Energy Dissipated in an RL Circuit

In summary, the conversation discussed a make-before-break switch in a circuit and the calculation of power dissipated in a resistor after the switch is thrown. The correct percentage of initial energy dissipated in the 90 Ω resistor in 1.4 ms was found to be 42.375%.
  • #1
Drakkith
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Homework Statement


In the circuit shown in (Figure 1) , the switch makes contact with position b just before breaking contact with position a. As already mentioned, this is known as a make-before-break switch and is designed so that the switch does not interrupt the current in an inductive circuit. The interval of time between "making" and "breaking" is assumed to be negligible. The switch has been in the a position for a long time. At t = 0 the switch is thrown from position a to position b.

What percentage of the initial energy stored in the inductor is dissipated in the 90 Ω resistor 1.4 ms after the switch is thrown from position a to position b?

Figure 1:
pr_7-3.jpg


Homework Equations


##i(t) = 0.5e^{-500t}##
##i(0) = 0.5 A##
##p(t) = i(t)^2R##
##W_i=\frac{1}{2}LI^2##

The Attempt at a Solution


I found the equation for the current after ##t≥0##. To find the power dissipated in the 90 ohm resistor, I thought I was supposed to find the integral of the power from 0 to 1.4 ms.

Initial energy is ##W_i = (0.5)(0.32)(0.5^2) = 0.04375 J##

Power is:
##p(t) = i(t)^2R##
##p(t) = (0.5e^{-500t})^2(90) = 22.5e^{-1000t}##

The integral of power is: ##\int_0^{1.4ms} 22.5e^{-1000t}\, dt = 22.5 \int_0^{1.4ms} e^{-1000t}\, dt = \left. \frac{-22.5}{1000}(e^{-1000t}) \right|_0^{0.0014} = -0.0225(e^{-1.4}-e^0) = -0.0225(-0.7534) = .01695 J##
##W(1.4ms) = 0.01695 J##

The percentage of initial energy is: ##\frac{0.01695}{0.04375}(100) = 38.75##

Unfortunately it appears that answer is incorrect. Any ideas where I might have gone wrong?
 

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  • #2
How did you get the .04375? 0.32 / 8 = 0.04, not 0.04375. The rest looks OK to me.
 
  • #3
phyzguy said:
How did you get the .04375? 0.32 / 8 = 0.04, not 0.04375. The rest looks OK to me.

Whoa... I have no idea. I must have typed something in incorrectly in my calculator.
Correcting that gives me 42.375%, which is the correct answer.

Thanks phyzguy.
 

What is an RL circuit?

An RL circuit is an electrical circuit that contains both a resistor (R) and an inductor (L). The inductor is a coil of wire that produces a magnetic field when an electrical current flows through it. In RL circuits, energy is stored in the magnetic field of the inductor.

What is energy dissipation?

Energy dissipation is the process of converting stored energy into heat or other forms of energy. In RL circuits, energy is dissipated when the magnetic field of the inductor collapses and releases energy as heat.

How is energy dissipated in an RL circuit?

Energy is dissipated in an RL circuit when the current through the inductor changes. This change in current causes the magnetic field of the inductor to collapse, releasing energy as heat. The rate of energy dissipation can be calculated using the formula P = I^2R, where P is power, I is current, and R is resistance.

What factors affect energy dissipation in an RL circuit?

The amount of energy dissipated in an RL circuit is affected by the resistance of the circuit, the inductance of the inductor, and the frequency of the current. A higher resistance will result in more energy dissipated, while a higher inductance or frequency will result in less energy dissipated.

What are some real-world applications of RL circuits and energy dissipation?

RL circuits and energy dissipation have many practical applications, such as in the power grid, electric motors, and electronic devices. They are also used in industries such as telecommunications, transportation, and healthcare. In these applications, understanding and controlling energy dissipation is crucial for efficient and safe operation of the circuits.

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