Finding potential at certain points in circuit

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Homework Help Overview

The discussion revolves around calculating the electric potential at specific points in a circuit, identified as V_1 and V_2, based on a provided circuit configuration. The subject area includes circuit analysis, specifically applying Ohm's law and Kirchhoff's laws.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods for determining the potentials, including superposition and mesh current analysis. Questions arise regarding the flow of currents and the assumptions made about their directions. Some participants express uncertainty about their results, particularly concerning the sign of V_2.

Discussion Status

Multiple interpretations of the circuit analysis methods are being explored, with some participants verifying their results against others. Guidance is offered regarding the assumptions of current direction in mesh analysis, and there is an ongoing exchange of ideas about the calculations involved.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the information shared. There is a focus on verifying calculations and understanding the underlying principles without reaching a definitive conclusion.

sodper
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[SOLVED] Finding potential at certain points in circuit

Homework Statement


Calculate the potential of [tex]V_1[/tex] and [tex]V_2[/tex]

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 [tex]V_2[/tex].
No matter how I try, I get a positive potential for [tex]V_2[/tex].

I've attached my solution (solution.jpg)

Any suggestions?
 

Attachments

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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 [tex]V_3[/tex], has the potential -6V.
The voltage over the 15 ohm resistance is [tex]u_3 = V_2 - V_3[/tex].
But we also have
[tex]u_3 = \frac{15 \Omega}{10 \Omega + 15 \Omega + 25 \Omega} 10V = \frac{150}{50}V = 3V[/tex]

And finally
[tex]V_2 = u_3 + V_3 = 3 + (-6) V = -3 V[/tex]

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 [tex]V_3[/tex], has the potential -6V.
The voltage over the 15 ohm resistance is [tex]u_3 = V_2 - V_3[/tex].
But we also have
[tex]u_3 = \frac{15 \Omega}{10 \Omega + 15 \Omega + 25 \Omega} 10V = \frac{150}{50}V = 3V[/tex]

And finally
[tex]V_2 = u_3 + V_3 = 3 + (-6) V = -3 V[/tex]

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 [itex]i_1[/itex] and took it to be clockwise, consistent with the voltage source in that loop. Similarly I called the current in the right mesh [itex]i_2[/itex] 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 [itex]i_1[/itex] and [itex]i_2[/itex]. That's all you need to get [itex]V_1[/itex] and [itex]V_2[/itex].
 
Ok, I see. Thanks for your help! I'll mark this as solved now.
 

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