Finding Zero Power at Inductor Terminals

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The discussion revolves around finding the time when power at the terminals of an inductor is zero, given specific current and voltage equations. Participants express confusion over the equations and initial conditions, particularly regarding the accuracy of the voltage function. There is a suggestion to derive the power function using the product of current and voltage, leading to a more manageable expression. The conversation highlights the importance of verifying the initial conditions against the provided equations, as discrepancies can lead to incorrect conclusions. Ultimately, the need for clarity in the problem statement and careful mathematical manipulation is emphasized.
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Homework Statement


Find the time, greater than zero, when the power at the terminals of the inductor is zero.
i(t) = .20e-500t + .1e-2000t A
v(t) = -5e-500t-10e2000t V
L = .050 H
v(

Homework Equations


P=VI

The Attempt at a Solution


It seems to me to be impossible to find a finite time to enter for the power to be 0 due to the exponentials, yet i don't think the online question service has a way of entering infinity as an answer. I am unsure how to proceed further. Please give me a hint
 
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Mathematically the power may never reach zero as the curve is asymptotic to zero. However, a there is a practical rule of thumb that it used in engineering that states that a circuit can be assumed to have reached steady state (all the interesting activity is done) after five time constants. Perhaps they're looking for you to determine the time constant that dominates the response and answer accordingly? Try expanding I*V and see if can be simplified in some beneficial way.
 
RMZ said:

Homework Statement


Find the time, greater than zero, when the power at the terminals of the inductor is zero.
i(t) = .20e-500t + .1e-2000t A
v(t) = -5e-500t-10e2000t V

Should the exponent in the last term of v(t) have a negative sign? If so, follow gneill's advice on expanding i*v. You will get a definite solution for t.
 
Thanks both of you. I am following your advice as best i can: I am trying to imagine that it is an inductor-resistor circuit, and using the initial conditions the problem gave [v(0) = 3V and i(0) = .12 A], I am trying to see if i can simulate the circuit it is in using a voltage source and a resistor so that i can find a value for R in tau = L/R. Ill let you know how that goes
 
RMZ said:
Thanks both of you. I am following your advice as best i can: I am trying to imagine that it is an inductor-resistor circuit, and using the initial conditions the problem gave [v(0) = 3V and i(0) = .12 A], I am trying to see if i can simulate the circuit it is in using a voltage source and a resistor so that i can find a value for R in tau = L/R. Ill let you know how that goes

Just reread your advice. I will try that too
 
Nothing has panned out..
 
I have no clue what to do. It did not give me a resistor value or capacitor value or anything like that.
 
The problem statement was not very specific. Did you state the problem exactly word for word?

Are we to interpret the current i(t) as the current in the inductor? Does v(t) represent the potential difference across the terminals of the inductor? Assuming the answer is yes to these two questions, how would you write an expression for the power delivered to the inductor as a function of time?
 
Yes, there is no information missing.
upload_2015-10-30_18-11-13.png
P = VI = Li(di/dt)
 
  • #10
OK. The voltage function v(t) represents the voltage across the terminals of the inductor.

Yes, P = IV. So, what mathematical expression do you get for the power as a function of time?
 
  • #11
RMZ said:
Im trying to imagine that it is an inductor-resistor circuit, and using the initial conditions the problem gave [v(0) = 3V and i(0) = .12 A]

Those values are not consistent with your statement of the problem:

RMZ said:
i(t) = .20e-500t + .1e-2000t A
v(t) = -5e-500t-10e2000t V

For example, when ##t = 0##, ##i = 0.3 \ \mathrm{A}## (edit) and ##v=-15 \ \mathrm{V}##.
 
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  • #12
Im sorry it will no let me edit the problem. the expression for voltage is supposed to contain e-2000t rather than e2000t
 
  • #13
I arrived at P = -e-4000t-2.5e-2500t-e-1000t
 
  • #14
RMZ said:
Im sorry it will no let me edit the problem. the expression for voltage is supposed to contain e-2000t rather than e2000t

You'll still get -15 volts at ##t=0##.
 
  • #15
Can you briefly explain how you determined the answer for part A?
 
  • #16
I treated i(t) as the general solution to a differential equation, with the initial conditions i(0) = .120A and Li'(0) = 3V. Using this approach i solved for A1 and A2. I then used this solution in v = L(di/dt) to find the expression or the voltage
 
  • #17
You have the initial condition i(0) = 0.12 A. But, as Mister T pointed out, your expression for i(t) gives i(0) = 0.30 A (edited) which does not agree with the initial condition i(0) = 0.12 A
 
  • #18
Wow. I am not sure why the v(0) yields 15 and i(0) = .3. i think it may have to do with the fact that there is a disconinuity in the current equation at t = 0, but my head hurts too much at this point to ponder it further. I am going to leave it alone for a few hours.
 
  • #19
When I read the OP I took the given expressions for voltage and current to be, well, given. Now that I know the original problem statement I can see that there's a problem with the expressions (and this has been pointed out by others checking the expressions against the given initial values). RMZ, you need to check your solution to the differential equation. In particular, your value of the a2 coefficient seems suspect to me.

Once that's sorted, check to see if either i(t) or v(t) has a zero.
 
  • #20
RMZ said:
Wow. I am not sure why the v(0) yields 15 and i(0) = .3. i think it may have to do with the fact that there is a disconinuity in the current equation at t = 0, but my head hurts too much at this point to ponder it further. I am going to leave it alone for a few hours.

There's no discontinuity there. Note that prior to ##t=0## the current is constant (steady state) allowing you to calculate the resistance of the inductor.
 
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