Question

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

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

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: \(P(t) = e(t)i(t) = E_O\sin(ct) I_O\sin(ct - \Phi)\)

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: \(P_{avg} = \frac{1}{T}\int_0^T P(t) dt\) \(P_{avg} = \frac{1}{\frac{2\pi}{c}}\int_0^{2\pi/c}\, E_I\sin(ct) I_O\sin(ct - \Phi) dt\)

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, \(P_{avg} = \frac{c}{2\pi}E_OI_O \int_0^{2\pi/c}\frac{1}{2}[\cos(0) - \cos(2ct - \Phi)]dt\) \(P_{avg} = \frac{cE_OI_O}{4\pi} \int_0^{2\pi/c}[\cos(0) - \cos(2ct - \Phi)]dt\)

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: \(P_{avg} = \frac{cE_OI_O}{4\pi}[2\pi/c - (\sin{(-\Phi)} - 0)] = \frac{1}{2}E_OI_O\cos\Phi\)

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).