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