In the following circuit all capacitors have
same capacitance and all are neutral except
capacitor 1, which has charge on it. When
switch is closed then in steady state it has
charge of 4pc then what was initial charge on
it (in uc)?
HE
HE
1
Answers
Answered by
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Answer:
Charge on capacitor at some time t is Q=Q
0
[1−e
RC
−t
]
⇒0.75Q
0
=Q
0
[1−e
RC
−t
]
⇒0.25=e
RC
−t
(i)
Q
0
=CV
0
=20μC
At this point of time, current in the circuit I=
dt
dQ
=
RC
Q
0
e
RC
−t
⇒1=
R
20
e
RC
−t
(ii)
Using (i) and(ii)
1=
R
20
(0.25)
R=5Ω
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