# Calculate the current

I was practicing for the exam suddenly I found this question the professor solved, but I think whether he or I made a mistake.

Part (III) in the image. His calculation is on the image he added the two current claiming that it is I of the source which is not true for me. What I have done is adding them together subtracting the value of the current moving in the resistor with Vx getting 26.5 but I realized that the voltage across the resistor with 4ohms is (-)....(+) So the current moving also to the node which is no true due to KCL.

Believe me it's not a homework but I'm solving this problem for practicing that is why I want to understand.

• The + and - signs on the 4 ohm resistor define the polarity of Vx that is used by the dependent current source to define the magnitude and direction of its current. It does NOT necessarily mean that this is the actual polarity of the voltage across the 4 ohm. – Chu Oct 19 '19 at 0:01
• I know that but there is 2v on the left and 8v on the right which means so the current should move to the left(out of the node ) but there is a problem the node has no entries @Chu – Mhd Ghd Oct 19 '19 at 8:35
• So $\small V_x=-6\:V$. What do you mean by 'no entries'? Are the 2V and 8V given, or have you calculated these and written them in? It's difficult to see what was printed in the original question. – Chu Oct 19 '19 at 17:46

Your professor's $$\V_x =-6V\$$ calculation is right. But $$\ Is \$$ is wrong.

simulate this circuit – Schematic created using CircuitLab

What you have to do first is - To mark current directions as per the sign conventions assumed for voltage drops on resistors. These directions may or may not be the actual directions of the currents, which will be clear after calculations based on the assumptions made.

Apply Kirchoff's current law at A: $$4V_x + I_s +I_{R3} = I_{R2}$$ $$4V_x + I_s +\frac{V_x}{4} = \frac{8}{2}$$

On solving, $$\I_s = 29.5A\$$

$$\I_{R3}\$$ is obtained as -1.5A. The meaning of this is:

Current through 4ohm resistor, $$\I_{R3}\$$ has a magnitude 1.5A. However its direction is actually in the opposite direction to what we assumed during calculations.