Voltage and Current in RC circuits

[stonecoldgen]

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$\Omega$ resistor. Then comes a junction, one way leads to a 400$\mu$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=$\tau$

Homework Equations

V=IR
V=V₀(1-e^-t/$\tau$)

The Attempt at a Solution

First I found the constant $\tau$ for the circuit by doing (2500)(400X10^-6)=1

Then found V at time $\tau$ by doing

V=80(1-e^{-$\tau$/$\tau$})=51




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

 

[gneill]

The voltage across the resistor is not the voltage across the capacitor.

HINT: You can modify your voltage formula slightly to give the voltage across the resistor.

 

[stonecoldgen]

The voltage across the resistor is not the voltage across the capacitor.

HINT: You can modify your voltage formula slightly to give the voltage across the resistor.

should i subtract/add the voltages?

 

[gneill]

should i subtract/add the voltages?

That's one way to do it; you know one end of the resistor is held at 80V and the other is determine by the capacitor voltage for which you've written an expression.

On the other hand, you should recall that all the voltages and currents in the circuit will be changing according to very similar exponential curves. The current through the resistor is going to start at some maximum value when the capacitor is initially empty, and then it will decay towards zero as the capacitor fills. So you should be able to write by inspection a function for the current.

 

[asteriatic]

I would like to clarify a few things about this circuit before providing a response. Firstly, I assume that the spark gap is not connected to the rest of the circuit, as that would create a short circuit and the capacitor would not discharge to the ground. Secondly, the value of \tau that you have calculated (1) seems to be a time constant, not a resistance. The resistance in this circuit is 2500\Omega.

With that being said, let's focus on the main question: finding the current through the resistor at time t=\tau. To do this, we can use Ohm's law, V=IR, where V is the voltage across the resistor, I is the current through the resistor, and R is the resistance. We know that the voltage across the resistor is equal to the voltage of the power supply, which is 80V. So, we can rearrange the equation to solve for I, giving us I=V/R. Plugging in the values, we get I=80V/2500\Omega, which simplifies to I=0.032A. This is the current through the resistor at time t=\tau.

I'm not quite sure what you meant by dividing 51 by 2500, but I hope this explanation helps. It's important to remember that the spark gap does not affect the current flowing through the resistor, as it is not connected to the rest of the circuit. The current through the resistor is determined by the voltage of the power supply and the resistance of the resistor.