(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I have the circuit that's attached as a .bmp image. I'm asked to find the initial current that passes trough the battery (V=60 V) knowing that the capacitor is initially charged with charge q_{0}=0.1 C. The resistances are R_1=R_2=R_3=1 Ohm and C=1/1000 F.

2. Relevant equations

The relevant equations are given by Kirchoff's Laws:

[tex]

\oint \vec{E}\cdot d\vec{l}=0

[/tex]

[tex]

I_3=I_1+I_2

[/tex]

3. The attempt at a solution

My attempt to the solution was to use Kirchoffs first law. Making two closed loops: One passing trough V, then resistance R_1, C and finally trough resistance R_3 and the other loop passing trough V, resistance R_2, and finally resistance R_3. With this we get the two equations:

[tex]

V-R_{1}I_{1}-q_{0}/C-R_{3}I_{3}=0

[/tex]

[tex]

V-R_{2}I_{2}-R_{3}I_{3}=0

[/tex]

From here we get:

[tex]

I_{1}=\frac{V-q_{0}/C-R_{3}I_{3}}{R_{1}}

[/tex]

[tex]

I_{2}=\frac{V-R_{3}I_{3}}{R_{2}}

[/tex]

And we finally use Kirchoffs second law, I_{1}+I_{2}=I_{3} and get:

[tex]

I_3=\frac{V-q_{0}/C-R_{3}I_{3}}{R_{1}}+\frac{V-R_{3}I_{3}}{R_{2}}

[/tex]

And you finally solve for I_3, wich gives I_3=6.7 Amperes. My question is that if you replace the data on I_1 you get I_1=-46.7 A. I'm unsure about the current going in the opposite way of what i tought. Anyways, I tought it would have some phyical meaning like all current flowing into the capacitor, but i'm not sure. What do you think?

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# Question about a paralell-like RC circuit

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