A network of four capacitors,each capacitance 30PF is connected across a baterry of 60V shown in figure. find the net capacitance and energy stored capictor
Answers
Answer:
Here C
1
,C
2
,C
3
are in series and the equivalent resistance for these is C
123
=30/3=10μF
Now C
123
,C
4
are in parallel. So the net capacitance C
net
=C
123
+C
4
=10+30=40μF
Now potential across C
123
and C
4
is same i.e, 60V
So energy stored in C
4
is U
4
=
2
1
C
4
V
2
=0.5×30×10
−6
×60
2
=0.054J=54mJ
As C
1
,C
2
,C
3
are in series so they have same charge is equal to Q
123
=C
123
V=10×60=600μC
Stored energy, U
1
=
2C
1
Q
123
2
=
2×30×10
−6
(600×10
−6
)
2
=6mJ
As C
1
,C
2
,C
3
have same charge and same capacitance so they store same energy.
Thus, U
1
=U
2
=U
3
=6mJ
Explanation: