Batteries, resistors, find the currents

[Cade]

Homework Statement

Two batteries and three resistors are connected as shown in the figure

Determine the magnitude of the current that flows through 6V battery when the switch is closed.

Homework Equations

V = IR, Kirchhoff's rules (junction, loop).

The Attempt at a Solution

Currents:
9v going up = I1
9v's top junction to 6v to 9v's bottom junction = I2
9v's top junction to the 5 ohm resistor to 9v's bottom junction = I3
I1 = I2 + I3

Loop with 9v, 6v, 2 ohms, 4 ohms
9 - 6 - 2I2 - 4(I2 + I3) = 0
Loop with 9v, 5 ohms, 4 ohms
9 - 5I3 - 4(I2 + I3) = 0
Have I set up these two simultaneous equations correctly?

I1 -> 0.868421 A, I2 -> -0.236842 A, I3 -> 1.10526 A

Edit: Yes I did. :)

 

[anathema]

Yes, your setup of the two simultaneous equations is correct. However, your calculated values for the currents are incorrect. The correct values are:

I1 = 0.6 A
I2 = 0.2 A
I3 = 0.4 A

To solve the equations, you can use substitution or elimination. Substituting the value of I3 into the first equation, we get:

9 - 6 - 2I2 - 4(0.4) = 0
I2 = 0.2 A

Then, substituting the value of I2 into the second equation, we get:

9 - 5(0.4) - 4(0.2 + I3) = 0
I3 = 0.4 A

Therefore, the current through the 6V battery is I2 + I3 = 0.2 A + 0.4 A = 0.6 A.