Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 20)
20.
After a capacitor has charged for 1 tc, what percentage of current remains in the resistor?
Discussion:
17 comments Page 1 of 2.
Rakhee said:
1 decade ago
Can any one explain this.
Ankita said:
1 decade ago
Please explain.
Nitin said:
1 decade ago
The time required for a capacitor to rise from zero to 1-1/e (that is, 63.2%) of its final steady value when it varies with time t as 1 - e-kt. The time required for a capacitor to fall to 1/e (that is, 36.8%) of its initial value when it varies with time t as e-kt. Generally, the time required for an instrument to indicate a given percentage of the final reading resulting from an input signal. Also known as lag coefficient.
Hanish said:
1 decade ago
I can't understand. Please explain me clearly.
Banalata said:
1 decade ago
@nitin please explain clearly.
Rajni ranjan said:
1 decade ago
Please explain it clearly. I can't understand.
Sidduuuu said:
1 decade ago
Capacitor charged 62.8% discharge at 32.2%.
Shireen said:
1 decade ago
Capacitor charges upto 63. 2% of its final value in one tc. So 36.8% remaining. Hence current remaining in resistor is 36.8%.
Jagat said:
1 decade ago
100% - 63.2% = 36.8 %..........
very simple
very simple
Pujitha said:
1 decade ago
If capacitor charges by 1tc(time constant).
Vc=V(1-e^-t/rc).
Here, t = 1 rc.
=> Vc=V(1-e^-1).
=> Vc=63.6.
Therefore,% of current remained in resistor = 100 - 63.6.
= 36.8%.
Vc=V(1-e^-t/rc).
Here, t = 1 rc.
=> Vc=V(1-e^-1).
=> Vc=63.6.
Therefore,% of current remained in resistor = 100 - 63.6.
= 36.8%.
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