Energy Stored in a Magnetic Field — Question 9

Section 30-9 · Physics pages 4928–4929

Question 9 Constant-power energy extraction makes the required inductor voltage grow as current falls An ideal inductor L initially carries I0 and supplies constant positive power P to a controlled load. Assume no other loss and fixed geometry. The load is stopped before the current reaches zero. t i Schematic; labels, not drawing lengths, specify the data. (a) Find i(t) and the magnitude of the inductor terminal voltage. (b) If that voltage must not exceed Vmax, find the smallest allowed current and maximum extractable energy, assuming the initial current is sufficient.
Show solutionHide solution
Question 9 - Solution Constant-power energy extraction makes the required inductor voltage grow as current falls t i (a) Method and calculation d dt(1 2Li2) = −P ⇒i(t) = √ I2 0 −2Pt/L . vi = −P ⇒|v| = P/i . (b) Interpretation and further calculation P/i ≤Vmax ⇒imin = P/Vmax . Wextracted,max = 1 2L[I2 0 −(P/Vmax)2] . Physical check Constant power is incompatible with finite terminal voltage all the way to zero current; the formula states the permitted endpoint explicitly.

Question 9

Constant-power energy extraction makes the required inductor voltage grow as current falls

An ideal inductor L initially carries I0 and supplies constant positive power P to a controlled load. Assume no other loss and fixed geometry. The load is stopped before the current reaches zero.

Diagram shown in the original SVG worksheet.

  1. Find i(t) and the magnitude of the inductor terminal voltage.

  2. If that voltage must not exceed Vmax, find the smallest allowed current and maximum extractable energy, assuming the initial current is sufficient.

Question 9 - Solution

Constant-power energy extraction makes the required inductor voltage grow as current falls

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ddt(12Li2)=−P⇒i(t)=I02−2Pt/L.\frac d{dt}(\tfrac12Li^2)=-P\Rightarrow\boxed{i(t)=\sqrt{I_0^2-2Pt/L}}.vi=−P⇒|v|=P/i.v i=-P\Rightarrow\boxed{|v|=P/i}.

(b) Interpretation and further calculation

P/i≤Vmax⇒imin=P/Vmax.P/i\le V_{\max}\Rightarrow\boxed{i_{\min}=P/V_{\max}}.Wextracted,max=12L[I02−(P/Vmax)2].\boxed{W_{\mathrm{extracted,max}}=\tfrac12L[I_0^2-(P/V_{\max})^2]}.

Physical check

Constant power is incompatible with finite terminal voltage all the way to zero current; the formula states the permitted endpoint explicitly.