# Where has this equation come from?

Discussion in 'Homework Help' started by StuckStudent123, Oct 9, 2016.

1. ### StuckStudent123 Thread Starter New Member

Sep 23, 2016
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I wonder if anyone can tell me why there is Vo=V1-6v in the solution on the 6th line in the PDF attached.
How come it doesn't include the 12k resistor and how come it's minus and not add?
Thanks

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
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The answer is simple. They first write a nodal equation for node 1 (V1). And solve for V1 voltage.
And the voltage at the node 1 (V1) is 6V. But they need Vo voltage, the voltage across 12kΩ resistor.
The voltage on the left side of a 12kΩ resistor is V1 = 6V and one the right side the voltage is also equal to 6V (right voltage source).
Therefore the voltage across 12kΩ resistor is a potential differences between V1 and the right voltage source ---->Vo = V1 - Vsource = 6V - 6V = 0V

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3. ### ErnieM AAC Fanatic!

Apr 24, 2011
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V1 is not indicated very precisely, it is between the lower Node junction (-) and the upper node junction (+). If you follow the loop to the right you have two elements, The 12K marked as having Vo across it and a 6v source.

By inspection of the voltage loop there are three voltages where:
V1 = Vo + 6V

Solve that for Vo and you get that negative 6V. To "see" why mark where V1 is in the circuit and follow around the loop: you see you enter V1 thru the + terminal and the 6V thru the negative, so one is added and the other subtracted.

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