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31. Alternating Current
Power in AC Circuits
Problem 30.78
Textbook Question
Textbook QuestionAn ac voltage source V = Vo sin (ωt + 90°) is connected across an inductor L and current I = Io sin (ωt) flows in this circuit. Note that the current and source voltage are 90° out of phase. (a) Directly calculate the average power delivered by the source over one period T of its sinusoidal cycle via the integral P = ∫₀ᵀ V I dt/T. (b) Apply the relation P = Iᵣₘₛ Vᵣₘₛ cos Φ to this circuit and show that the answer you obtain is consistent with that found in part (a). Comment on your results.
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1
Step 1: Express the voltage and current functions in terms of sine and cosine. Given V = V_0 \sin (\omega t + 90^\circ), you can rewrite this as V = V_0 \cos (\omega t) because \sin (\theta + 90^\circ) = \cos (\theta). The current is given as I = I_0 \sin (\omega t).
Step 2: Set up the integral for the average power over one period T. The average power P is given by P = \frac{1}{T} \int_0^T V I \, dt. Substitute the expressions for V and I into the integral: P = \frac{1}{T} \int_0^T (V_0 \cos (\omega t))(I_0 \sin (\omega t)) \, dt.
Step 3: Simplify the integral using trigonometric identities. The product of sine and cosine can be expressed as \cos (\omega t) \sin (\omega t) = \frac{1}{2} \sin (2\omega t). Thus, the integral becomes P = \frac{V_0 I_0}{2T} \int_0^T \sin (2\omega t) \, dt.
Step 4: Evaluate the integral over one period. The integral of \sin (2\omega t) over one period from 0 to T is zero, because the integral of sine over its full period is zero. Therefore, the average power P = 0.
Step 5: Verify using the RMS values and phase difference. Calculate the RMS values of the voltage and current, V_{rms} = \frac{V_0}{\sqrt{2}} and I_{rms} = \frac{I_0}{\sqrt{2}}. The phase difference \Phi between the voltage and current is 90^\circ, so \cos(\Phi) = \cos(90^\circ) = 0. Using the formula P = I_{rms} V_{rms} \cos(\Phi), we find P = \left(\frac{V_0}{\sqrt{2}}\right)\left(\frac{I_0}{\sqrt{2}}\right)(0) = 0. This result is consistent with the integral calculation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
AC Voltage and Current
In alternating current (AC) circuits, voltage and current vary sinusoidally with time. The voltage source given as V = Vo sin(ωt + 90°) indicates that the voltage reaches its maximum value a quarter cycle (90°) before the current I = Io sin(ωt) does. This phase difference is crucial for understanding how power is transferred in AC circuits, particularly in inductive loads where the current lags behind the voltage.
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Average Power in AC Circuits
The average power delivered by an AC source over one complete cycle can be calculated using the integral P = ∫₀ᵀ V I dt/T. This formula accounts for the instantaneous power (product of voltage and current) over the period T. For sinusoidal functions, this average power can also be expressed in terms of root mean square (RMS) values and the phase difference between voltage and current, which is essential for accurate calculations in AC systems.
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Power Factor and RMS Values
The power factor, represented as cos Φ, is the cosine of the phase angle between the voltage and current waveforms. In this case, with a 90° phase difference, the power factor is zero, indicating that no real power is delivered to the circuit. The relation P = Iᵣₘₛ Vᵣₘₛ cos Φ highlights how RMS values of current and voltage are used to calculate average power, reinforcing the understanding of energy transfer in AC circuits.
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