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Homework Help: Archived Voltage Divider (in a circuit)

  1. Apr 29, 2010 #1
    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!!
     

    Attached Files:

  2. jcsd
  3. Feb 5, 2016 #2

    gneill

    User Avatar

    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##.
    upload_2016-2-6_0-32-23.png

    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##
     
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