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KVL and finding circuit currents

  1. Jan 27, 2017 #1
    1. The problem statement, all variables and given/known data


    The diagram shown below shows a bi-directional opto coupler input
    interface circuit. When a supply voltage of 20 V is applied the LED
    carries a current and 2 V is dropped across it. Calculate the value of
    the LED current and the value of current through the 3 kΩ resistance.

    Q6.png

    2. Relevant equations


    3. The attempt at a solution

    As the voltage in a parallel circuit is the same in each leg, for a voltage drop of 2V across the LED, the 470Ohm resistor must also have a voltage drop of 2V.

    A6a.png
    So the voltage drop across the 3kohm resistor using KVL is:
    E - V1 - V2 = 0
    V1 = E - V2
    V1 = 20 - 2 = 18V

    So I now have V and R for the 3k resistor, so using Ohms Law:
    I = V / R
    I = 18 / 3000 = 6x10^-3A or 6mA
    6mA is the current flowing through the 3k resistor

    A6b.png

    As the 470 resistor has a voltage drop of 2V, I have V and I

    I = V / R
    I1 = 2 / 470 = 4.255x10^-3A = 4.255mA

    Current in a parallel circuit is the sum of the current in each leg, so

    I = I1 + I2

    So
    I2 = I - I1
    I2 = 6 - 4.255 = 1.745mA

    The answers are the current flowing through the LED is 1.745mA and the current flowing through the 3k resistor is 6mA

    Is this looking somewhere close. I struggled with this

    Kr
    Craig
     
  2. jcsd
  3. Jan 27, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    Your method and result values look good. You might want to round the final results to appropriate numbers of significant figures.
     
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