Ch. 30 - Inductance, Electromagnetic Oscillations, and AC Circuits

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo

Giancoli Douglas 5th edition

Ch. 30 - Inductance, Electromagnetic Oscillations, and AC Circuits

Problema 66

Capitolo 29, Problema 66

Show that the power delivered by a three-phase ac source equals a constant P = 3Vo²/2R, by combining the four equations in Section 30–11.

Guida verificata passo dopo passo

Start by recalling the key equations for a three-phase AC source. Each phase delivers a voltage described by $ V(t) = V_0 \sin(\omega t + \phi) $, where $ V_0 $ is the peak voltage, $ \omega $ is the angular frequency, and $ \phi $ is the phase angle. For a balanced three-phase system, the phase angles are separated by $ 120^\circ $ (or $ 2\pi/3 $ radians).

The instantaneous power delivered by one phase is given by $ P(t) = \frac{V^2(t)}{R} $, where $ R $ is the resistance of the load. Substituting $ V(t) = V_0 \sin(\omega t + \phi) $, we get $ P(t) = \frac{V_0^2 \sin^2(\omega t + \phi)}{R} $.

For a three-phase system, the total instantaneous power is the sum of the powers from all three phases. Using the phase angles $ \phi_1 = 0 $, $ \phi_2 = -120^\circ $, and $ \phi_3 = -240^\circ $, the total power is $ P_{\text{total}}(t) = \frac{V_0^2}{R} [\sin^2(\omega t) + \sin^2(\omega t - 120^\circ) + \sin^2(\omega t - 240^\circ)] $.

Use the trigonometric identity $ \sin^2(x) = \frac{1 - \cos(2x)}{2} $ to simplify each term. After simplification, the cosine terms cancel out due to symmetry, leaving $ P_{\text{total}}(t) = \frac{3V_0^2}{2R} $.

Conclude that the total power delivered by the three-phase AC source is constant and equals $ P = \frac{3V_0^2}{2R} $, as the time-dependent terms have been eliminated through the summation of the three phases.

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Three-Phase AC Power

Three-phase AC power systems use three alternating currents that are offset in phase by 120 degrees. This configuration allows for a more efficient power delivery compared to single-phase systems, as it provides a constant power transfer and reduces the amount of conductor material needed. The total power in a three-phase system can be calculated using the formula P = √3 × V_L × I_L, where V_L is the line voltage and I_L is the line current.

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05:37

Power in AC Circuits

Voltage and Resistance Relationship

In electrical circuits, the relationship between voltage (V), current (I), and resistance (R) is described by Ohm's Law, which states V = I × R. This fundamental principle allows us to understand how voltage drops across resistors and how it affects the current flow in the circuit. In the context of power calculations, the power dissipated in a resistor can be expressed as P = V²/R, highlighting the importance of both voltage and resistance in determining power.

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RMS Current and Voltage

Power Calculation in AC Circuits

The power delivered by an AC source can be calculated using the root mean square (RMS) values of voltage and current. For a resistive load, the average power can be expressed as P = I_RMS² × R or P = V_RMS²/R. In three-phase systems, the total power is the sum of the power in each phase, leading to the formula P = 3V_o²/2R when considering the phase voltage and the resistive load, which is derived from the combination of the relevant equations.

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Power in AC Circuits

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