Electronic Circuits - Calculate voltage

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To calculate voltage v1 using Kirchhoff's Current Law (KCL), it's essential to express the current I3 correctly. The discussion highlights the need to apply both KCL and Kirchhoff's Voltage Law (loop rule) to derive the necessary equations. The equation I1 + I2 = I3 must be complemented by the loop rule to solve for v1 accurately. The final equation derived from the loop rule is v - I1R - v1 - v1 = 0, leading to a comprehensive understanding of the circuit. Mastering these principles is crucial for solving similar electronic circuit problems effectively.
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TL;DR Summary: Electronic Circuits

How do I calculate the voltage v1 using KCL?

1711693746026.png

I don't understand how I should express the current I3 to get the equation right.

1711693762988.png



I already watched this video but I don't know what to do when there are two resistors.
 
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This is the equation I should be getting but I don't know how to get there
1711695014513.png
 
The answer they give has the form of, I3 + I1 - I2 = 0 ;
Your I1 has the opposite circulation to their assumption.
Vx = 2 * V1 ;
I1 = ( 2 * V1 - V) / R ;
I2 = α * V1 ;
I3 = V1 / R .
 
Normally for HW type problems, we don't like to just give answers without some teaching.
So now that we have @Baluncore's suggestions, I suggest you look at each of his steps and compare to the schematic to see why/how he concluded each item. For example, why does Vx = 2 * V1, etc? Then see how those combine to give the equation you were supposed to derive.

Let us know if you are still confused. Do you think you could solve a similar problem yourself now?
 
agatha said:
This is the equation I should be getting but I don't know how to get there
1711695014513.png
It is impossible to solve this problem using only two rules, the junction rule expressed by $$ I_1 + I_2 = I_3 $$ and the rule expressed by $$ v_1 = I_3 R $$ which is called Ohm’s law. Here must be applied one more rule, the loop rule (https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws). The equation $$ v - I_1 R - v_1 - v_1 = 0 $$ presents the loop rule for the left loop in the schematic diagram and after including it the final equation $$ \frac{v_1}{R} + \frac{2v_1-v}{R} - \alpha v_1 = 0 $$ will be got.
 
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