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:
Best Answer
Good first step.
Knowing U6, you have U7 by inspection - U6 and U7 are identical as these voltage variables are across the same two nodes.
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.
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.