What is the Voltage and Power in a Circuit with Current Sources?

  • Thread starter Mosaness
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In summary, the voltage v can be determined by simplifying the circuit and applying KVL and KCL. By doing so, the power supplied by each current source can be calculated. However, it is important to note that the assumption of the same current flowing through both resistors is incorrect and KCL should be used to express i1 in terms of ix. Additionally, KVL should be applied to the blue loop in the circuit to obtain the correct value for ix.
  • #1
Mosaness
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1. Determine the voltage v as labeled in Fig. 3.72, and calculate the power supplied by each current source.

Homework Equations



KVL and Ohms Law

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: [itex]\frac{1}{1 + \frac{1}{3}} [/itex]. 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) = -14/3The 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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  • #2
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
 

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  • #3
ehild said:
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

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
Am I on the correct path?
 
  • #5
Mosaness said:
Based on your diagram i1 = -ix.
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
 

1) What is voltage and why is it important to determine?

Voltage is a measure of the electric potential difference between two points in a circuit. It is important to determine because it tells us the strength of the electric field and the amount of energy available to move charges through a circuit.

2) How do you calculate voltage?

Voltage can be calculated by dividing the change in energy (in joules) by the amount of charge (in coulombs) that moves through a circuit.

3) What are the units of voltage?

The units of voltage are volts (V).

4) Can voltage be negative?

Yes, voltage can be negative. This indicates that the direction of the electric field is opposite to the direction of the current flow.

5) How is voltage measured in a circuit?

Voltage can be measured using a voltmeter, which is connected in parallel to the component or circuit being measured.

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