KCET 2018 Physics Solved Paper

If Q is charge on capacitor and C is capacitance then, potential difference V=

Q

C

When two capacitors are connected in series, then equivalent capacitance is
‌

1

C

=‌

1

C1

+‌

1

C2

=‌

1

3µF

+‌

1

6µF

⇒C=‌

6×3

3+6

=2µF
Now, Q = C × V
Given, V = 900 V
Therefore, Q=2×900=1800μC
When two capacitors are connected in parallel, then equivalent capacitance is
Therefore V=‌

Q

C

=‌

1800µC

9µF

=200V
Thus, potential difference across the combination is 200 V