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Current in RL circuit

  • Thread starter semc
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  • #1
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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?
 

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  • #2
vela
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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?
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.
 
  • #3
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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?
 
  • #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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