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An 11 V pulse of 10 µs duration is applied to the circuit shown in Figure. Assuming that the capacitor is completely discharged prior to applying the pulse, the peak value of the capacitor voltage is
11 V
5.5 V
6.32 V
0.96 V
Since voltage pulse is applied to 
. So capacitor will change up to . So capacitor voltage will be at peak 
and then the capacitor will discharge upto 
Before applying voltage pulse,
Equivalent Resistance across Capacitor Terminals,
At steady state capacitor acts as open circuit, so voltage can be determined by Potential Divider.
Capacitor voltage as a function of time is,
; 
This equation has been written assuming a constant 11V source at the terminals but this equation is only valid upto 10μs till the source is connected and after that discharging RC circuit will be used.
For peak voltage
