# Currents - RC circuit

## 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,

$$q \ = \ CE(1-e^{-t/\tau}) + Q_oe^{-t/\tau}$$

differentiating it will give current,

$$i \ = \ \frac{Ee^{-t/\tau}}{R} \ - \ \frac{Q_oe^{-t/\tau}}{\tau}$$

putting values give 6

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SammyS
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Is the 1Ω shorting wire temporary? -- that is to say: is it removed before the switch is closed?

No ... but the question asks use for t=0 ... i.e. when capacitor behaves as a wire ... so it will short circuit the resistor ... :)

SammyS
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Yup, at t=0, capacitor A has zero potential across it.

At time t=0, what's the potential across capacitor B ?

SammyS
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Sorry for the delay. The problem is written in an unfamiliar way (at least for me).

I also came up with 6 amps @ t=0.

so that means that the answer in book is wrong?

SammyS
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