Verified answer

Call the left battery V1 and the right battery V2.

Call the upper resistor R1, the middle resistor R2, and the bottom resistor R3.

Call currents through each resistor I1, I2, and I3, corresponding to the resistor with the same label

Define current I1 to be positive when flowing rightward, current I2 to be positive when flowing down, and current I3 to be positive when flowing right.

Use Kirchoff's current branch law about the upper junction of 3 wires, and relate the currents.

I2 = I1 + I3

For KVL, the voltage loop law, pick any loop and trace it in the assumed direction of current.

Voltage sources add voltage, resistors subtract voltage (via V=I*R). It is opposite when going against the direction of current or against the marked direction of the voltage source.

Now use the Kirchoff's voltage loop law for the left loop:

V1 - R1*I1 - R2*I2 = 0

Use Kirchoff's voltage loop law for the right loop:

Note on your drawing: I noticed last minute that the minus terminal of V2 is on top. You really need to draw these more carefully if you want everyone to think it is negative. It is also more clear to draw the four line battery symbol.

Anyways, construct KVL:

-R2*I2 - R3*I3 - V2 = 0

We have three equations and our three unknowns are the currents

Solve system in your favorite way (substitution, addition, elimination, matrix, etc). I prefer to use a software package (Maple) to do this and skip the steps of algebra.

You will end up with the following:

I1 = (R2*V2+R3*V1+R2*V1)/(R2*R1+R3*R1+R3*R2)

I2 = (R3*V1-R1*V2)/(R2*R1+R3*R1+R3*R2)

I3 = -(R1*V2+R2*V2+R2*V1)/(R2*R1+R3*R1+R3*R2)

Plug in data:

R1:=30 ohms; R2:=8 ohms; R3:=30 ohms; V1:=6 V; V2:=4V;

Results (notice that I shifted the decimal 3 places and added the metric prefix "milli", which means "thousandth"):

I1 = 188.4 milliamps

I2 = 43.478 milliamps

I3 = -144.9 milliamps

Since you need to do this on a test, I will show you how to solve by hand:

I2 = I1 + I3

V1 - R1*I1 - R2*I2 = 0

-R2*I2 - R3*I3 - V2 = 0

Substitute the expression for I2 in the other equations:

V1 - R1*I1 - R2*(I1 + I3) = 0

-R2*(I1 + I3) - R3*I3 - V2 = 0

Solve second KVL equation for I3:

-R2*(I1 + I3) - R3*I3 - V2 = 0

-R2*I1 - R2*I3 - R3*I3 - V2 = 0

V2 + R2*I1 = -I3*(R2 + R3)

I3 = -(R2*I1+V2)/(R2+R3)

Substitute in to 1st KVL equation:

V1 - R1*I1 - R2*(I1 + I3) = 0

V1 - R1*I1 - R2*(I1 - (R2*I1+V2)/(R2+R3)) = 0

Expand:

V1 - R1*I1 - R2*I1 + 1/(R2+R3)*R2^2*I1 + R2/(R2+R3)*V2 = 0

Gather terms containing I1 to the left, and those that don't to the right:

-R1*I1 - R2*I1 + R2^2/(R2+R3)*I1 = -V1 - R2/(R2+R3)*V2

Factor left side:

I1*(-R1 - R2 + R2^2/(R2+R3)) = -V1 - R2/(R2+R3)*V2

Isolate I1:

I1 = (-V1 - R2/(R2+R3)*V2)/(-R1 - R2 + R2^2/(R2+R3))

Simplify:

I1 = (V1*R2 + V1*R3 + R2*V2)/(R1*R2 + R1*R3 + R2*R3)

Substitute in to previous expressions to find other currents

Kirchhoff's contemporary regulation, additionally oftentimes going on as Kirchhoff's Junction regulation and Kirchhoff's First regulation, defines the way that electric powered contemporary is allotted while it crosses by way of a junction - a element the place 3 or greater conductors meet. extremely, the regulation states that: The algebraic sum of contemporary into any junction is 0. considering contemporary is the bypass of electrons by way of a conductor, it won't be able to accumulate at a junction, because of the fact of this contemporary is conserved: what's equipped in might desire to pop out. while performing calculations, contemporary flowing into and out of the junction commonly have opposite indicators. this helps Kirchhoff's contemporary regulation to be restated as: The sum of contemporary right into a junction equals the sum of contemporary out of the junction.