Three Simple Circuits — Question 7

Section 31-7 · Physics pages 5104–5105

Question 7 A nonzero mean voltage prevents periodic current in a lossless inductor An ideal inductor is driven by v = V0 + Vp cos ωt with nonzero V0. Its initial current is zero. ∼ v i L Schematic; labels, not drawing lengths, specify the data. (a) Find i(t). (b) Find the current change per source period and the condition required for any periodic inductor current.
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Question 7 - Solution A nonzero mean voltage prevents periodic current in a lossless inductor ∼ v i L (a) Method and calculation i(t) = 1 L ∫t 0 (V0 + Vp cos ωu)du. i(t) = V0t/L + (Vp/ωL) sin ωt . (b) Interpretation and further calculation i(t + T) −i(t) = 1 L ∫t+T t v dt = V0T/L. ∆iT̸ = 0; periodicity requires ∫t+T t v dt = 0 . Physical check A zero-mean AC component cannot cancel cumulative current growth caused by a DC voltage.

Question 7

A nonzero mean voltage prevents periodic current in a lossless inductor

An ideal inductor is driven by v=V0+Vpcos⁡ωtv=V_0+V_p\cos\omega t with nonzero V0. Its initial current is zero.

Diagram shown in the original SVG worksheet.

  1. Find i(t).

  2. Find the current change per source period and the condition required for any periodic inductor current.

Question 7 - Solution

A nonzero mean voltage prevents periodic current in a lossless inductor

Diagram shown in the original SVG worksheet.

(a) Method and calculation

i(t)=1L∫0t(V0+Vpcos⁡ωu)du.i(t)=\frac1L\int_0^t(V_0+V_p\cos\omega u)du.i(t)=V0t/L+(Vp/ωL)sin⁡ωt.\boxed{i(t)=V_0t/L+(V_p/\omega L)\sin\omega t}.

(b) Interpretation and further calculation

i(t+T)−i(t)=1L∫tt+Tvdt=V0T/L.i(t+T)-i(t)=\frac1L\int_t^{t+T}v\,dt=V_0T/L.ΔiT≠0;periodicity requires∫tt+Tvdt=0.\boxed{\Delta i_T\ne0;\quad\text{periodicity requires}\ \int_t^{t+T}v\,dt=0}.

Physical check

A zero-mean AC component cannot cancel cumulative current growth caused by a DC voltage.