RL Circuit Transient Analysis: Simple Circuit Homework with Equations

In summary, the conversation discussed solving a circuit problem involving a switch, resistors, and an inductor. The solution involved using KVL and a first-order transient equation for the inductor to find the voltage across the resistor after the switch is opened. The conversation also touched on the importance of not memorizing equations and instead understanding the underlying principles in order to solve more complex circuit problems.
  • #1
galaxy_twirl
137
1

Homework Statement


dokil4.jpg


Homework Equations



iL = (iL(0) + Vs/R)e(-t/τ + Vs/R

The Attempt at a Solution



When the switch is closed before t=0, v(t) = 100V as the first 100Ohm resistor is shorted and L acts as a short wire in DC steady state.

When the switch opens, v(t) = 50V by voltage divider rule, since current flows through both 100Ohm resistors.

However, I have trouble putting the above in a v(t) expression, as v(t) is the voltage across R. I learned how to write first-order transient equations for L and C but can they be applied here?

τ = L/R (τ is tau, the time constant)

iL = (iL(0) + Vs/R)e-t/τ + Vs/R
 
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  • #2
I would recommend not memorizing such equations. It really makes things complicated when presented with problems that don't fit them. There are several ways you could solve the problem. Do you know the equation for KVL for after the switch is opened?
 
  • #3
GreenPrint said:
I would recommend not memorizing such equations. It really makes things complicated when presented with problems that don't fit them.

Oh dear. >< I see.

There are several ways you could solve the problem. Do you know the equation for KVL for after the switch is opened?

Assuming i is the current in the circuit, is the KVL equation 100 = i(100) + i(100) ? By V=IR, i = 0.5A.
 
  • #4
What about the voltage across the inductor?
 
  • #5
x
galaxy_twirl said:
I learned how to write first-order transient equations for L and C but can they be applied here?
Sure they can. You're already halfway to the answer because this is what you have worked out: the inductor current starts from a steady level of 1A and [once the switch is opened] it moves toward a final value of 0.5A on a 1st order exponential with a time-constant set by 200Ω and 1H.

In any general first order system, the exponential is always described by either e-t/τ or (1 - e-t/τ) and you choose which to fit the situation you are needing to describe.

I use only one general equation, preferring to perform additional algebra rather than memorize more than one.

contd on next post ...
 
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  • #6
IL = Iinitial + (Ifinal - Iinitial) (1 - e-t/τ)

Now, just fill in the numbers. When you have it finished, and tidied up, then simply multiply by 100 to convert from resistor current to resistor voltage.

That last equation I wrote is a handy form to memorize, as it reads aloud precisely how you'd express it in words, making it easy to write 'on the fly' without error, once you get the hang of it.
 
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  • #7
My advive would be to follow GreenPrint's advice.
You don't learn much by memorizing equations. If you don't analyze the simpler circuits you will be in big trouble with the more complex ones.
 
  • #8
NascentOxygen said:
IL = Iinitial + (Ifinal - Iinitial) (1 - e-t/τ)

Now, just fill in the numbers. When you have it finished, and tidied up, then simply multiply by 100 to convert from resistor current to resistor voltage.

That last equation I wrote is a handy form to memorize, as it reads aloud precisely how you'd express it in words, making it easy to write 'on the fly' without error, once you get the hang of it.

Sorry for the late reply. >< I was busy revising for my exams tomorrow. :(

I see. Thank you for the equation. :) I managed to solve the question by converting it to Thevenin and applying my formula, which can only be used if I have my circuits in Thevenin form.
 
  • #9
GreenPrint said:
What about the voltage across the inductor?

Is the voltage 0? This is because at DC steady state, inductor acts as short wire and hence, ##VL = 0##.
 
  • #10
rude man said:
My advive would be to follow GreenPrint's advice.
You don't learn much by memorizing equations. If you don't analyze the simpler circuits you will be in big trouble with the more complex ones.

I think the formula my teacher gave was to prevent us by solving first order differential equations which I still don't know how to do at the moment, though I did learn a little at A levels.
 

1. How does a simple RL circuit work?

A simple RL circuit consists of a resistor (R) and an inductor (L) connected in series with a voltage source. When the voltage source is turned on, it creates a changing magnetic field in the inductor, which in turn induces a voltage across the inductor. This induced voltage creates a current flow through the circuit, causing a voltage drop across the resistor. The current in the circuit will continue to flow until the magnetic field in the inductor reaches equilibrium.

2. What is the role of a resistor in a simple RL circuit?

The resistor in a simple RL circuit acts as a limiting factor for the current flow. It helps to control the rate at which the current increases and prevents it from becoming too high, which can damage the circuit components. Additionally, the resistor also causes a voltage drop in the circuit, which is necessary for the inductor to create a changing magnetic field.

3. How does the inductor store energy in a simple RL circuit?

The inductor in a simple RL circuit stores energy in the form of a magnetic field. When a current flows through the inductor, it creates a magnetic field around it. This magnetic field contains energy, and when the current stops flowing, the magnetic field collapses, releasing the stored energy back into the circuit.

4. What happens when the power source in a simple RL circuit is turned off?

When the power source in a simple RL circuit is turned off, the current in the circuit stops flowing, and the magnetic field in the inductor begins to collapse. This collapsing field creates a voltage spike, which can damage the circuit components. To prevent this, a diode is often added in parallel with the inductor to allow the current to flow in a loop and dissipate the energy from the collapsing field.

5. What are some real-world applications of simple RL circuits?

Simple RL circuits have many practical applications, including inductors used in electronic devices such as speakers, motors, and generators. They are also used in power supplies, where they help to regulate the voltage and smooth out fluctuations. Additionally, RL circuits are used in signal processing, such as in filters and amplifiers, and in circuits that need a time delay, such as in timers and oscillators.

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