- #1
cupid.callin
- 1,132
- 1
Homework Statement
The Attempt at a Solution
As at t=0 the capacitor behaves as a wire ... so the circuit at t=0 will be like:
q at any time for this circuit will be,
[tex]q \ = \ CE(1-e^{-t/\tau}) + Q_oe^{-t/\tau}[/tex]
differentiating it will give current,
[tex]i \ = \ \frac{Ee^{-t/\tau}}{R} \ - \ \frac{Q_oe^{-t/\tau}}{\tau}[/tex]
putting values give 6
but answer is 7