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a. find the 3 currents in the figure
b. find Vab
Given: R1=1ohm R2=2ohms E1=2v E2=E3=4v

Hello ryan,

There are different ways to solve this. My preferred method is with Thevenin voltage loops. For my purposes, I need to have 2 currents that both flow through the path from a to b. So I edited your drawing showing the currents I1 and I2. As drawn then, the current through R2 is the difference between I1 and I2. When working with the I1 path, the current through R2 is I1-I2 and when working with the I2 path, the current through R2 is I2-I1.

Starting from point c, writing the loop equation for I1, we have

E1 -I1R1 - I1R2 + I2R2 - E2 - I1R1 = 0
Now I will put in the voltage values.
2 -I1R1 - I1R2 + I2R2 - 4 - I1R1 = 0
Note that the last equation starts with a 2 without units. Since this is a voltage loop, assume the unit Volts to be understood. The other terms in the equation are currents multiplied by resistance, which would be voltage. Now I will substitute 2R1 for R2.
2 -I1R1 - 2I1R1 + 2I2R1 - 4 - I1R1 = 0
Next several steps, rearranging and solving for I1
4I1R1 = -2 + 2I2R1
I1 = (-2 + 2I2R1)/(4R1) = (-1 + I2R1)/(2R1)

Some discussion about the signs on voltage drops across resistors in these equations, mainly because I made a few sign errors before I got a result that stood up to verification: I circulate around the loop in the direction I defined for the current that goes around the loop. Starting at c and circulating clockwise in this analysis, the first circuit element is E1. That is an increase in voltage. If the current, I1, actually circulates in the direction of the arrows I drew, the voltage across a resistor, R1, is a decrease in voltage, so I give it a negative sign. Notice R2, I2 flows through it in the direction opposite that of I1, so I2R2 gets a + sign in the above equations.

Starting from point b, writing the loop equation for I2, we have
E2 + I1R2 - I2R2 - I2R1 - E3 - I2R1 = 0
-2I2R1 + I1R2 - I2R2 = 0
Now I am plugging in the expression for I1 above.
-2I2R1 + [(-1 + I2R1)/(2R1)]R2 - I2R2 = 0
-2I2R1 + [(-1 + I2R1)/(2R1)]2R1 - 2I2R1 = 0
-2I2R1 + (-1 + I2R1) - 2I2R1 = 0
I2(-2R1 + +R1 - 2R1) = 1
I2 = 1 / (-3R1)
Since R1 is 1 Ohm,
I2 = -1/3 A

Going back to the expression for I1
I1 = (-1 + I2R1)/(2R1) = -1 V/(2R1) + (I2/2) = -1/2 A - 1/6 A = -0.667 A

a. So, you need i1, i2, and i3. Your variable names for i1 and i3 are equivalent to my I1 and I2. Your i2 is my I1-I2.

b. Vab = -i2R2 - 4 V = ____ V

I did this with paper and pencil -- and eraser. In writing it up I may have an error. Check my work.

I hope this helps,


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Steve Johnson


I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.


I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

BS Physics, North Dakota State University
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