Solve Currents Homework: Kirchhoff's Laws

[Angie K.]

Homework Statement

Determine the magnitudes of the currents through each resistor in the above circuit. The batteries have emfs of E1 = 9.2 V and E2 = 13.8 V, and the resistors have values of R1 = 11.8Ω, R2 = 29.3Ω, and R3 = 34.3Ω. Use positive numbers for currents to the right, and negative for currents to the left.

Homework Equations

Kirchhoff's Laws (junction rule and loop rule)
V=I

The Attempt at a Solution

I found three equations to figure out one of the currents.
Loop 1 --> 9.2-29.3(I2)-11.8 (I1)=0
Loop 2 --> 29.3(I2)-13.8+34.3=0
Junction Rule --> (I1)+(I2)=(I3)

I chose to solve for I2 first and I got 23/59.6 A
for I1 I got -2.1070 A
for I I got 2.4929 A

Neither of those are correct and I am not sure where I went wrong.

 

[gneill]

Loop 1 --> 9.2-29.3(I2)-11.8 (I1)=0
Loop 2 --> 29.3(I2)-13.8+34.3=0
Junction Rule --> (I1)+(I2)=(I3)

Presumably that 34.3 in the Loop 2 equation is meant to be multiplied by I3?

I suspect that you're not being consistent with your assumed current directions. Given the signs of the terms in your equations it would appear that you are summing potential drops moving counter-clockwise around the loops. That would make your assumed current directions as depicted here:

Now, is your KCL equation (Junction rule) consistent with this?

 

[Hesch]

Loop 1 --> 9.2-29.3(I2)-11.8 (I1)=0
Loop 2 --> 29.3(I2)-13.8+34.3=0

You must draw two loopcurrents, for example a loop, L1, in upper square clock-wise and a loop, L2, in lower square counter-clock-wise:

Equations:

L1: E1 - L1*R1 - L1*R2 - L2*R2 = 0 ( Both L1 and L2 passes R2 )
L2: E2 - L2*R3 - L2*R2 - L1*R2 = 0 ( do. )

Now reduce the equations and solve L1 and L2.

You find I1, I2, I3 by (look at your drawing, and the drawing with red arrows in #2):

I1 = L1
I2 = L1+L2
I3 = L2

 

[gneill]

You must draw two loopcurrents, for example a loop, L1, in upper square clock-wise and a loop, L2, in lower square counter-clock-wise:

I doubt that Angie K. has covered mesh current analysis yet. Most likely they are still using individual branch currents at this point, so bare-bones KVL and KCL with branch currents.

 

[Hesch]

I doubt that Angie K. has covered mesh current analysis yet. Most likely they are still using individual branch currents at this point, so bare-bones KVL and KCL with branch currents.

I don't know. I've tried to look at AK's profile: No information.

But I think that my "version" of KVL is the correct one if you ask Kirchhoff.