# Solving a circuit with Kirchoff/Ohms Law

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)

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Minor correction: the symbol for the battery is shown the wrong way around. The higher potential (+ve) terminal is the longer of the two parallel lines & must correspond with the arrow 'head' when indicating the polarity of $U_{0}$. – theo Nov 21 '12 at 2:37

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.

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So I can substitute $I_1, I_2, I_3, I_4$ through $I_5$ because they are parallel to each other? I accidently calculated $I_5$ twice then in order to recieve the value for $I_8$ which then results in a rather unpleasent value for $R_8$. Is the Voltage in a parallel serie always the same for each branch? Thank you very much. – optional Nov 21 '12 at 9:14
Look at the node between R5 and and the parallel resistors R1 through R4. See that 4 currents enter (II through I4) and one current exits (I5). KCL says that the sum of currents entering a node must equal the sum of currents exiting the node. For parallel circuit elements, the voltage across each element is identical. It sometimes helps to see this be redrawing the circuit like I've done here: physics.stackexchange.com/questions/31079/… – Alfred Centauri Nov 21 '12 at 12:14