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In the circuit shown in figure, the switch is closed at time t = 0 , while the capacitor is initially charged to – 5V [i.e. VC (0) = -5V ]
The time after which the voltage across the capacitor becomes zero is (Round off to 3 decimal places) ______ ms.
Correct answer is '0.1386'. Can you explain this answer?
Most Upvoted Answer
In the circuit shown in figure, the switch is closed at time t = 0 , ...
Given, initial value of voltage across capacitor
VC (0− ) =− 5 V
When switch is closed and circuit is in steady state (i.e. t =∞, and capacitor becomes open circuit) So circuit becomes as,
Apply KCL at node VC (∞) is,
Vc(∞) 5\3 V → Steady state voltage of capacitor 3
Now, for calculating time constant ( τ ), first we calculate Rth as shown below,
Apply KCL at node N,
Thus, time constant, て=Rth ×C
Apply transient equation for voltage across capacitor is,
If VC(t)=0, then equations (i) becomes as,
Hence, the correct answer of t is 0.1386 msec.
Queston_Type: 4
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Community Answer
In the circuit shown in figure, the switch is closed at time t = 0 , ...
Given, initial value of voltage across capacitor
VC (0− ) =− 5 V
When switch is closed and circuit is in steady state (i.e. t =∞, and capacitor becomes open circuit) So circuit becomes as,
Apply KCL at node VC (∞) is,
Vc(∞) 5\3 V → Steady state voltage of capacitor 3
Now, for calculating time constant ( τ ), first we calculate Rth as shown below,
Apply KCL at node N,
Thus, time constant, て=Rth ×C
Apply transient equation for voltage across capacitor is,
If VC(t)=0, then equations (i) becomes as,
Hence, the correct answer of t is 0.1386 msec.
Queston_Type: 4
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In the circuit shown in figure, the switch is closed at time t = 0 , while the capacitor is initially charged to – 5V [i.e. VC (0) = -5V ]The time after which the voltage across the capacitor becomes zero is (Round off to 3 decimal places) ______ ms.Correct answer is '0.1386'. Can you explain this answer?
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