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The question in a study book says to find the current through R2. I have very little experience dealing with dependent current and voltage sources, so I get thrown off when performing loop current (mesh) analysis. Here's the work so far, but the unknown of the voltage I1 makes too many unknowns.

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I then attempted to do node analysis thinking that I could find the nodes between R2/R3 and R1/R2 then do a simple Ohm's calculation (Va-Vb/R2 = I2). I've searched the forums for similar questions, which there definitely are...but I'm not understanding how to do this since the current source is not expressed in terms of another branch current or some voltage coefficient.

I appreciate any help steering me in the right direction.

  • 3
    \$\begingroup\$ There are no dependent sources here; only fixed. Simple KCL at Node A contains one (solvable) unknown variable, which is VA. Which should then simplify your analysis. \$\endgroup\$
    – pat
    Commented Mar 3, 2019 at 4:50
  • \$\begingroup\$ Use a supermesh. \$\endgroup\$
    – Mr. Snrub
    Commented Mar 3, 2019 at 5:52

3 Answers 3


The first thing you should notice is the \$R_1\$ resistor is connected directly in parallel with \$V_{S2}\$ voltage source.

All this means that we already know the \$I_A\$ current value.

$$I_A = \frac{V_{S2}}{R_1} = \frac{12V}{6k\Omega} = 2mA $$

And this current will have no effect on the remaining part of the circuit because the voltage at node \$B\$ is fixed and it is equal to \$12V\$.

Therefore your circuit will now look like this: enter image description here

Loop B KVL

$$V_{S1} - I_B R_2 - (I_B + I_C)R_3 - V_{S2} - (I_B + I_C)R_6 = 0$$

The KVL equation for loop C is not needed because we have a current source in this

loop hence \$I_C = I_1 = 1mA\$

Solving this will give us the answer:

$$I_B = I_{R2} = -4.4mA$$

And this minus sign tells us the \$I_B\$ current is flowing in the opposite direction to the direction I have marked on the diagram.

We can do the nodal analysis also.

We know that the voltage at node B is equal to 12V. Hence the nodal equation for node A can look like this:

$$ \frac{V_A - V_{S1}}{R_2} + \frac{V_A - V_{S2}}{R_3 + R_6} - I_1 = 0$$

And the solution is $$V_A = 20.8V$$

And we are done.

  • \$\begingroup\$ Very much appreciated. I guess I was way over thinking it! I was stuck thinking I needed IA to get back to finding IB for the mesh method. When I did the node analysis, I came up with (Vs2-Va)/(R3+R6) + I1 - (Va-Vs1)/R2 = 0. I must have been doing some math wrong because my result was incorrect (I checked and Multisim showed 4.4mA and 20.8V). The way you did it definitely helps me understand how chosing the current directions (except for the given I1) is arbitrary. Thanks again. \$\endgroup\$
    – T. Paul
    Commented Mar 3, 2019 at 15:41

The circuit simplifies to the following schematic. Note what I did on the left side. I swapped the relative positions of \$V_2\$ and \$R_6\$ (which doesn't affect the current in \$R_2\$.) That's it for the left side below:


simulate this circuit – Schematic created using CircuitLab

On the right side, I threw away \$R_1\$ and \$R_4\$ since they do not matter at all when figuring the current in \$R_2\$. The current in those two resistors are entirely determined and you should be able to see why their current doesn't impact the current in \$R_2\$.

The right side is the simplified version of the remaining important part of the left side schematic. Here, you should be able to construct the Thevenin equivalent for the portion within the dashed box on the right side. Knowing \$V_\text{TH}\$ and \$R_\text{TH}\$ (representing that box), you can work out the node voltage quite easily. Knowing the node voltage, you can work out the current in \$R_2\$.

Given that enough time has gone by and G36 already provided an answer, I'll follow up now.

