Sorry, no picture my bad.(adsbygoogle = window.adsbygoogle || []).push({});

Here is the circuit, everything is in PARALLEL,

First its a 1mA source, then its a switch, then its a resistor of 10k-ohm then its a capacitor 1uF. Sorta like below:

----------------------------------

| ...........| ............. |............. |

O.........Switch ........ R ............ C

|.............| .............|.............|

----------------------------------

initial conditions, t=0, switch is CLOSED.

I said that the closed switch shorts out the rest of the circuit, so the inital current/voltage across the resistor is zero.

Then when its opened, I said that its like a current source, in parallel with the resistor and the capacitor. But the current source and the resistor is the norton circuit with a load of a capactior. So I changed it to a thevenin eq. circuit, by using Vt=InRt = 10V

Now my circuit is a 10V source, in series with a 10k-ohm resistor, and a load of the capacitor, and this is charging the circuit

Then the voltage across the resistor as a function of time, (For the inital circuit!) is the same as that of the capacitor,

Vr(t)= 10-10e^(-t/tau)

Yes, no? Maybe so?

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# Homework Help: RC circuit

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