Electrical Engineering - Time Response of Reactive Circuits - Discussion
Discussion Forum : Time Response of Reactive Circuits - General Questions (Q.No. 3)
3.
Referring to Problem 5, how long will it take the capacitor to discharge if the internal resistance of the pulse source is 100 
?
?300 
s
s600 s
s900 s
s1.5 ms
Discussion:
16 comments Page 2 of 2.
Ajit Kumar Yadav DTU said:
5 years ago
Time constant = R*C.
Where, R = 100 ohm.
Where, And C=1.5 uF.
Where, R = 100 ohm.
Where, And C=1.5 uF.
Hafiz aziz said:
4 years ago
How? Please explain the solution in detail.
Habib said:
4 years ago
Please explain the answer in detail.
Sachin Kumar said:
3 years ago
Anyone, explain this in detail.
(1)
Tanzi said:
2 years ago
Kindly Explain with formula.
Anonymous said:
2 years ago
Final capacitor voltage=0 volt, after complete discharge.
Vc(t) = Vc(infinity)+(Vc(initial)-Vc(infinity))e^-t/RC---------eq(1)
Vc(t) = 0, Vc(infinity)=0,Vc(initial)=some volts calculated from question no 5, RC=100*1.5*10^-6.
Put all values in eq(1) and then find t.
Vc(t) = Vc(infinity)+(Vc(initial)-Vc(infinity))e^-t/RC---------eq(1)
Vc(t) = 0, Vc(infinity)=0,Vc(initial)=some volts calculated from question no 5, RC=100*1.5*10^-6.
Put all values in eq(1) and then find t.
(1)
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