# Two voltage sources

Discussion in 'Homework Help' started by JamesCA, Sep 15, 2013.

1. ### JamesCA Thread Starter New Member

Sep 4, 2013
2
0
I'm seeing myself having this problem once in while when i get lost in circuits with two voltage sources. Lets take the circuit below for example.

How would i find the voltage over V0? There is no mention of refrence point. Also how would i find the the current? Can i draw the circuit in some other way to make it more clearer visually?

2. ### #12 Expert

Nov 30, 2010
16,252
6,750
Sure. Just add 5 volts to each label, work the math, and subtract 5 volts from the answer.

Re-drawing a circuit should be the first trick you learn when working with schematics. After all, "ground" is just a method of convenience.

Jul 18, 2013
10,507
2,367
It would appear that V0 is the +5- 0- -5v supply voltage reference point, or are you intended to find the voltage at V0 when only +5 & -5v is applied?
Not made very clear from the drawing.
Max.

4. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
Sure there's a mention of a reference point. The top node is at +5V relative to the reference point and the bottom is at -5V relative to the reference point.

Let's break the problem down into two pieces:

IF you knew what the value of the current I was, could you find Vo?

If yes, then focus on finding I. If no, then pause at this point and consider the bottom resistor in which you would know the voltage at one end, the value of the resistor, and the current flowing through that resistor.

For the second part, namely finding I, use KVL, which basically says that the voltage difference between two points is independent of the path taken to go between them.

Since the voltage difference, Vab, between any two nodes, 'a' and 'b', is the voltage at 'a', known as Va, minus the voltage at 'b', known as Vb. So

Vab = Va - Vb

If 'a' is your top node and 'b' is your bottom node, what is Vab?

Now, in terms of I, what is the voltage from 'a' to 'b' going through the two resistors and the diode?

What model are you using for your diode (i.e., when it is conducting, what voltage drop are you assuming is across it)?

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