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## Homework Statement

A circuit includes a spark gap that connects to the ground. The width of the gap is adjusted so that the capacitor discharges to the ground each time the voltage reaches 99% of the maximum.

(The circuit is basically a simple series circuit with an 80v power supply, connected to a 2500[itex]\Omega[/itex] resistor. Then comes a junction, one way leads to a 400[itex]\mu[/itex]F capacitor, and the loop starts again in the power supply. The other way leads to the grounded spark gap).

Find the current flowing through the resistor at time t=[itex]\tau[/itex]

## Homework Equations

V=IR

V=V

_{0}(1-e

^{-t/[itex]\tau[/itex]})

## The Attempt at a Solution

First I found the constant [itex]\tau[/itex] for the circuit by doing (2500)(400X10

^{-6})=1

Then found V at time [itex]\tau[/itex] by doing

V=80(1-e

^{-[itex]\tau[/itex]/[itex]\tau[/itex]})=51

Then I thought that simply dividing 51 by 2500 would do it, but apparently not, any ideas?

Thanks...