# 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.

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?

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

• Please show us what you have done so far. Commented Oct 20, 2019 at 16:13
• Welcome to the site. I'm sure you know that this is not a free design house, homework-answering service or an on-line technical encyclopedia, copied out to users on demand. People will help you take the next step if your question shows you've done as much as you possibly could on your own - which yours doesn't, I'm afraid. Please edit your question and greatly improve it. Show your work and findings so far in considerable detail. The better the quality of question, the better the quality of the answers you will attract. Again, a warm welcome. Commented Oct 20, 2019 at 16:15
• With multiple sources a good choice might be using superposition. Commented Oct 20, 2019 at 16:50
• Your mesh equations are not correct. Ignore $I_S$ and focus on $I_1$ and $I_2$. Don't use labels of both $I_2$ and $i_2$ for different currents. Clearly show us your equations for both of the meshes. Commented Oct 20, 2019 at 18:06
• You can do this one without mesh analysis (but it is good practice to do it anyway.) You know you have 6V at the output thus 6V over 3K which gives you the current though the 3K. From there you can work your way to Vs. Commented Oct 20, 2019 at 19:02

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

simulate this circuit – Schematic created using CircuitLab

For $$\\text{R}_\text{th}\$$, we get:

simulate this circuit

So:

$$\text{R}_\text{th}=\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}\tag1$$

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

simulate this circuit

So:

$$\text{I}_\text{th}=\text{I}_{\text{o}_1}+\text{I}_{\text{o}_2}=\frac{\text{V}_\text{in}}{\text{R}_1}+\text{I}_2\tag2$$

So, we also know:

$$\text{V}_\text{th}=\text{I}_\text{th}\cdot\text{R}_\text{th}=\left(\frac{\text{V}_\text{in}}{\text{R}_1}+\text{I}_2\right)\cdot\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}\tag3$$

Using the given values, gives:

$$6=\left(\frac{\text{V}_\text{in}}{3000}+3\cdot10^{-3}\right)\cdot\frac{3000\cdot3000}{3000+3000}\space\Longleftrightarrow\space\text{V}_\text{in}=3\space\text{V}\tag4$$

The simple way- using superposition. Make the supply 0V and calculate the contribution from the current source:

From the 3mA source we have 3mA * (3K || 3K) = 4.5V

Now we need 1.5V more so open the current source and the output voltage is Vs/2 = 1.5V, so Vs must be 3V.

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.

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.