Circuit Analysis : Using Kirchoffs

In summary, the given circuit has a 6V source and the solution for Vo was calculated to be 428.57V using KCL and KVL equations. However, the answer given by the professor was -50V, which may be incorrect.
  • #1
gatsbycollege
19
0

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)
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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 ?
 
  • #5
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?
 
  • #6
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:
 
  • #7
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?).
 
  • #8
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:
 

FAQ: Circuit Analysis : Using Kirchoffs

What is Kirchoff's Circuit Analysis?

Kirchoff's Circuit Analysis is a method used to analyze electrical circuits by applying two laws: Kirchoff's Voltage Law (KVL) and Kirchoff's Current Law (KCL). These laws help to calculate the current and voltage at different points in a circuit, and can be used to solve complex circuit problems.

What is Kirchoff's Voltage Law (KVL)?

Kirchoff's Voltage Law states that the sum of all voltages in a closed loop in a circuit must equal zero. This means that the voltage drops in a series circuit must add up to the voltage source, and in a parallel circuit, the voltage drops must be equal to the voltage source.

What is Kirchoff's Current Law (KCL)?

Kirchoff's Current Law states that the total current entering a node (or junction) in a circuit must equal the total current leaving the node. In other words, the amount of current flowing into a point must be equal to the amount of current flowing out of the point.

How do I apply Kirchoff's Circuit Analysis?

To apply Kirchoff's Circuit Analysis, you must first draw a circuit diagram and label all the components and their values. Then, apply KVL and KCL to write equations for the different loops and nodes in the circuit. Finally, solve the equations simultaneously to find the unknown currents and voltages.

What are some common applications of Kirchoff's Circuit Analysis?

Kirchoff's Circuit Analysis is commonly used in the design and analysis of electrical circuits, such as in electronic devices, power systems, and communication networks. It is also useful in troubleshooting circuits and predicting the behavior of complex circuit systems.

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