Energy Stored in a Magnetic Field — Question 2

Question 2

Source energy during RL charging divides between field storage and heat

A series RL circuit is connected to a constant source V from zero current. Set $I_\infty=V/R$ and $\tau=L/R$.

1. Find source work and stored energy at an arbitrary time t.

2. Find dissipated heat at the instant the current reaches half its final value.

Question 2 - Solution

Source energy during RL charging divides between field storage and heat

(a) Method and calculation

$$i=I_\infty(1-e^{-t/\tau})\Rightarrow W_s=\int_0^tVi\,dt=\frac{V^2}{R}[t-\tau(1-e^{-t/\tau})].$$$$\boxed{U(t)=\tfrac12LI_\infty^2(1-e^{-t/\tau})^2}.$$

(b) Interpretation and further calculation

$$i=I_\infty/2:\ t=\tau\ln2,\quad W_s=LI_\infty^2(\ln2-1/2),\quad U=LI_\infty^2/8.$$$$\boxed{H=W_s-U=LI_\infty^2(\ln2-5/8)>0}.$$

Physical check

Counting all source work as magnetic energy would omit resistive heating during the current buildup.