Alternating Current — Question 3

Section 31-5 · Physics pages 5056–5057

Question 3 A DC offset adds in quadrature to the sinusoidal rms contribution A voltage is v(t) = 6.00 + 8.00 cos ωt volts and is applied to a 10.0-ohm resistor. t v Schematic; labels, not drawing lengths, specify the data. (a) Find its mean, total rms, and AC-only rms values. (b) Find mean resistor power and determine whether the current reverses.
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Question 3 - Solution A DC offset adds in quadrature to the sinusoidal rms contribution t v (a) Method and calculation ⟨v⟩= 6.00 V, ⟨v2⟩= 62 + 82/2 = 68 V 2. Vrms = √ 68 = 8.24621 V, Vac,rms = 8/ √ 2 = 5.65685 V . (b) Interpretation and further calculation ⟨P⟩= 68/10 = 6.80 W . vmin = −2.00 V < 0 < vmax = 14.0 V ⇒current reverses . Physical check An AC-coupled meter excludes the DC contribution and would understate total resistor heating.

Question 3

A DC offset adds in quadrature to the sinusoidal rms contribution

A voltage is v(t)=6.00+8.00cos⁡ωtv(t)=6.00+8.00\cos\omega t volts and is applied to a 10.0-ohm resistor.

Diagram shown in the original SVG worksheet.

  1. Find its mean, total rms, and AC-only rms values.

  2. Find mean resistor power and determine whether the current reverses.

Question 3 - Solution

A DC offset adds in quadrature to the sinusoidal rms contribution

Diagram shown in the original SVG worksheet.

(a) Method and calculation

⟨v⟩=6.00V,⟨v2⟩=62+82/2=68V2.\langle v\rangle=6.00\,V,\quad\langle v^2\rangle=6^2+8^2/2=68\,V^2.Vrms=68=8.24621V,Vac,rms=8/2=5.65685V.\boxed{V_{\mathrm{rms}}=\sqrt{68}=8.24621\,V,\quad V_{{\mathrm{ac}},\mathrm{rms}}=8/\sqrt2=5.65685\,V}.

(b) Interpretation and further calculation

⟨P⟩=68/10=6.80W.\boxed{\langle P\rangle=68/10=6.80\,W}.vmin=−2.00V<0<vmax=14.0V⇒current reverses.v_{\min}=-2.00\,V<0<v_{\max}=14.0\,V\Rightarrow\boxed{\text{current reverses}}.

Physical check

An AC-coupled meter excludes the DC contribution and would understate total resistor heating.