The Thevenin voltage is \$V_\text{TH}=\frac{12\:\text{V}\cdot\left(R_3+R_6=8\:\text{k}\Omega\right)+48\:\text{V}\cdot\left(R_2=2\:\text{k}\Omega\right) }{\left(R_2=2\:\text{k}\Omega\right)+\left(R_3+R_6=8\:\text{k}\Omega\right)}=19.2\:\text{V}\$ and the Thevenin resistance is \$R_\text{TH}=\frac{\left(R_2=2\:\text{k}\Omega\right)\cdot\left(R_3+R_6=8\:\text{k}\Omega\right)}{\left(R_2=2\:\text{k}\Omega\right)+\left(R_3+R_6=8\:\text{k}\Omega\right)}=1.6\:\text{k}\Omega\$.

The circuit now reduces to:


simulate this circuit

It's now obvious that \$V_\text{X}=19.2\:\text{V} + 1\:\text{mA}\cdot 1.6\:\text{k}\Omega= 20.8\:\text{V}\$. Now you know the voltages on both sides of \$R_2\$: \$12\:\text{V}\$ on one side and \$20.8\:\text{V}\$ on the other side. The magnitude of the current is then obviously just \$\frac{20.8\:\text{V}-12\:\text{V}}{R_2=2\:\text{k}\Omega}=4.4\:\text{mA}\$.

You get to decide the sign, depending upon what the problem question is asking of you.

  • \$\begingroup\$ This really, REALLY helps me understand the circuit logic! Seeing how the circuit simplifies to the image on the right, I would've had no problems solving the current in R2 using a KCL and Ohm's law substitutions. I'm not understanding how Thevenizing the resistors in the dashed box would help. Wouldn't it simplify down to an equivalent circuit based on Norton's theorem though? R2||(R3+R6) would be in parallel with the current source I1? But I still don't understand how we could find the voltage between R2 and (R3+R6) if it were simplified to an equivalent circuit. \$\endgroup\$
    – T. Paul
    Commented Mar 3, 2019 at 15:47
  • \$\begingroup\$ @T.Paul With a simple voltage on one side, \$V_\text{TH}\$, and the thevenin resistance with one milliamp through it, it is very easy to work out the voltage across that resistance. With the drop added to the thevenin voltage, you have the node voltage. And with that, you can look back at the original circuit and compute the current that G36 gave you. Try it. You'll see. \$\endgroup\$
    – jonk
    Commented Mar 3, 2019 at 16:05

No time to draw nice schematics. Repeatedly applying Norton/Thévenin equivalents reduces the circuit to a single mesh. Here's the outline.

First get rid of R1, R4, as jonk said. They are in parallel to a voltage source and so they don't contribute at all.

Then simplify (merge) R3 with R6 (they are in series, even if V2 is in-between), so you get V2 in series with an \$8k\Omega\$ resistance.

This branch can be now converted to a Norton equivalent, i.e. an \$8\,k\Omega\$ resistance in parallel with a current source (polarity upwards) of \$ \frac{48V}{8\,k\Omega} = 6\,mA \$, both in parallel with I1.

Since they are in parallel, you can merge that \$ 6\,mA \$ current source with I1 and get a single current source of 7mA, still in parallel with that \$ 8\,k\Omega \$ resistance.

Now convert that source+resistance back into into a Thévenin equivalent, getting an \$ 8\,k\Omega \$ resistance in series with a \$ 8\,k\Omega \cdot 7\,mA = 56\,V \$ voltage source.

Now the whole circuit has been reduced to a single mesh:

V1, R2, an 8k resistor and a voltage source of 56V (polarity going against V1 in the mesh). Apply direct KVL and you get for the unknown current Ix (assumed flowing leftwards):

$$ Ix = \frac{56\,V - V1} {R2 + 8\,k\Omega} = \frac{44\,V}{10\,k\Omega} = 4.4\,mA $$

Extremely simple and no complicated equations, only back-of-the-envelope calculations! Easy peasy!


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