Finding potential at certain points in circuit

AI Thread Summary
The discussion focuses on calculating the potentials V_1 and V_2 in a given circuit configuration using Ohm's law and Kirchhoff's laws. The initial attempt involved superposition, but the user encountered issues with the sign of V_2, arriving at a positive value despite expectations. After further analysis, the user determined V_2 to be -3V, corroborated by another participant who used Mesh Current Analysis to arrive at the same result. The conversation emphasizes the importance of assuming current directions in circuit analysis and the simplicity of using mesh equations for solving such problems. The thread concludes with the user feeling confident in the solution, marking the issue as resolved.
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[SOLVED] Finding potential at certain points in circuit

Homework Statement


Calculate the potential of V_1 and V_2

See attached circuit configuration (circuit2.gif)


Homework Equations


Ohms law: U = RI
Kirchhoff's laws


The Attempt at a Solution


I tried to solve it with superposition but my teacher says I've got the wrong sign for V_2.
No matter how I try, I get a positive potential for V_2.

I've attached my solution (solution.jpg)

Any suggestions?
 

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I think I've found a solution to this problem, but I'm still not convinced about how the currents flow in this circuit configuration.

My solution:
Point 1 obviously has the potential 7V. The node beneath the 15 ohm resistance, which I call V_3, has the potential -6V.
The voltage over the 15 ohm resistance is u_3 = V_2 - V_3.
But we also have
u_3 = \frac{15 \Omega}{10 \Omega + 15 \Omega + 25 \Omega} 10V = \frac{150}{50}V = 3V

And finally
V_2 = u_3 + V_3 = 3 + (-6) V = -3 V

Could somebody verify this for me?
 
sodper said:
I think I've found a solution to this problem, but I'm still not convinced about how the currents flow in this circuit configuration.

You don't need to know a priori what the current directions are. Just assume a direction for the current in each loop. If you're wrong, then you'll pick up a minus sign. No worries.

My solution:
Point 1 obviously has the potential 7V. The node beneath the 15 ohm resistance, which I call V_3, has the potential -6V.
The voltage over the 15 ohm resistance is u_3 = V_2 - V_3.
But we also have
u_3 = \frac{15 \Omega}{10 \Omega + 15 \Omega + 25 \Omega} 10V = \frac{150}{50}V = 3V

And finally
V_2 = u_3 + V_3 = 3 + (-6) V = -3 V

Could somebody verify this for me?

I got -3V too, but I did it by Mesh Current Analysis (which is pretty easy in this problem, as each mesh equation only has one variable).
 
Tom Mattson said:
I got -3V too, but I did it by Mesh Current Analysis (which is pretty easy in this problem, as each mesh equation only has one variable).

I'm curious about how you used mesh current analysis in this problem. Would you mind describing it to me?
 
Last edited:
Assume a direction for each mesh current. I called the current in the left mesh i_1 and took it to be clockwise, consistent with the voltage source in that loop. Similarly I called the current in the right mesh i_2 and took it to be counterclockwise. Then I wrote down mesh equations (that's just KVL for each loop in this case) and solved them for i_1 and i_2. That's all you need to get V_1 and V_2.
 
Ok, I see. Thanks for your help! I'll mark this as solved now.
 
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