# Solving a linear circuit

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1. Jan 19, 2017

### diredragon

1. The problem statement, all variables and given/known data
This is one of the problem i had had some help in solving but some things remain unclear to me and i will highlight them by posting the exact question in the brackets next to the line of solution so that you can see what bothers me. Hopefully you can help me better understand this. Thanks :)

2. Relevant equations
3. The attempt at a solution

$P_{R4}=0$ the power of the element is zero so the current through that element must be zero. Basically that branch is cut of. Conclusion : $I_4=0$
Now the part i cant seem to grasp the idea of:

$RT=\frac{R2R1}{R2+R1}$ [I understand thevenin's circuit equvalent but how is it used here?]
$ET=\frac{R2E1}{R1+R2}$ [If the branch with R4 doesnt exit how can we find the thevenin's if R2 isn't next to E1?]
$I3=\frac{-ET}{RT+R3}$ [And why isn't E5 and R5 included, when we look for ET where do we cut the circuit here? I thought since I3 is left we cut on that branch and leave open circuit but then E5 must be included? Could you explain?]
Here is the rest of the solution:
$U_{ab}=I3R3=-I2R2+I4R4$
$I2=\frac{-U_{ab}}{R2}$
$U_{ab}=-I_2(R2+R5)+E_5$
and to get E5 is easy from here.
The only problem with this problem is in the brackets above. :)

2. Jan 19, 2017

### Staff: Mentor

The tricky bit is to realize that since $I4 = 0$ there is no potential drop across $R4$. That means that the right end of $R2$ must have the same potential as node A. So the effect of $E5$ and $R5$ must be to maintain this zero potential difference (reminiscent of the way an op-amp with feedback maintains a zero potential difference between its inputs).

3. Jan 19, 2017

### diredragon

I get that I4 is zero but cant understand why when they decided to turn everything into thevenin to calculate the I3, didn't take into consideration the part of the circuit with R5, and just took R2.
Its like they erased the branch with R4 and short circuited the E5 branch. Could they have done that? Why could the branch with E5 be short circuited? I dont understand that.
What comes to mind is that since there is no potential drop in the R4 branch then there is none across E5 as well. But still lacks fundamental understanding. One path leads us to conclude that there is no potential drop so all the others must as well and that branch during the thevenin transformation is only a wire, short circuited?

4. Jan 19, 2017

### Staff: Mentor

Since I4 is zero the potential difference across R4 is zero. That means you can replace R4 with a wire (short circuit) without changing how the rest of the circuit operates. So as you say, they erased R4 and short circuited the E5 branch!