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Let I1 be the current through R6, V3, and R7 Let I2 be the current through R5, and V2 Let I3 be the current through R4

Suppose I did not know V2 = 36.4V, but I know the voltage across R7 is 15V.

How do I figure out V2 ?

Here is my KCL =

I2 = I1 + I3

KVL

30I1+25 + 50I1-V2 + 20I2 = 0

-20I2 + V2 -75I3 = 0

When I solve the above 3 equations I get V2 = 25, not 36.4. What am I doing wrong?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I1 is not the same for R6, V3 and R7. The sum of currents for V3 and R7 is equal the current through R6. \$\endgroup\$ – vini_i Mar 25 '16 at 1:16
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Your equations are wrong. The \$V_3\$ branch is not considered since the switch is open. If the switch was open, you would then have 3 loops in total, and \$I_1\$ would be split.

Following your assumptions about the currents:

$$I_2 = I_1 + I_3$$

\$I_1\$ can be gotten right away since you somehow know that the voltage across \$R_7\$ (top to bottom) is \$15 V \$.

$$I_1 = 15/50 = 0.3 A$$ $$I_2 = 0.3 + I_3 ............(1)$$

The equation for the first loop is:

$$-20I_2 - V_2 - 75I_3 = 0$$ $$20I_2 + V_2 = -75I_3..........(2)$$

The equation for the second loop is:

$$80I_1 + V_2 + 20I_2 = 0$$ $$24 + V_2 + 20I_2 = 0$$ $$20I_2 + V_2 = -24............(3)$$

Equating (2) and (3) gives:

$$-24 = -75I_3$$ $$I_3 = 0.32 A$$

From (1):

$$I_2 = 0.3 + 0.32 = 0.62 A$$

From (3):

$$V_2 = -24 - 20I_2 = -24 - 12.4 = -36.4 V$$

Meaning, the polarity of \$V_2\$ in your schematic is the other way round.

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