Basic Circuit Problems - KVL and KCL

  1. 1. The problem statement, all variables and given/known data
    Find [tex]R_2[/tex] in the circuit below.


    2. Relevant equations
    Kirchoff's Current Law:
    [tex]\sum {I_{IN} } = \sum {I_{OUT} }[/tex]

    Kirchoff's Voltage Law:
    [tex]\sum {V = 0}[/tex]

    3. The attempt at a solution

    Using KCL for the node joining the 8ohm resistor, R2 and the 1 ohm resistor:
    [tex]i_1 = i_2 + 2[/tex]

    Now using KVL for the shortest route down the 1 ohm resistor from the voltage source.
    [tex] - 10 + 8i_1 + 2 = 0\; \Rightarrow \;i_1 = 1\;A[/tex]

    From the first equation;
    [tex]i_2 = i_1 - 2 = 1 - 2 = - 1\;A[/tex]

    Until this point, everything seems logical, and I think it is right.

    Now in order to solve for [tex]R_2[/tex] I must now use [tex]V = IR[/tex] correct?

    But, am I right in saying that because the voltage is flowing in the same direction and has changed by 2V, that the potential difference = 2V. Therefore [tex]R_2 = \frac{2}{{ - 1}} = - 2\;\Omega[/tex].
    Resistance can't be negative though can it?

    Can someone explain the situation in simple terms or what laws govern this?
    I'm happy to read further if some people can give me some hints in the right direction.

    Last edited: Mar 9, 2008
  2. jcsd
  3. What is the voltage across the 1 ohm resistor?
    R2 has one side connected with the 1 ohm resistor and the other connected to the 12V source. What is the voltage drop across it?
  4. V=IR = (2)(1) = 2V

    The Voltage drop is therefore 12 - 2 = 10V?
  5. Knowing the voltage drop and the current through R2, you can calculate its value.
  6. excellent cheers mate.

    I was also shown another method today which involved more dependability on the sign conventions, but is more politically correct, yet yields the same result.
  7. resistance can never be negative
    but it coming negative because direction of current taken is opposite

    you can call this beauty of kvl and kcl
  8. Of course the resistance cannot be negative, but the problem is resolved by realizing the direction of the current is into the common node of the resistors, not away from it.
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