Alternating Current — Question 10

Section 31-5 · Physics pages 5070–5071

Question 10 A measured delay fixes relative phase only after the period is known A voltage v = Vp cos ωt has period 8.00 ms. A sinusoidal current of the same frequency reaches its positive maximum 1.00 ms after the voltage maximum. Assume the delay is within one cycle. ωt v/Vp, i/Ip solid v; dashed i Schematic; labels, not drawing lengths, specify the data. (a) Find the current phase lag and write i(t). (b) Find current at a voltage maximum and at the next voltage zero crossing.
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Question 10 - Solution A measured delay fixes relative phase only after the period is known ωt v/Vp, i/Ip solid v; dashed i (a) Method and calculation ϕ = ω∆t = 2π(1/8) = π/4. i(t) = Ip cos(250πt −π/4) . (b) Interpretation and further calculation i(0) = Ip cos(−π/4) = Ip/ √ 2. i(T/4) = Ip cos(π/2 −π/4) = Ip/ √ 2 . Physical check Equal instantaneous current values can occur while its slope has opposite signs.

Question 10

A measured delay fixes relative phase only after the period is known

A voltage v=Vpcos⁡ωtv=V_p\cos\omega t has period 8.00 ms. A sinusoidal current of the same frequency reaches its positive maximum 1.00 ms after the voltage maximum. Assume the delay is within one cycle.

Diagram shown in the original SVG worksheet.

  1. Find the current phase lag and write i(t).

  2. Find current at a voltage maximum and at the next voltage zero crossing.

Question 10 - Solution

A measured delay fixes relative phase only after the period is known

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ϕ=ωΔt=2π(1/8)=π/4.\phi=\omega\Delta t=2\pi(1/8)=\pi/4.i(t)=Ipcos⁡(250πt−π/4).\boxed{i(t)=I_p\cos(250\pi t-\pi/4)}.

(b) Interpretation and further calculation

i(0)=Ipcos⁡(−π/4)=Ip/2.i(0)=I_p\cos(-\pi/4)=I_p/\sqrt2.i(T/4)=Ipcos⁡(π/2−π/4)=Ip/2.\boxed{i(T/4)=I_p\cos(\pi/2-\pi/4)=I_p/\sqrt2}.

Physical check

Equal instantaneous current values can occur while its slope has opposite signs.