Question

A capacitor of capacitance `C` has potential difference `E/2` and another capacitor of capacitance C is uncharged. They are joined to form a closed circuit as shown in the figure.

a. Find the current in the circuit at t =0.
b. Find the charge on C as a function of time.

Answer 1

Correct Answer - B::C
a. At `t=0,` capacitor is equivalent to a battery of emf `E/2`
Net emf of the circuit `=E-E/2=E/2`
Total resistance is R
Therefore, current in the circuit at `t=0`
would be `i=(E/2)/R=E/(2R)`
b. Let in steady state there is total `q` charge on `C` initial charge on C was `CE/2`. Therefore `(CE)/(2-q)` with polarities as shown. This is because net charge on lower late of `C` and of upper plate on `2C` should remain constant. Applying looplaw in the circuit in steady state, we have.
`E-q/C+(CE/2-q)/(2C)=0`
`:. q=5/6CE`

Therefore charge of `C` increases from `q_i=(CE)/2 to q_f=(5CE)/6` exponentially.
Equivalent time constant would be
`tau_C=((Cxx2C)/(C+2C))R=2/3CR`
Therefore charge as function of time would be
`q=q_i+(q_f-q_i)1-e^(-t/tau_C))`
`(CE)/2+(CE)/3(1-e^(-(3t)/(2CR)))`