How Do You Calculate Time-Dependent Current in an RL Circuit?

[semc]

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?

 

[vela]

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.

 

[semc]

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]

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.