A variable capacitor is adjusted to its lowest capacitance `C_(0)` and is connected with a source of constant voltage `V` for a long time. The resistance of connecting wires is `R`. At `t = 0`, its capacitance starts to increase so that a constant current `I` starts to flow through the circuit. Calculate at time `t`,
(i) Power supplied by the source
(ii) thermal power generated in the connecting wire
(iii) rate of increase of electrostatic energy stored in capacitor
(iv) What do you infer from the above three results?

Explanation

Since voltage `V` of the source is constant and the circuit draws constant current `I` from it , power supplied by the source is `P = VI`.
(ii) Thermal power generated in connecting wires is `H= I^(2) R`.
(iii) Since the initial capacities of the capacitor was equal to `C_(0)` and it was connectws with the source for long time , initial charge on capacitor was `q_(0) = C_(0)V`. ,brgt Since a constant current `I` starts to flow at `t = 0`, at time t, charge on capacitor becomes `q = (C_(0)V + It)` . At time `t` , the circuit will be as shown in `Fig. 7.22` . Potential difference across the capacitor is
` V_( c ) = V_(A) - V_(B) = (V - IR) rarr constant`
Therefore , electrostatic energy in capacitor at this instant is
`U = (1)/(2) q V_(c)`
Rate of increases of electrostatic energy is
`(dU)/(dt) = (1)/(2) V_(c ) (dq)/( dt) = (1)/(2) ( V - IR)I`
`= (1)/(2) (VI - I^(2)R)`
But power acting across the capacitor at this instant is
`P_(0) = P - H = (VI - I^(2) R )`
while the rate of increase of electrostatic energy in the capacitor is half of it.
(iv) In fact, a force of attraction exists between the plates of the capacitor . When these surfaces move toward each other , capacitance increases. Hence , the remaining part of the paper acting across the capacitor is used to increase kinetic energy of surface ( plate) of the capacitor.