LR Circuit Analysis: Solving Equations w/ Kirchhoff's Rule

[christang_1023]

Homework Statement

For the circuit shown in the figure, the inductors have no appreciable resistance and the switch has been open for a very long time.
(a) The instant after closing the switch, what is the current through the 60.0-Ω resistor?
(b) The instant after closing the switch, what is the potential difference across the 15.0-mH inductor?
(c) After the switch has been closed and left closed for a very long time, what is the potential drop across the 60.0-Ω resistor?

Relevant Equations

Equations have been obtained based on Kirchhoff's rule

Above is the figure of the question.

According to Kirchhoff's Rule, I have obtained three equations
$$\varepsilon-iR_{10}-L_{40}\frac{di_1}{dt}=0$$
$$L_{40}\frac{di_1}{dt}-i_2R_{60}=0$$
$$i_2R_{60}-R_{30}(i-i_1-i_2)-L_{15}\frac{d(i-i_1-i_2)}{dt}=0,$$
where $\varepsilon$ stands for emf=100V, $i, i_1,i_2$ stand for the current passing through the 10Ω resistor, the 40mH inductor and the 60Ω resistor respectively.

Although I have got equations based on Kirchhoff's rule, I am totally lost about dealing with them.
I also wonder if there is any shortcut; because an inductor and a resistor that are in series with an emf is a known model, where $$I(t)=\frac{\varepsilon}{R}(1-e^{-t/\tau}).$$

 

[Paul Colby]

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
This is a random set of notes I found with google (Paul is no relation, well that I know of) by searching on "method of constant coefficients". This is a method of solution for linear differential equations with constant coefficients. It's a very useful one cause these types of equation crop up a lot.

 

[Orodruin]

Note that you do not need to solve the differential equations to answer the questions you have been asked. It is sufficient to know how the inductors behave when you suddenly put a potential across them versus how they behave when the system has been allowed to reach equilibrium.

 

[christang_1023]

Note that you do not need to solve the differential equations to answer the questions you have been asked. It is sufficient to know how the inductors behave when you suddenly put a potential across them versus how they behave when the system has been allowed to reach equilibrium.

Thank you for your useful suggestion.
I know that at t=0, the inductor can be regarded as an open circuit. However, I don't know why.
Could you please explain it for me?

 

[Orodruin]

I know that at t=0, the inductor can be regarded as an open circuit. However, I don't know why.
Could you please explain it for me?

At $t = 0$ the inductor can be regarded as having infinite resistance (i.e., an open circuit) because it takes time for the current to build up inside the inductor. That is what inductance is, it is an inertia in the system to changes in current. Therefore, for early enough times, the current in the inductors will be equal to zero, which is the same as saying that all the current that goes through the battery goes through the 60 Ω resistor.

Note that you will have a much easier time seeing this in your equations if you instead take the currents through the three rightmost branches as your unknowns as this will isolate a Kirchoff loop with only the voltage source and two resistors and you can put two of your currents to zero immediately for small times (those going through the inductors).

So what does regarding the inductors being treated as open circuits tell you about (a) and (b)?

Also, how can you treat the inductors in the stationary limit when $t \to \infty$?

 

[christang_1023]

At $t = 0$ the inductor can be regarded as having infinite resistance (i.e., an open circuit) because it takes time for the current to build up inside the inductor. That is what inductance is, it is an inertia in the system to changes in current. Therefore, for early enough times, the current in the inductors will be equal to zero, which is the same as saying that all the current that goes through the battery goes through the 60 Ω resistor.

Note that you will have a much easier time seeing this in your equations if you instead take the currents through the three rightmost branches as your unknowns as this will isolate a Kirchoff loop with only the voltage source and two resistors and you can put two of your currents to zero immediately for small times (those going through the inductors).

So what does regarding the inductors being treated as open circuits tell you about (a) and (b)?

Also, how can you treat the inductors in the stationary limit when $t \to \infty$?

Such treatment simplifies (a) a lot, such that $I=\frac{\varepsilon}{R_{60}}=1.43A.$
Meanwhile, in (b), $\varepsilon_L=-\varepsilon_{R60}$, where $\varepsilon_{R60}$ represents the potential drop of the resistor of 60Ω, only when satisfying this condition, can the 15mH inductor be treated as an open circuit.

Inductors can be treated as normal circuit conducting wires(i.e. no particular effects in the circuit) when $t \to \infty$

Thank you very much!

 

[christang_1023]

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
This is a random set of notes I found with google (Paul is no relation, well that I know of) by searching on "method of constant coefficients". This is a method of solution for linear differential equations with constant coefficients. It's a very useful one cause these types of equation crop up a lot.

Thank you for your sharing.

 

[Orodruin]

Careful now. There are other components relevant at t=0 than the battery and the 60 ohm resistor. Also be careful with how you define the sign of the voltage over the inductors.