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Power in RL circuits

  • Thread starter Redd
  • Start date
  • #1
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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 in to 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:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
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.
 
  • #3
47
0
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.
 
  • #4
rl.bhat
Homework Helper
4,433
5
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
 
  • #5
47
0
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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