I am currently studying some old exams for the upcoming one and I got stuck on this question, I seem to get the right answer by eyeballing the circuit but I want to be able to calculate the value.

The question is as follows: Calculate the resistance of R2.

What I tried doing was to calculate the voltage in the middle using Kirchoff's current law but I do not know where to go from there.

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  • \$\begingroup\$ Ohms law + KCL. What id the voltage across R1 (trick question)? What can you do to simplify the schematic? \$\endgroup\$ – winny Aug 1 '19 at 10:20
  • \$\begingroup\$ Annoyingly, circuit lab won't work for me so I can't make a proper answer just yet. Either way, re-draw the circuit so it is more logical. You can then see which resistors are in series and parallel. You can then use this information to use KCL and Ohm's Law to solve \$\endgroup\$ – MCG Aug 1 '19 at 10:26
  • \$\begingroup\$ I assume those A1 A2 boxes are 'ideal' current meters ? In which their resistance is zero. Hence R1 is irrelevant and total current is "8m". Now this is divided between R2 & R3 in parallel, where R2 is passing "2m". Hence rest must be through R3 = (8-2) 6m. So voltage across R3 = 2K x 6m = 12 K.m. Which is the same across R2. Hence R2 = (12 K.m / 2m) = 6K \$\endgroup\$ – Kripacharya Aug 1 '19 at 10:52
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    \$\begingroup\$ @Kripacharya As you have given the solution away, why have you put it as a comment? If you are going to answer, put it in the answer section. Also, as this is a studying question (like a homework question) we would tend to explain how to get to the correct answer, rather than just give it away. This encourages the OP to solve for themselves so they can learn. Also, since when was K.m. a unit of voltage? \$\endgroup\$ – MCG Aug 1 '19 at 12:42
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    \$\begingroup\$ @Kripacharya it wasnt criticism really. You answered the question. So it should be an answer instead of a comment. And I was only explaining how people here usually deal with homework or study questions. Yes, the math is trivial to some, but to people studying, it may not be. Hence the practise of guiding people towards an answer, rather than giving it away. None of that was a criticism. If there was any, it was the units you used. Just use V for volts and A or mA for current. Makes it easier to understand. \$\endgroup\$ – MCG Aug 2 '19 at 15:56

I seem to have gotten the right answer now, what I did was that I used KCL as follows:

8mA - 2mA(From the unknown resistor) - ( 0 - Vu / R3) = 0

I then moved the equation around as following:

-8mA + 2mA = Vu * ( - 1/R3)


-6mA / (- 1/R3) = Vu = 12 V

hen, by using Ohm's law I calculated the resistance:

R = V / I = 12 / 2mA = 6000 Ohm's

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