# Question about Finding unknown resistance by Ohms and Kirchoff's Laws

1. Jan 27, 2014

### Orellanal

1. The problem statement, all variables and given/known data

Using Ohm's and Kirchoff's Rules find the unknown resistance Rx

2. Relevant equations

V = IR

Sum of currents at a node is zero
Sum of voltages around a loop is zero

3. The attempt at a solution

For the 2 ohm resistor I found the voltage to be 4V by the formula V = IR or V = 2A * 2ohm = 4 V
Therefore for the Rx resistor the voltage must also be 4 V as is it connected to the 2 ohm resistor in parallel.

The 4 and the 8 ohm resistors on top are connected in parallel and therefore must have the same voltage. 12 V - 8 V on the bottom two resistors gives 4 V for them which means that the voltage through each of the top resistors must be 2 V. Knowing this I then took Kirchoff's Current law at the nodes.

Node 2 (Right)

2V/ 8 ohm = 0.25 A (going in to the node)

- 0.25 A + 2A = -1.75 A (going into the node)

Node 1 (Left)

-.5 A (into the node) + 1.75 A (going out of the node) = 1.25 A (going out of the node)

I then used ohm's law to find the resistance x by V = IR --> R = V/I or 4V/ 1.25 A = 3.2 ohms
which is wrong.

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2. Jan 27, 2014

### Staff: Mentor

Good so far!

You'll want to revisit that logic. Yes, the two upper resistors are in parallel. Being in parallel they will share the same potential drop across them (voltage is across a component, current goes through a component). You determined that the voltage across the two lower resistors is 4 V. What does that leave (of the supply potential's 12 V) to appear across both of those upper resistors?