How can I use nodal analysis to solve for V in this circuit?

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The discussion focuses on using nodal analysis to solve for voltage V in a circuit. The initial attempt involved setting up equations based on Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL), but the user struggled with the calculations. After several iterations and corrections, the user arrived at the equations I1 - I2 = 0.02A and 240I1 + 100I2 = 9V. Ultimately, the user determined the currents I1 and I2, leading to the calculation of voltage across R3 as 4.06V. The conversation highlights the importance of correctly applying KCL and KVL in circuit analysis.
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Homework Statement


solve for voltage V with polarity as shown.
1ethjm.jpg



Homework Equations


KCL - current into a node = current out


The Attempt at a Solution


I tried to do nodal analysis using the top center node as v.

v/240 + (v-9)/100 - 20mA - v/100 = 0

=> v=26.4V

Obviously, I'm doing something wrong. Any tips to get me back on the right track?
 
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I think current leaving the node is positive (+20mA) and the current through the middle resistor must be the same as that current source since it's the same wire
 
The top node is not at V.

The current through the middle leg is constrained to 20 ma however, and you know that the currents coming in the top must sum to 20 ma.

I1 + I2 = 20 ma.

Hence construct your Voltage loops.

I think that will resolve into the current I2 that you need to determine the voltage across R3.
 
I'm confused again. How do I solve for the i1 and i2?

sum of the voltages in a loop = 0, so for loop 1 with current i1 on the eft:

v1 + v2 = 0
i1*r1 + i1*r2 = 0
i1*240 + i1*100 = 0

loop 2 with current i2 on right:

v2 + v3 = 0
i2*r2 + i2*r3 = 0
i2*100 + i2*100 = 0


I know I'm not calculating the loops correctly, what am I doing wrong?
 
240Ω*I1 = 20ma*100Ω

9v + 100Ω*I2 = 20ma*100Ω

and as before

I1 + I2 = 20ma
 
What's going wrong now?

240Ω*I1 = 20ma*100Ω

==> I1 = (20mA*100Ω)/240Ω
I1 = 8.33mA

9v + 100Ω*I2 = 20ma*100Ω

==> I2 = [(20mA*100Ω)-9]/100Ω
I2 = -70mA


I1 + I2 =/= 20
 
Here we go, I think I got it.

the two equations:
I1 - I2 = .02A (KCL top center node)
240I1 + 100I2 = 9V (KVL outer loop)

=> I1 = 20.59mA
=> I2 = 40.59mA

=> VR3 = I2 * R3 = 40.59 * 100 = 4.06V
 
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