Question

three of the three microfarad capacitor are included in the category then the capacitor capacitance will be​

Answer 1

Explanation:

Given : C

1

=3μF C

2

=6μF C

3

=6μF

Equivalent capacitance for series combination

C

eq

1

=

C

1

1

+

C

2

1

+

C

3

1

C

eq

1

=

3

1

+

6

1

+

6

1

⟹C

eq

=1.5μF

Thus the charge flowing through the circuit q=C

eq

V

q=1.5μF×120 volt =180μC

Potential difference across 3μF, V

1

=

C

1

q

=

3μF

180μC

=60 volt