Solving a circuit with Kirchoff/Ohms Law

Question

Good evening,

I haven't had physics since year 7 and now I need to use elementary things in university. Since I lack a lot of basics I am now trying my best to fill these holes. Currently I am stuck at solving circuits and I hope one of you can help me with the following circuit. I've also attached an image at the bottom of the post (large image).

Given:
$R_1 = 20\Omega$
$R_2 = 100\Omega$
$R_3 = 50\Omega$
$R_6 = 100\Omega$
$U_0 = 2V$
$U_5 = 900mV$
$U_6 = 1V$
$I_5 = 10mA$
$I_7 = 80mA$

What needs to be calculated: $U_3, I_4, R_5, R_8, U_8, I_8$

My attempt on solving the circuit:

1) Since $U_5$ and $I_5$ are given we can use Ohm's law to calcualte $R_5 = 90\Omega$.
2) Knowing $U_6$ we can use the Mesh-Current Method to determine $U_7 = 1V$.
3) Again I used Ohm's law to calculate $I_7 = 80mA$ with $U_7 = 1V$ from 2)
4) Ohm's law to determine $I_6 = 10mA$ since we have $U_6, R_6$ given.
5) Now I used the Mesh-Current Method between $U_1, U_5 and U_6$. Since $U_1, U_2, U_3, U_4$ are parallel to each other I assumed that they must be equal (? not quite sure on this one). However this assumption lead to the following result for $U_1, U_2, U_3, U_4 = 100mV$.
6) Knowing $R_1, R_2, R_3$ and now $U_1, U_2, U_3$ I calculated $I_1 = 5mA, I_2 = 1mA, I_3 = 2mA$ with Ohm's law. 7) Now this is the point where I am stuck at. I'm not sure if it is allowed to reduce the circuit to only a specific area. For Instance I reduced the circuit to the upper left corner to determine $I_4 = 2mA$ with the Branch-Current Method. Having that given allows me to easily calculate $R_4 = 50\Omega$ with the help of Ohm's law. 8) Knowing that $U_0 = 2V$ I used the Mesh-Current Method for the whole outer part of the circuit to calculate $U_8 = 1V$. 9) Using the Branch-Current Method I determined the value of $I_8 = 110mV$ 10) Now I used Ohm's law to get the value of $R_8 = 9.09091\Omega$.

$U_3 = 100mV, I_4 = 2mA, R_5 = 90\Omega, R_8 = 9.09091\Omega, U_8 = 1V, I_8 = 110mV$

I somehow cannot attach any images yet so I hope it is okay if I post an outisde link:

https://www.dropbox.com/s/bfj64pkm0kg36ht/01.jpg (link to the circuit, large file, same as shown below)

Answer 1

1) Since U5 and I5 are given we can use Ohm's law to calcualte R5=90Ω.

Good first step.

2) Knowing U6 we can use the Mesh-Current Method to determine U7=1V.

Knowing U6, you have U7 by inspection - U6 and U7 are identical as these voltage variables are across the same two nodes.

3) Again I used Ohm's law to calculate I7=80mA with U7=1V from 2)

I7 is given as 80mA. No calculation required. What you should do here instead is calculate R7 = 1V / 80mA = 12.5 ohms if you need it.

4) Ohm's law to determine I6=10mA since we have U6,R6 given.

Correct.

Now, you know I8 since, by KCL, I8 = I5 + I6 + I7 = 100mA.

And now the dominoes topple. By KVL, U0 = U6 + U8 -> U8 = 1V.

The value of R8 is given by Ohm's Law: R8 = 1V / 100mA = 10 ohms.

Again, by KVL, U0 = U1 + U5 + U8 -> U1 = 0.1V

Now, since U1 = U2 = U3 = U4, by inspection, we know I1, I2, and I3 using Ohm's law.

By KCL, I5 = I1 + I2 + I3 + I4. Solve for I4 to get I4 = 2mA -> R4 = 50 ohms.