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Kirchhoff's Laws and current through resistors

by Vashti
Tags: current, kirchhoff, laws, resistors
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Vashti
#1
Feb25-08, 08:47 PM
P: 4
Question:

The diagram below shows a circuit where; R1 = 5.00 Ω, R2 = 6.00 Ω, R3 = 1.00 Ω, V1 = 4.500 V, V2 = 20.00 V, and V3 = 6.00 V. (In solving the problems that follow, initially pick the current directions as shown. If the actual current turns out to be in the opposite direction, then your answer will be negative).
What is the value of I1? I2? I3?

Attempts:

So I know that I need three equations to solve it as I have three unknowns. I tried to make one using the junction where I2 meets I1 and I3, but they all come together and kind of crash, so I don't know how to put that into equation form any more :( I'm assuming I need to do loops to get the other two equations but again I run into the issue of knowing how to set them up. If I try to make points on either side of each resistor they won't 'flow' really....
In summary I am just very confused, so any help is welcome!

Thanks!
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Astronuc
#2
Feb26-08, 06:59 AM
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P: 21,870
There are two loop equations for the voltages around each loop, and there is one current equation.

Sum if currents (net) into a node is zero, so I1+I2+I3=0, because what goes in must come out, otherwise there is a net charge accumulation. If all current arrows go into a node, then one must be negative, i.e. the current is opposite the arrow.


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