Potential difference between two points (VA - VB)

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SUMMARY

The voltage difference between points A and B (VA - VB) in the given circuit is calculated to be 6.018 V. The analysis involves applying Kirchhoff's Junction Rule and Loop Rule to determine the currents I1 and I2, which are found to be 0.161 A and 0.222 A, respectively. The calculations for VA and VB incorporate the resistances of 15Ω and 22Ω, leading to the final voltage difference. The solution confirms the accuracy of the approach taken to derive the voltage difference.

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Homework Statement


Q. With the switch closed, what is the voltage difference, VA-VB ?

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Homework Equations


Junction rule: I3 = I1 + I2
Loop rule(s)

The Attempt at a Solution


I solved for the currents in this circuit when the switch is closed.

Top loop(clockwise): +10V1 - 15*I1 - 47*I1 + (0V since I'm assuming no Voltage drop across switch) = 0
Bottom loop(counter clockwise): -5*I2 + 6V2 - 22*I2 + 0V = 0

For the top loop: I isolated I1 and got 10/62 = 0.161 = I1
For the bottom loop: I got 6/27 = 0.222A = I2
I1 + I2 = 0.161 + 0.222 = 0.383A.= I3

for the voltage difference VA - VB, I tried doing this with the guidance of a friend.
Va = 10 - I1*15 = 7.585 V.
Vb = Va - Vc - 6 (Vc being the voltage at node C where I3 flows through.)
Then Vc = Va - i1*47 = 7.585 - (7.567) = 0.018 V.
Vb = Va - Vc - 6 = 7.585 - 0.018 - 6 = 1.567 V.
Finally Va - Vb = 7.585 - 1.567 = 6.018 V

I am not sure whether I did this part correct.
 
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