Question

13. Energy stored in a system consisting of two capacitors in series and connected across
4kV line is 8J. When the same two capacitors are in parallel across the same line,
energy stored is 36J. Find the capacitance of the capacitors. ​

Answer 1

Explanation:

Let the capacitance of two capacitors be C

1

and C

2

.

Line voltage V=4000 volts

Parallel connection :

Equivalent capacitance of series connection C

p

=C

1

+C

2

Energy stored E

p

=

2

1

C

p

V

2

∴ 36=

2

1

(C

1

+C

2

)(4000)

2

⟹ C

1

+C

2

=4.5μF

We get C

2

=4.5μF−C

1

Series connection :

Equivalent capacitance of series connection C

s

=

C

1

+C

2

C

1

C

2

=

4.5

C

1

C

2

μF

Energy stored E

s

=

2

1

C

s

V

2

∴ 8=

2

1

(

4.5

C

1

C

2

)(4000)

2

⟹ C

1

C

2

=4.5μF

Or C

1

(4.5−C

1

)=4.5

Or C

1

2

−4.5C

1

+4.5=0

Solving we get C

1

=1.5μF

⟹ C

2

=4.5−1.5=3.0μF

hope this can help you

Answer 2

Explanation:

hope it's helps you...