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Homework Help: Current in RL circuit

  1. Mar 31, 2010 #1
    Using Kirchhoff's rules for the instantaneous currents and voltages in the two-loop circuit, find the current in the inductor as a function of time.

    I used Kirchhoff's rule on the left loop consisting of the battery and the right loop which consist of the inductor and gotten the following equations.

    E-I1R1 - I3R2 = 0
    L dI2/dt - I3R2 = 0

    Combining both the equations give me
    E-I1R1 - L dI2/dt =0

    So from here how do we express I1 into I2 and R to get the differentiate equation form?
     

    Attached Files:

  2. jcsd
  3. Mar 31, 2010 #2

    vela

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    Your equation for the second loop needs a negative sign in front of the first term. When you draw the currents in, put a plus sign where the current enters a capacitor, resistor, or inductor, and a minus sign where it leaves. That's the normal convention for the voltage drop across that element.

    Note that you have three unknowns but only two equations. You need a third equation to solve this system. It comes from applying Kirchoff's current law to the circuit.
     
  4. Apr 1, 2010 #3
    Do you mean the right loop when you say the second loop? Well current I3 passes through the load R2 so the potential drop -I3R2 and potential rise when it pass the inductor so +LdI2/dt? So the last equation is I1+I2=I3? So how do we proceed from here?
     
  5. Apr 1, 2010 #4

    vela

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    Given the direction you went around the loop and the direction you assumed for I2, you get a potential drop through the inductor.

    You have three equations and three unknowns. It's just algebra to get it down to an equation in one variable.
     
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