Power in AC Circuits: Practice Problem & Solution
In an a.c. circuit with phase voltage V and current I, the power dissipated is (1997)
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:
The average power dissipated in an A.C. circuit is $P = V_{rms} I_{rms} \cos\phi$.
Here, $\phi$ is the phase angle between the voltage and the current.
The term $\cos\phi$ is known as the power factor.
Therefore, power dissipated depends on the phase angle between V and I.
Detailed Solution (Best of Understanding Complex Concepts)
Concepts Involved
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Alternating Nature of AC Circuits: Unlike DC circuits where voltage and current are constant, in an AC circuit, the voltage and current vary sinusoidally with time. They can be represented as:
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$V(t) = V_m \sin(\omega t)$
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$I(t) = I_m \sin(\omega t - \phi)$Where $V_m$ and $I_m$ are peak values, $\omega$ is the angular frequency, and $\phi$ is the phase angle difference between the voltage and the current.
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Instantaneous vs. Average Power: The instantaneous power at any given moment is the product of voltage and current at that instant ($p = V(t) \cdot I(t)$). However, for practical purposes, we are interested in the average power dissipated over a complete cycle.
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Power Factor ($\cos\phi$): In AC circuits, the power dissipated is not just the product of the magnitudes of voltage and current. It is scaled by a factor called the "power factor," which is the cosine of the phase angle between the voltage and current.
Step-by-Step Solution
Step 1: Write the formula for instantaneous power.
The power dissipated at any instant $t$ is:
$$p(t) = V(t) \cdot I(t)$$
$$p(t) = [V_m \sin(\omega t)] \cdot [I_m \sin(\omega t - \phi)]$$
Step 2: Determine the average power dissipated over one complete cycle.
Using trigonometric identities and integrating over a full cycle (time period $T$), the average power $P$ is found to be:
$$P_{avg} = \frac{V_m}{\sqrt{2}} \cdot \frac{I_m}{\sqrt{2}} \cdot \cos\phi$$
Step 3: Substitute RMS values.
The terms $\frac{V_m}{\sqrt{2}}$ and $\frac{I_m}{\sqrt{2}}$ represent the Root Mean Square (RMS) values of voltage ($V_{rms}$) and current ($I_{rms}$), respectively. Often, when a problem states "voltage $V$ and current $I$" without specifying, it implies their RMS values.
$$P_{avg} = V_{rms} \cdot I_{rms} \cdot \cos\phi$$
Step 4: Analyze the dependence.
From the final formula $P_{avg} = V_{rms} I_{rms} \cos\phi$, we can see that the power dissipated is not simply the product of voltage and current ($V \cdot I$). It is directly multiplied by $\cos\phi$.
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If the circuit is purely resistive, $\phi = 0^\circ$ and $\cos\phi = 1$, making $P = VI$.
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If the circuit is purely inductive or capacitive, $\phi = 90^\circ$ and $\cos\phi = 0$, making the average power dissipated $0$.
Therefore, for a general AC circuit, the power dissipated inherently depends on the phase angle ($\phi$) between the voltage and the current.
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