A capacitor of capacitance `2 mu F` is initially connected to a battery of emf `10` volt and steady state is reached. Now, `10 V` battery is removed and another battery of emf `20 V` is connected with like polarities together. Find the amount of heat energy developed (after the steady state is reached).

Options

A. `10^(-4) J`
B. `2 xx 10^(-4) J`
C. `3 xx 10^(-4) J`
D. Zero

Answer

A

Explanation

When `10` volt battery is connected,
`q_(1)=20muC U_(1)=(1)/(2)xx2xx10^(-6)xx100 =10^(-4)J`
When `20 V` battery is connected, `q_(2) = 40 muC`
`U_(2)=(1)/(2)xx2xx10^(-6) xx 400 =4 xx10^(-4)J`
`Delta q=40-20 =20 muC` W.D by the battery `= 20 xx Delta q=20 xx 20 xx10^(-6) =4 xx 10^(-4)J`
Heat developred `= U_(1) -U_(2) = W.D` by battery `=10^(-4)-4 xx10^(-4) +4 xx10^(-4) =10^(-4) JAm`.