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Response Details:
Okay for this problem. You can first do a thevanin equivalent with the 12Ω resistor and the 8A current source by doing V = IR, V = (8A)(12Ω) = 96V, so therefore the 8A current source becomes a 96V voltage source. Now the 12Ω resistor becomes in parallel with the 40Ω resistor as well as the 96 V voltage source. You can add the two voltage sources (V = 96V and V = 40V) since one voltage source goes one way (+ to - means the current goes up) and the other voltage source goes the other way , they collide into one another; therefore you can subtract the two (V = 96V-40V = 56V). Now, you should have a circuit with a 50V source with the 10Ω resistor, the 40Ω resistor in parallel with the 12Ω resistor and the 56V voltage source together in series with the 20Ω resistor. Now you can set the area between the 40Ω and the 12Ω resistors as V with a ground and use a node voltage equation to find Vx. Using node voltage method, you should get:
(V-50)/(10) + (V)/(40) + (V)/(12) + (V-56)/(20) = 0
V[0.1+0.025+0.083333+0.05] = 7.8
V = 7.8/0.258333 = 30.1936
Now that we have found V, we need to find Ix. Ix will be the point where Vx is located, so it is V/12. So we get
Ix = 30.1936/12 = 2.52A
Finally, we can get Vx by using sign convention: Vx = IxR = (2.52A)(12Ω) = 30.24V
Hope this helps.
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