Power in RL Circuits: Solving for Time Constant

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Homework Help Overview

The problem involves analyzing an RL circuit where a battery is connected to a resistor and an inductor. The focus is on determining the time at which the rate of energy dissipation in the resistor equals the rate of energy storage in the inductor's magnetic field, given a specific inductive time constant.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss setting the power equations equal to each other and simplifying the resulting expressions. There are attempts to derive the time variable from the equations involving current and its derivative.

Discussion Status

Some participants have provided alternative expressions for the derivative of current and have engaged in checking the validity of each other's equations. There is a recognition of a common result among participants, but no explicit consensus has been reached regarding the correctness of the approaches taken.

Contextual Notes

Participants express confusion over the source of certain equations and the application of the time constant in their calculations. There are indications of differing interpretations of the equations involved, particularly regarding the terms used for current and its rate of change.

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Homework Statement



At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 42.1 ms, at what time (in ms) is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?

Homework Equations



power dissapated by resistor: P=(I^2)R

Power stored by inductor: P=IL(di/dt)

Current in an RL cuircuit: I=V/R(1-exp(-t/T))

where T is the time constant tau = (L/R)

The Attempt at a Solution



I just set them equal, then simplified.

(I^2)R = IL(di/dt)

I=(L/R)(di/dt)

replace L/R with Tau and Current with above equation

V/R(1-exp(-t/T)) = T (di/dt)

take derivative of current

V/R(1-exp(-t/T)) = T (V/R)(1/T)*exp(-t/T)

cancelations

1-exp(-t/T)=exp(-t/T)

add exp(-t/T) to both sides and take natural logs

Ln(1/2) = -t/T

so t = -T*Ln(1/2)

I plug this into the online homework and it keeps telling me I'm wrong, but I can't see
what I'm doing incorrectly, any ideas?

(sorry I don't know LaTex yet :P)

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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The instantaneous rate at which the source delivers energy to the circuit is P = V*I. This is equal to
V*I = I^2*R + LI*dI/dt
Put I^2*R = LI*dI/dt . So
V*I = 2*LI*dI/dt
dI/dt = V/L*e^-t/T
Put this in the above equation and solve for t.
 
Hi, I am a little confused. Where did you get the equation:

dI/dt = V/L*e^-t/T? My book says it is (V/R) instead of (V/L)?

Also I plugged your equation for di/dt into the equation above it.

And it reduced to:

1/2= exp(-t/T)

or

-T*Ln(1/2) = t, which is what I also got.
 
Instantaneous current is given as
i = V/R*[1 - e^-(R/L)t]
di/dt = V/R*(R/L)*e^-(R/L)t
= V/L*e^-(R/L)t
 
ah sorry I've been so late, heh and it seems I can't even take a derivative anymore.

thanks for the help.
 

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