# Archived Voltage Divider (in a circuit)

1. Apr 29, 2010

### Katfazack

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

I'm looking at a larger circuit, but part of the problem is clearly an application of a voltage divider. The input voltage goes through a resistor (let's call it R1, it's value is 3kohms) and then it hits a node at which point it can go through R2 or R3 (12 and 6kohms respectively). If I want to know the voltage that goes through R3, how do I calculate this?

I've attached an image of the problem and listed my prior assumptions below.

2. Relevant equations

3. The attempt at a solution

I've made assumptions in this problem, starting with a source transformation (to make my current source a voltage source) and from there I'm assuming that if I use a voltage divider to get the voltage across the 6kohm resistor (further to the left) I can find the voltage at that node and ultimately find ix (the desired value). But I'm very stuck!! Please help!!

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2. Feb 5, 2016

### Staff: Mentor

A complete solution is offered.

Let's analyze the circuit as-is. The op-amp is configured as a voltage follower (buffer) so both its inputs will be at the same potential as the output. Also, since ideal op-amp inputs have infinite impedance, the same current $i_x$ flows through both of the 6 kΩ resistors and thus they form a voltage divider. If the junction of those two resistors is at $v_o$, then the top of the divider must be at $2 v_o$.

Now we're in a position to write a node equation for the $2 v_o$ node. Summing currents leaving the node:

$-4 mA + \frac{2 v_o}{3 k} + \frac{2 v_o - v_o}{6 k} + \frac{2 v_o - v_o}{12 k} = 0$

$v_o = \frac{48}{11} V = 4.364 V$

The current through the 6 kΩ resistors is then:

$i_x = \frac{2 v_o - V_o}{6 kΩ} = \frac{8}{11} mA = 727.3 μA$