# Current in RL circuit

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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## Answers and Replies

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.

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?

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.