pulse, Vc = 10(1 - e^-1) = 6.32 V.