Power in RL Circuits: Solving for Time Constant

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In an RL circuit with a time constant of 42.1 ms, the problem involves finding the time when the power dissipated in the resistor equals the power stored in the inductor. The relevant equations include the power dissipated by the resistor (P = I^2R) and the power stored in the inductor (P = LI(dI/dt)). The solution approach involves setting these powers equal and simplifying, leading to the equation 1 - exp(-t/T) = exp(-t/T). Confusion arises regarding the derivative of current and the correct form of the equation for dI/dt, with participants verifying their calculations and seeking clarification on the steps. The discussion highlights the complexity of solving RL circuit problems and the importance of correctly applying calculus in circuit analysis.
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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