Power in RL Circuits: Solving for Time Constant

In summary, the equation for the instantaneous current is given by i = V/R*[1 - e^-(R/L)t] and di/dt = V/R*(R/L)*e^-(R/L)t.
  • #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 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

 
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  • #2
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
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
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
ah sorry I've been so late, heh and it seems I can't even take a derivative anymore.

thanks for the help.
 

FAQ: Power in RL Circuits: Solving for Time Constant

1. What is the time constant in a RL circuit?

The time constant (τ) in a RL circuit is a measure of how quickly the current changes in response to a sudden change in voltage. It is equal to the product of the resistance (R) and inductance (L) in the circuit, and is measured in seconds.

2. How do I calculate the time constant of a RL circuit?

The time constant of a RL circuit can be calculated using the formula τ = L/R, where L is the inductance in henries and R is the resistance in ohms. Alternatively, if you know the current (I) and voltage (V) in the circuit, you can use the formula τ = L/(R*I).

3. What is the significance of the time constant in a RL circuit?

The time constant is important because it determines how quickly the current will reach its steady state value after a change in voltage. It also affects the rate of energy dissipation in the inductor, which can be important in certain applications.

4. How do I solve for the time constant in a RL circuit?

To solve for the time constant in a RL circuit, you will need to know the values of the inductance and resistance in the circuit. Once you have these values, you can plug them into the formula τ = L/R to calculate the time constant.

5. Can the time constant be changed in a RL circuit?

Yes, the time constant in a RL circuit can be changed by altering the values of the inductance or resistance. For example, increasing the resistance will result in a longer time constant, while decreasing the inductance will result in a shorter time constant.

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