Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - True or False (Q.No. 11)
11.
In the given circuit, the voltage across the capacitor 1 ms after the switch is closed is 11.61 V.
Discussion:
5 comments Page 1 of 1.
Pritamnutil said:
4 years ago
@All.
You don't need to subtract 3.387 to 15. 3.387 is the value of voltage in the capacitor after 1ms. Just follow the equation Vc = E(e^-(t/T)) for a discharging capacitor.
You don't need to subtract 3.387 to 15. 3.387 is the value of voltage in the capacitor after 1ms. Just follow the equation Vc = E(e^-(t/T)) for a discharging capacitor.
Pritam said:
6 years ago
TRUE.
After closing switch, the capacitor starts discharging.
Vc=Vs*e^(-t/T).
Vc=15*e^(-10^-3/(.56*10^-6*1.2*10^3))=3.387V.
15-3.387=11.61V.
After closing switch, the capacitor starts discharging.
Vc=Vs*e^(-t/T).
Vc=15*e^(-10^-3/(.56*10^-6*1.2*10^3))=3.387V.
15-3.387=11.61V.
Shambu said:
6 years ago
Thanks for the explanations.
Ranjith Garuda said:
1 decade ago
When the switch is closed the capacitor starts discharging through resistor. The discharge equation is. Vc=Vs*e^ (-t/rc). Substituting for Vc=15v, t=1ms, r=1. 2k & c=0. 56uf. We arrive at Vo=3. 387V.
Here 1ms specified is the elapsed time after the switch is closed, and not the time constant to check with.
Here 1ms specified is the elapsed time after the switch is closed, and not the time constant to check with.
Nagashree said:
1 decade ago
T=RC
So wil get T=0.672ms
1ms=15v
for 0.672ms wil get 10.08v
So wil get T=0.672ms
1ms=15v
for 0.672ms wil get 10.08v
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