# Parallel Circuit 2 batteries, solving for current of resistor

## Homework Statement

For the circuit shown in the figure below, calculate
the current I4 through the 2.18 resistor. As shown,
R1 = 4.20
R2 = 8.12
R3 = 6.10
R4 = 2.18

I1 = 3.16A and E1 = 18V

I couldnt get the picture on here but I will describe it:

It is a parallel circuit, with three lines from top to bottom,
on the top line from left to right is R1 (with I1 pointing right) and R2
on the middle line is R3 and E1 (battery) (from left to right it is + to -)
on the bottom line is E2 (unknown voltage, and is + to - from right to left) and R4 (with I4 pointing left).

V=IR
Kirchoff's rules

## The Attempt at a Solution

I determined 2 loop equations,

for the upper loop I thought that R1 and R2 would have the same current being in series so, i have: -I1R1-I1R2+18V+I3R3=0

and for the lower loop I have -I3R3-18V-I4R4+E2=0

For the first loop equation I solved for I3=3.43A then I thought (just a guess) that each line would add up to the same total voltage in relation to each other line. The upper two lines both came out to 38.9V so I thought I was right in guessing that. So I figured E2+I4R4=38.9V also.

So I used my -I3R3-18V-I4R4+E2=0 equation and substituted the E2 for 38.9V-I4R4 and solved for I4. It came out very close to zero, -.005275A to be exact. Does this seem right?

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gneill
Mentor
Does this picture describe your circuit?

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yes but the positive and negative are switched on e2

gneill
Mentor
yes but the positive and negative are switched on e2
Okay. I was just trying to follow your text description. If you think about it, the current I1 in the top branch sets the voltage between the left and right vertical rails. V1 = I1(R1 + R2).

You should be able to work out the currents in each of the other branches by knowing the voltage that must be across them (V1).

Okay. I was just trying to follow your text description. If you think about it, the current I1 in the top branch sets the voltage between the left and right vertical rails. V1 = I1(R1 + R2).

You should be able to work out the currents in each of the other branches by knowing the voltage that must be across them (V1).
Okay so solving for V1 i get 3.16A(4.20+8.12)=38.9312V

Then I use that and do what I said on the bottom horizontal line, set E2+I4R4=38.9312V and set it so E2=38.9312V-I4R4 and replace that into my bottom loop equation so..

-I3R3-18V-I4R4+38.9312V-I4R4=0

Then I solve for I4 and I get 0 which was not right.

gneill
Mentor
I think that would be E2 - I4R4 = V1. The current I4 causes a right-to-left voltage drop.

But if I may suggest, You can easily solve for I3, the current though R3, because you have the voltage V1 across the branch. You also know that I4 = I1 + I3. So that gives you I4 as a known quantity. Finding E2 should be a piece of cake after that.

Ohhhh that is far simpler! Okay so I solved for I3 and had gotten 3.43A so I3+I1 would just be 3.43A+3.16A= 6.59A do you think this is correct? It makes sense to me because of the junction rule correct? Thank you for all your help btw!

gneill
Mentor
Looks good from where I'm sitting!

Be careful of the polarity of E2 when you go for its value.

Cheers.