Power in AC Circuits: Practice Problem & Solution
In an AC circuit the e.m.f (e) and the current (i) at any instant are given respectively by $e = E_0 sin(\omega t)$, $i = I_0 sin(\omega t - phi)$. The average power in the circuit over one cycle of a.c. is: (2008)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Power in AC Circuits. Understanding the underlying formula is key to arriving at the correct answer below:
Average power is given by $P_{avg} = E_{rms} I_{rms} \cos\phi$.
Substituting RMS values: $E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.
$P_{avg} = \left(\frac{E_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) \cos\phi$.
$P_{avg} = \frac{E_0 I_0}{2} \cos\phi$.
Detailed Solution (Best of Understanding Complex Concepts)
Here is the detailed step-by-step solution for the average power in an AC circuit based on the question in your editor.
Concepts Involved
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Instantaneous Power ($P$): The power in an AC circuit at any specific instant of time is the product of the instantaneous voltage ($e$) and the instantaneous current ($i$).
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Average Power ($P_{avg}$): Since the instantaneous power fluctuates with time, we calculate the average power over one complete cycle (time period $T$). This is done by integrating the instantaneous power over one cycle and dividing by the time period.
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Trigonometric Identity: The calculation utilizes the product-to-sum formula: $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$.
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Average of Sinusoidal Functions: The average value of any purely sinusoidal function (like $\sin(\theta)$ or $\cos(\theta)$) over one complete cycle is always zero.
Step-by-Step Solution
Step 1: Write the given equations for instantaneous voltage and current.
The instantaneous e.m.f. and current are given by:
$$e = E_0 \sin(\omega t)$$
$$i = I_0 \sin(\omega t - \phi)$$
Step 2: Calculate the instantaneous power.
The instantaneous power $P$ is the product of $e$ and $i$:
$$P = e \cdot i$$
$$P = [E_0 \sin(\omega t)] \cdot [I_0 \sin(\omega t - \phi)]$$
$$P = E_0 I_0 \sin(\omega t) \sin(\omega t - \phi)$$
Step 3: Apply the trigonometric identity.
Multiply and divide the right side by 2 to apply the formula $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$:
$$P = \frac{E_0 I_0}{2} [2 \sin(\omega t) \sin(\omega t - \phi)]$$
Here, $A = \omega t$ and $B = (\omega t - \phi)$:
$$P = \frac{E_0 I_0}{2} [\cos(\omega t - (\omega t - \phi)) - \cos(\omega t + \omega t - \phi)]$$
$$P = \frac{E_0 I_0}{2} [\cos(\phi) - \cos(2\omega t - \phi)]$$
Step 4: Find the average power over one cycle.
To find the average power ($P_{avg}$), we take the time average of the instantaneous power equation over one full cycle:
$$P_{avg} = \text{Average of } \left( \frac{E_0 I_0}{2} [\cos(\phi) - \cos(2\omega t - \phi)] \right)$$
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The first term, $\cos(\phi)$, is a constant with respect to time, so its average over one cycle is simply $\cos(\phi)$.
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The second term, $\cos(2\omega t - \phi)$, is a time-varying sinusoidal wave with twice the angular frequency. The average value of a cosine wave over any complete cycle is zero.
Therefore:
$$P_{avg} = \frac{E_0 I_0}{2} [\cos(\phi) - 0]$$
$$P_{avg} = \frac{E_0 I_0}{2} \cos(\phi)$$
Step 5: Express in terms of RMS values (Optional but common).
Since the Root Mean Square (RMS) values for peak voltage and current are $E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$, the formula can also be written as:
$$P_{avg} = \frac{E_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cos(\phi)$$
$$P_{avg} = E_{rms} I_{rms} \cos(\phi)$$
The correct mathematical answer in the format of your options is $\frac{E_0 I_0}{2} \cos \phi$.
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