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Branch current analysis (Part 3 where I get scared)

Physics
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In previous notes, I have introduced Ohm's law, and its generalizations KVL and KCL. I will now show you a simple example where parallel-series analysis will not apply. Example problem 1.
Determine the currents in i_1, i_2, and i_3. Remember to do this with reference to the direction indicated by the arrows. It should be quite evident that the resistors are not connected in parallel or series - there are two power sources. We shall apply Kirchoff's laws to determine the answer regardless. There are a number of ways to do this, but a straightfoward way uses KCL, KVL, and Ohm's law- you'll find that you can formulate systems of equations with regards to the current of the junctions. Note that although you might not know the current's themselves, you can express them using Ohm's law, I=V/R. Step 1: Choose a reference node to start the problem from. Let's use node 1. Step 2: Label the current paths to and from this reference node (I_1,I_2,I_3)
Step 3: Formulate your first equation using KCL. In this case it would be: I_1-I_2+I_3=0 Step 4: Formulate subsequent equations using KVL around closed loops. The twist to this is that we use Ohm's law to express the unknown voltage drops across resistors (remember that R1_v=6*I_1). Thus we have on the left side -12+6*I_1+4*I_2=0, because -B1_v+R1_r*I_1+R2_r*I_2=0 We also have 6-4*I_2-2*I_3=0 We then have a system of 3 equations in 3 variables. I_1-I_2+I_3=0 -6-4*I_2-2*I_3=0 -12+6*I_1+4*I_2=0 Solving for this yields I_1=24/11 I_2=-3/11 I_3=-27/11 The negative sign means that the current is flowing in the direction opposite to the arrow indicated. Thus, we have calculated the currents flowing in and out of Node 1. Further application of Ohm's law allows one to obtain values for voltage drops and currents.

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