Resistive Circuits Homework

Thread Starter

sqrt(Arthur)

Joined Oct 7, 2020
4
Hi guys!!

I´m an electrical engineering student taking my first circuits subject!! I´m super hyped but kinda stucked at this problem.


By KVL on left side i got that I_1 = 10 A , and by the right side i get to V_3+Vi=12V
and using KCL that I_2+I_3= 15 A
Since yesterday i got stucked on this, tried to use R3*(15-I2)+Vi=12 but yet still unconclusive.

I feel like i`m running in circles with this problem, always finding new equations although they are linear dependent asasd.PNG

Answer is 0.8 Ω btw
 

WBahn

Joined Mar 31, 2012
26,320
How do you get I_2 + I_3 = 15 A ?????

Look at that junction. You have I1 going into it and I2 and I3 coming out of it. What does KCL tell you about the relationship between the currents going into a node and the currents coming out of that node?
 

Thread Starter

sqrt(Arthur)

Joined Oct 7, 2020
4
How do you get I_2 + I_3 = 15 A ?????

Look at that junction. You have I1 going into it and I2 and I3 coming out of it. What does KCL tell you about the relationship between the currents going into a node and the currents coming out of that node?
I got it due to the fact that the current that flows by R3 is (I3-I)A and I is 5
 

WBahn

Joined Mar 31, 2012
26,320
I got it due to the fact that the current that flows by R3 is (I3-I)A and I is 5
Look at the diagram and ask if this makes sense?

Let's say that I3 turned out to be 8 A. Would you really claim that 8 coulombs of charge is flowing from out of the right side of R3 every second while 5 coulombs of charge is flowing up out of the current source ever second?

Isn't that dumping a total of 13 coloumbs of charge on that wire every second? Is that reasonable?

By definition, I3 is the current that flows through R3 (from left to right). The only way that the current flowing through R3 can be (I3-I) is if I is identically zero.
 

WBahn

Joined Mar 31, 2012
26,320
Finally understand it, much appreciated.

But i´m still in doubt, why can we ignore R3, I3, V_I?
You're not ignoring I3. Because the current source is in series with I3, the current source forces the value of I3 to be consistent with the current output of the source. By the nature of what an ideal current source does, by definition, V_I will be whatever it takes in order to push 5 A into the right-hand terminal of R3. Change R3 and V_I will change to whatever it takes to push 5 A into the right-hand terminal of R3. As far as the circuit in the left hand mesh is concerned, it might as well be connected direction to the 5 A source because it sees no difference. Now, if you were to want to know how much power is dissipated in R3 or how much is supplied (or absorbed) by the source, that's a different matter.
 
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