Consider once again the RL circuit shown in Figure 32.6. Recall that the current in the right-hand loop decays exponentially with time according to the expression I = I 0e^-t/τ, where I0 = ε/R is the initial current in the circuit and τ =L/R is the time constant. Show that all the energy initially

Question

Consider once again the RL circuit shown in Figure 32.6. Recall that the current in the right-hand loop decays exponentially with time according to the expression [latex] I=I_{0} e^{-t / au}, [/latex], where [latex] I_{0}= \varepsilon / R [/latex] is the initial current in the circuit and τ =L/R is the time constant. Show that all the energy initially stored in the magnetic field of the inductor appears as internal energy in the resistor as the current decays to zero.

Accepted Answer

The rate dU/dt at which energy is delivered to the resistor (which is the power) is equal to I²R, where I is the instantaneous current:

[latex] \frac{d U}{d t}=I^{2} R=\left(I_{0} e^{-R t / L} ight)^{2} R=I_{0}^{2} R e^{-2 R t / L} [/latex].

To find the total energy delivered to the resistor, we solve for dU and integrate this expression over the limits [latex] t=0 ext { to } t ightarrow \infty [/latex] . (The upper limit is infinity because it takes an infinite amount of time for the current to reach zero.)

(1) [latex] U=\int_{0}^{\infty} I_{0}^{2} R e^{-2 R t / L} d t=I_{0}^{2} R \int_{0}^{\infty} e^{-2 R t / L} d t [/latex].

The value of the definite integral can be shown to be L/2R (see Problem 34) and so U becomes

[latex] U=I_{0}^{2} R\left(\frac{L}{2 R} ight)=\frac{1}{2} L I_{0}^{2} [/latex].

Note that this is equal to the initial energy stored in the magnetic field of the inductor, given by Equation 32.12, as we set out to prove.

[latex] U=\frac{1}{2} L I^{2} [/latex] (32.12).

Question 32.4: Consider once again the RL circuit shown in Figure 32.6. Rec...

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