Circuit Analysis : Using Kirchoffs

AI Thread Summary
The discussion revolves around solving a circuit analysis problem using Kirchhoff's laws to find the output voltage Vo. The initial confusion stemmed from a misinterpretation of the voltage source, which was corrected from 50V to 6V. After recalculating, the participant arrived at Vo = 428.57V but questioned the accuracy of the professor's given answer of -50V. Other contributors confirmed that the calculations appeared correct for the circuit presented, suggesting a possible error in the professor's response. The conversation highlights the importance of verifying circuit parameters and calculations in circuit analysis.
gatsbycollege
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Homework Statement


http://img202.imageshack.us/img202/939/95788682.jpg
Find Vo
Vo = -50V (given answer by my prof)

Homework Equations


KVL
(Series) V=V1+V2
(Parallel) V=V1=V2
KCL
Current toward the node = Current Away from the node
V=IR

The Attempt at a Solution



i tried to solve for the i in the left ckt
i=(6v+0.004Vo)/2k <-- 6v is in series with 0.004Vo (i think)

by KCL at the right
25i - (Vo/40k) - (Vo/10k) = 0 <--- 40k and 10k is in parallel so Voltage would be equal
the substitute terms

25((6v+0.004Vo)/2k) - (Vo/40k) - (Vo/10k) = 0
Vo = 1kV <--- actually is not right XD
so there it is,, what did i miss?

Edit : (50v change to 6v)
 
Last edited by a moderator:
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Where does the 6V come from in your equations? The diagram shows a 50V source.

Also, the potential across the 2k resistor appears to be 50V - 0.004Vo. "Walk" around the loop in the direction of the current and it goes: 50V - i*2k - 0.004Vo = 0.
 
gneill said:
Where does the 6V come from in your equations? The diagram shows a 50V source.

Also, the potential across the 2k resistor appears to be 50V - 0.004Vo. "Walk" around the loop in the direction of the current and it goes: 50V - i*2k - 0.004Vo = 0.
Aw.. my mistake,, 50v at the left should be 6v Sorry :-p

Edit: So there it is i changed the circuit.
 
by referring to what you said sir,

i= (6v-0.004Vo)/ 2k

BY KCL
then 25(i)-Vo/40k - Vo/10k = 0
then substitute
25((6v-0.004Vo)/2l)) -Vo/40k -Vo/10k = 0
Therefore

Vo = 428.57V

Hmm :confused: still not the right answer ?
 
gatsbycollege said:
by referring to what you said sir,

i= (6v-0.004Vo)/ 2k

BY KCL
then 25(i)-Vo/40k - Vo/10k = 0
then substitute
25((6v-0.004Vo)/2l)) -Vo/40k -Vo/10k = 0
Therefore

Vo = 428.57V

Hmm :confused: still not the right answer ?

Or perhaps it's the right answer to the wrong problem :smile:

Your calculations appear to be sound for the circuit shown. Can you verify that the circuit matches the problem given?
 
gneill said:
Or perhaps it's the right answer to the wrong problem :smile:

Your calculations appear to be sound for the circuit shown. Can you verify that the circuit matches the problem given?

i double check the given here in my copy.. the circuit is right.. but not so sure about my computation :-p

so what do you think.. maybe the answer given by my prof is wrong? Vo = -50V :confused:
 
gatsbycollege said:
i double check the given here in my copy.. the circuit is right.. but not so sure about my computation :-p

so what do you think.. maybe the answer given by my prof is wrong? Vo = -50V :confused:

As I said, your computation looks correct for the circuit given. It is possible that your Prof was incorrect (or was misinterpreted?).
 
gneill said:
As I said, your computation looks correct for the circuit given. It is possible that your Prof was incorrect (or was misinterpreted?).

Ahm ok :approve: Thx for Helping me :smile:
 

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