Question

2
(c) Find accumulated charges on both the
capacitors in the given circuit.
4uF 6uF
HE H ㅔ
B
A
522
422
www
192
WWW
E = 20 volt​

Answer 1

Answer:

32μJ

The capacitance 4 and (3,5,1) are in parallel so potential across (3,5,1) will be V=24V.

The equivalent capacitance of (3,5,1) is C

eq

=

3+(5+1)

3(5+1)

=2μF

and Q

eq

=C

eq

V=2×24=48μC

As 3 and (5,1) are in series so charge on 3 and (5,1) is equal to Q

eq

.

Thus potential across (5,1) is V

51

=

C

51

Q

eq

=

5+1

48

=8V

As 5 and 1 are in parallel so potential difference of capacitors will be same i.e 8V

Now energy stored in 1 is =

2

1

C

1

V

51

2

=

2

1

×10

−6

×8

2

=32μJ

Explanation:

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