Two capacitors of `3muF and 6 muF` are connected in series and a potential difference of 900 V is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the combination is

Options

A. zero
B. 100 V
C. 200 V
D. 400 V

Answer

D

Explanation

Let `V_ 1 and V_2` be the voltages across `3muF and 6 mF` capacitors respectively as shown in the figure . Then
`V_1 +V_2 = 900 V " "...(i)`
As `3muF and 6 muF` capacitors are series , charges on each is the same.
`:.C_1V_1=C_2V_2`
or `V_1/V_2=C_2/C_1=(6muF)/(3muF)=2`
or `V_1 = 2V_2" "...(ii)`
Substituting this value of `V_1` in eqn. (i) , we get
`2V_2+V_2=900V`
or `V_2=(900V)/3 = 300 V`
From eqn. (i)
`V_1 = 900 V - V_2 = 900 V - 300 V = 600 V`
Now they are disconnected and reconnected in parallel.
Let the potential difference across the combination be V. Then
`V=(C_1V_1+C_2V_2)/(C_1+C_2)`
`=((3muF)(600V)+(6muF)(300V))/(3muF+6muF)`
`=3600/9 V = 400 V`