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What's the value of the supply voltage in this circuit?

I’ve been trying to find out the answer to this question. I know that the final answer is 3V but I can’t seem to prove that.

I always get confused when there’s two or more supplies on the circuit.

enter image description here

I’ve already tried the kirchhoff laws but I’ve never reached to the same result.
Can someone give me a hand and let me know if I at least pointed the directions of the current properly? And tell me the best approach that I should try next?

Thank you!

enter image description here

EDIT: Here are my tries in the meshes and nodes equations. Ignoring Is I am not sure what the I2 equation would be.

enter image description here

Electrical Engineering Asked by RockLife on December 31, 2020

2 Answers

2 Answers

Mesh Analysis

By sensible inspection of the circuit you can say that there must be 2mA through the bottom 3k resistor to produce the output voltage of 6V. That leaves 1mA through the top resistor as 3mA - 2mA = 1mA. 1mA through the top resistor drops 3V from the 6V output voltage leaving 3V as Vs.

Here's a bit more complicated example I created last night. See if you can follow how the mesh equations were derived.

Mesh analysis example

It doesn't matter which way around you draw the mesh current arrows. They can be either way and can even be in different directions to each other. It is OK to step around the loop in either direction to create the mesh equations, the important thing is that you get the signs correct. Voltage rises are positive, voltage drops are negative in sign

Ignore the down marker. Some people become unpleasant when they think you shouldn't be helped so much and they see help being provided.

Answered by James on December 31, 2020

Well, using Thevenin and Norton we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

For $text{R}_text{th}$, we get:

schematic

simulate this circuit

So:

$$text{R}_text{th}=frac{text{R}_1text{R}_2}{text{R}_1+text{R}_2}tag1$$

And for $text{I}_text{th}$, we get:

schematic

simulate this circuit

So:

$$text{I}_text{th}=text{I}_{text{o}_1}+text{I}_{text{o}_2}=frac{text{V}_text{in}}{text{R}_1}+text{I}_2tag2$$

So, we also know:

$$text{V}_text{th}=text{I}_text{th}cdottext{R}_text{th}=left(frac{text{V}_text{in}}{text{R}_1}+text{I}_2right)cdotfrac{text{R}_1text{R}_2}{text{R}_1+text{R}_2}tag3$$

Using the given values, gives:

$$6=left(frac{text{V}_text{in}}{3000}+3cdot10^{-3}right)cdotfrac{3000cdot3000}{3000+3000}spaceLongleftrightarrowspacetext{V}_text{in}=3spacetext{V}tag4$$

Answered by Jan on December 31, 2020

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