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## Homework Statement

I am looking for the voltage of each resistor as well as the current passing through each one.

## Homework Equations

Kirchoff's Voltage/Current Laws: The sum of any potential differences across a closed path is zero; current (charge) is conserved, so any current exiting a node is equal to the sum of the currents entering it.

[tex]V = IR[/tex]

[tex]I = \frac VR[/tex]

## The Attempt at a Solution

I know [tex]R_{eq}=339\Omega[/tex]. Thus, [tex]I = \frac VR = \frac{43}{339} = .127 A[/tex].

[tex]V_5 = I*R_5=.127*246=31.2 V[/tex]

[tex]V_1+V_2+V_3-V_4=0[/tex]. This is Kirchoff's voltage equation.

[tex]I_1=I_2=I_3[/tex] b/c current is conserved, so [tex]\frac{V_1}{70}=\frac{V_2}{144}=\frac{V_3}{237}[/tex]. I solve for [tex]V_1[/tex] and [tex]V_2[/tex] in terms of [tex]V_3[/tex]: [tex]V_1=\frac{70V_3}{237}[/tex] and [tex]V_2=\frac{144V_3}{237}[/tex].

I know the current through [tex]V_3[/tex] and [tex]V_4[/tex] should sum to .127, so: [tex]\frac{V_3}{237}+\frac{V_4}{117}=.127[/tex].

[tex]V_4=117\left(.127-\frac{V_3}{237}\right)[/tex].

[tex]V_4=V_1+V_2+V_3[/tex] from above, so plugging in everything I get:

[tex]117\left(.127-\frac{V_3}{237}\right)=\frac{70V_3}{237}+\frac{144V_3}{237}+V_3[/tex].

I solve for [tex]V_3[/tex] and then plug that in to everything else.

My final answers are: [tex]V_1 = 1.83[/tex], [tex]V_2=3.77[/tex], [tex]V_3=6.2[/tex], [tex]V_4=12.8[/tex], [tex]V_5=31.2[/tex]. However, these answers are incorrect, so I was hoping to find out what I'm doing wrong.