Engineering Basic Circuit Problems - KVL and KCL

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The discussion revolves around solving for R_2 in a circuit using Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL). The user initially calculates the current through the circuit and finds a negative value for R_2, leading to confusion since resistance cannot be negative. Clarifications are provided regarding the voltage drop across the resistors, with the voltage drop across R_2 determined to be 10V. It is emphasized that the negative result arises from the assumed direction of current flow, which can be adjusted to align with KCL and KVL principles. Ultimately, understanding the direction of current flow resolves the issue of the negative resistance calculation.
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Homework Statement


Find R_2 in the circuit below.

http://img214.imageshack.us/img214/6396/circuithh3.jpg

Homework Equations


Kirchoff's Current Law:
\sum {I_{IN} } = \sum {I_{OUT} }

Kirchoff's Voltage Law:
\sum {V = 0}

The Attempt at a Solution



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

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

From the first equation;
i_2 = i_1 - 2 = 1 - 2 = - 1\;AUntil this point, everything seems logical, and I think it is right.

Now in order to solve for R_2 I must now use V = IR 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 R_2 = \frac{2}{{ - 1}} = - 2\;\Omega.
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.

Cheers
Steven.
 
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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?
 
CEL said:
What is the voltage across the 1 ohm resistor?

V=IR = (2)(1) = 2V

CEL said:
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?

The Voltage drop is therefore 12 - 2 = 10V?
 
Knowing the voltage drop and the current through R2, you can calculate its value.
 
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.
 
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
 
steven10137 said:
Resistance can't be negative though can it?

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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