A capacitor of capacitance C has charge Q. it is connected to an identical capacitor through a resistance. The heat produced in the resistance is

Options

A. `(Q^2)/(2C)`
B. `(Q^2)/(4C)`
C. `(Q^2)/(8C)`
D. dependent on the value of the resistance

Answer

B

Explanation

In the steady state, no current is passing throughl the capacitor.
Let the charge on each capacitor be q. Since the current through galvanometer is zero,
`I_(1) = I_(2)`
The potential difference between the ends of the galvnometer
will be zero. Therfore
`V_(A) -V_(B) V_(A) - V_(D)`
`I_(1)R_(2) = (q)/(C_(1)) (i)`
Similarly, `V_(B) - V_(C) = V_(D) - V_(C)`
`I_(2)R_(2) = (q)/(C_(2)) (ii)`
Dividing Eq. (i) by Eq. (ii) we get
`(I_(1)R_(1))/(I_(2)R_(2)) = (q//C_(1))/(q//C_(2)) = (C_(2))/(C_(1)) or (C_(1))/(C_(2)) = (R_(2))/(R_(1))`