# Determine the voltage V

1. Sep 9, 2012

### Mosaness

1. Determine the voltage v as labeled in Fig. 3.72, and calculate the power supplied by each current source.

2. Relevant equations

KVL and Ohms Law

3. The attempt at a solution

The first step is to try and simplify the circuit so that it gives a resistance of 0.75Ω, which is obtained by combining the two resistors that are in parallel: $\frac{1}{1 + \frac{1}{3}}$. We thus now have a current dependent current source, 3ix, a current source of 2A and a resistor of 0.75Ω, with the total current flowing counterclock-wise.

Using this, KCL can be applied:

-3ix - 2 = 0 so that ix = -2/3

V is then:

(1)(-2/3) + (1)(-3)(-2/3) + (-2)(3) = -14/3

The power supplied by each current source then should be:

P(3ix) = (14/3)(2/3) = 28/9

P(2A) = (14/3)(-2) = -28/9

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• ###### Figure 3.72.png
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Last edited: Sep 9, 2012
2. Sep 10, 2012

### ehild

You assumed that the same current flows through both resistors which is not true. Use KCL for the node N to express i1 in terms of ix, and KVL for the blue loop in the attached picture.

ehild

#### Attached Files:

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3. Sep 10, 2012

### Mosaness

Based on your diagram i1 = -ix.

As for the inner loops,

Do we combine the two resistors? They are in parallel after all and should give 0.75 ohms no?

And if we know that,

Then by KVL,

0.75(-3ix) -1.50 = 0 and we can solve for ix?

4. Sep 10, 2012

### Mosaness

Am I on the correct path?

5. Sep 10, 2012

### ehild

NO, they are different. The loop is not a physical loop, only a closed path in the circuit, to get the sum of potential differences, which should be zero. How does the potential change across the 3 Ω and the 1 Ω resistors?

ehild