Solving Circuit with Three Pathways

[pdeco1]

Homework Statement

This circuit has three pathways, consisting of three parallel horizontal wires, connected by two vertical wires on each end.

Moving from left to right:
The top row has a 3 ohm resistor.

The middle row has a 12V battery with the + end on the left side, followed by a 2 ohm resistor.

The bottom row has a 4 ohm resistor followed by a 12V battery with the - end on the left side.

The left vertical wire has a 2uF capacitor between the middle and top row with the + on the north end.

The question asks to determine the currents and charge on the capacitor.

Homework Equations

V = Q/C
V = IR

The Attempt at a Solution

IF the rows are labeled I(top), I(middle) and I(bottom) then Equation 1: I am = It + Ib

Starting the loop from the left middle intersection and looping through the bottom loop.

-12V - Im2 -12V - Ib4 then reduced

Equation 2: -12V -Im - Ib2

Starting from the same left middle intersection and looping up to the top loop.

-12V - 2Im -3It -Q/2uF

If max Q = CV, then Q = (2uF)(12V) = 24uC

Equation 3 = -24V -2Im - 3It

I am not sure if Equation 3 is correct. As I understand the book, the current through a capacitor is zero due to the capacitor absorbing all of the current, but the voltage is conserved throughout?

If the equations I derived are correct, I am = 84/13, It = 120/13, and Ib = -36/13

Thanks

 

[gneill]

Unless you're trying to evaluate the transient response rather than the steady state, I might suggest a simplification...

At steady state the current in the capacitor will be zero. That leaves you with a single loop (the bottom loop) with current flowing. You've got a pretty simple voltage divider situation that will then tell you both the current and the voltage across the capacitor...