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Chapter 14: Problem 2078

In an ac circuit the emf (e) and the current (i) at any instant core given respectively by \(\mathrm{e}=\mathrm{E}_{\mathrm{O}}\) sin $\operatorname{ct}, \mathrm{I}=\mathrm{I}_{\mathrm{O}} \sin (\cot -\Phi)$. The average power in the circuit over one cycle of ac is. (a) \(\left[\left\\{E_{O} I_{O}\right\\} / 2\right] \cos \Phi\) (b) \(\mathrm{E}_{\mathrm{O}} \mathrm{I}_{\mathrm{O}}\) (c) \(\left[\left\\{E_{Q} I_{Q}\right\\} / 2\right]\) (d) \(\left[\left\\{E_{O} I_{O}\right\\} / 2\right] \sin \Phi\)

Short Answer

Expert verified
The short answer is: The average power in the circuit over one cycle of AC is \(\left[\frac{E_O I_O}{2}\right] \cos \Phi\).

Step by step solution

01

Recall the Instantaneous Power Formula

In an AC circuit, the instantaneous power (P) can be given by the product of the voltage (e) and the current (i). So, we have: \(P(t) = e(t)i(t) = E_O\sin(ct) I_O\sin(ct - \Phi)\)
02

Calculate the Average Power

To find the average power over one cycle, we need to integrate P(t) over one cycle and then divide it by the duration of the cycle, T. We know that the period of the sinusoidal functions is T = \(\frac{2\pi}{c}\), which are their common duration of one cycle. So, the average power (P_avg) can be calculated as follows: \(P_{avg} = \frac{1}{T}\int_0^T P(t) dt\) \(P_{avg} = \frac{1}{\frac{2\pi}{c}}\int_0^{2\pi/c}\, E_I\sin(ct) I_O\sin(ct - \Phi) dt\)
03

Use Trigonometry Identity to Simplify the Integral

We can simplify the integral using the trigonometry product-to-sum identity: \(\sin\alpha \sin\beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)]\). So, \(P_{avg} = \frac{c}{2\pi}E_OI_O \int_0^{2\pi/c}\frac{1}{2}[\cos(0) - \cos(2ct - \Phi)]dt\) \(P_{avg} = \frac{cE_OI_O}{4\pi} \int_0^{2\pi/c}[\cos(0) - \cos(2ct - \Phi)]dt\)
04

Integrate the Simplified Expression and Find P_avg

Now, we can integrate the simplified expression and substitute the limits to find P_avg: \(P_{avg} = \frac{cE_OI_O}{4\pi} [\int_0^{2\pi/c}dt - \int_0^{2\pi/c}\cos(2ct - \Phi)dt ]\) \(P_{avg} = \frac{cE_OI_O}{4\pi}[t |^{2\pi/c}_0 - \frac{c}{2c}\sin(2ct - \Phi) |^{2\pi/c}_0]\) Since \(\sin(0) = 0\) and \(\sin(2\pi - \Phi) = \sin(-\Phi)\), we can simplify the expression to: \(P_{avg} = \frac{cE_OI_O}{4\pi}[2\pi/c - (\sin{(-\Phi)} - 0)] = \frac{1}{2}E_OI_O\cos\Phi\)
05

Compare with Given Choices

Comparing the P_avg value to the given answer choices, we can see that (a) matches the calculated average power: (a) \(\left[\frac{E_O I_O}{2}\right] \cos \Phi\) Hence the correct answer is (a).

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Most popular questions from this chapter

In a circuit \(\mathrm{L}, \mathrm{C}\) and \(\mathrm{R}\) are connected in series with an alternating voltage source of frequency \(\mathrm{f}\). The current leads the voltage by \(45^{\circ}\). The value of \(c\) is, (a) \([1 /\\{2 \pi \mathrm{f}(2 \pi \mathrm{fL}+\mathrm{R})\\}]\) (b) \([1 /\\{\pi f(2 \pi \mathrm{fL}+R)\\}]\) (c) \([1 /\\{2 \pi f(2 \pi f L-R)\\}]\) (d) \([1 /\\{\pi \mathrm{f}(2 \pi \mathrm{fL}-\mathrm{R})\\}]\)

In general in an alternating current circuit. (a) The average value of current is zero. (b) The average value of square of current is zero. (c) Average power dissipation is zero. (d) The phase difference between voltage and current is zero.

In a series resonant LCR circuit, the voltage across \(R\) is \(100 \mathrm{~V}\) and \(\mathrm{R}=1 \mathrm{k} \Omega\) with \(\mathrm{C}=2 \mu \mathrm{F}\). The resonant frequency 0 is \(200 \mathrm{rad} / \mathrm{s}\). At resonance the voltage across \(\mathrm{L}\) is. (a) \(40 \mathrm{~V}\) (b) \(250 \mathrm{~V}\) (c) \(4 \times 10^{-3} \mathrm{~V}\) (d) \(2.5 \times 10^{-2} \mathrm{~V}\)

Two identical circular loops of metal wire are lying on a table near to each other without touching. Loop A carries a current which increasing with time. In response the loop B......... (a) Is repelled by loop \(\mathrm{A}\) (b) Is attracted by loop \(\mathrm{A}\) (c) rotates about its centre of mass (d) remains stationary

In a LCR circuit capacitance is changed from \(\mathrm{C}\) to \(2 \mathrm{C}\). For the resonant frequency to remain unchanged, the inductance should be change from \(L\) to (a) \(4 \mathrm{~L}\) (b) \(2 \mathrm{~L}\) (c) \(L / 2\) (d) \(\mathrm{L} / 4\)
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