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

## Homework Statement

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

## Homework Equations

V = IR

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

## 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.

I thank you in advance for your help.

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• HW #2_8 Circuit.PNG
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gneill
Mentor

## Homework Statement

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

## Homework Equations

V = IR

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

## 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.
Good so far!

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.
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?

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.

I thank you in advance for your help.