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Homework Statement
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.
Homework 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]
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, which 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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