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Finding potential at certain points in circuit

  • Thread starter sodper
  • Start date
10
0
[SOLVED] Finding potential at certain points in circuit

1. Homework Statement
Calculate the potential of [tex]V_1[/tex] and [tex]V_2[/tex]

See attached circuit configuration (circuit2.gif)


2. Homework Equations
Ohms law: U = RI
Kirchhoff's laws


3. 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?
 

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Answers and Replies

10
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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?
 
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
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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.
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).
 
10
0
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:
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,474
20
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].
 
10
0
Ok, I see. Thanks for your help! I'll mark this as solved now.
 

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