The Question is from neet-physics alternating-current , and it was created by the author P.NARASIMHA

Question

An AC circuit with a non-ohmic resistive element has a current described by $i(t) = 5\sin(\omega t + \pi/3)$. If the voltage across the element is $v(t) = 10\sin(\omega t)$, which statement BEST describes the relationship between instantaneous power, average power, and RMS power?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: 2
## CONCEPT FRAMEWORK
 **Primary Concept**: AC power calculations, instantaneous power, average power, RMS power
 **Related Topics**:  Phase difference in AC circuits, non-ohmic elements
 **Prerequisites**: Understanding of sinusoidal functions, trigonometric identities, and power calculations in AC circuits.
## SOLUTION PATHWAY
**Given Information**:
  $i(t) = 5\sin(\omega t + \pi/3)$
  $v(t) = 10\sin(\omega t)$
**Method & Steps**:
 1. Calculate the instantaneous power.
  Reasoning: Instantaneous power is given by $p(t) = v(t)i(t)$. 
  Formula/Rule: $p(t) = v(t)i(t) = 10\sin(\omega t) 	imes 5\sin(\omega t + \pi/3) = 50\sin(\omega t)\sin(\omega t + \pi/3)$
 2. Determine if the instantaneous power is constant or fluctuates.
  Working: The expression for $p(t)$ contains a product of sine functions with different phases, indicating that the instantaneous power will fluctuate with time.
  Key Point: The product of two sinusoidal functions results in a waveform that is not purely sinusoidal, hence the fluctuation in power.
 3. Calculate the average power.
  Reasoning: Average power is given by $P_{avg} = \frac{1}{T}\int_{0}^{T} p(t)dt$, where T is the period.  Due to the phase difference between voltage and current, the average power will not be zero.  However, a more precise calculation is needed to confirm this.
  Working: Using trigonometric identities, we can rewrite $\sin(\omega t)\sin(\omega t + \pi/3)$ as a sum of cosine functions. The integral of the cosine terms over a full period is zero. Therefore, the average power will be non-zero due to the phase difference (π/3).
  Key Point: The average power is the DC component of the instantaneous power.
 4. Determine the RMS power.
  Reasoning:  RMS power is related to the RMS values of voltage and current.  $P_{RMS} = V_{RMS}I_{RMS}$ where $V_{RMS} = \frac{V_m}{\sqrt{2}}$ and $I_{RMS} = \frac{I_m}{\sqrt{2}}$.
  Working: $V_{RMS} = \frac{10}{\sqrt{2}}$ and $I_{RMS} = \frac{5}{\sqrt{2}}$. Therefore, $P_{RMS} = \frac{10}{\sqrt{2}}	imes\frac{5}{\sqrt{2}} = 25$. The average power will be less than the RMS power. 
  Result: The instantaneous power fluctuates, the average power is non-zero, and the RMS power is greater than the average power.
  Verification: The solution aligns with the general behavior of AC circuits with non-zero phase angles between voltage and current.
 ## OPTION ANALYSIS
A :x: Incorrect. Instantaneous power is not constant due to the phase difference.
B ✓: Correct. The instantaneous power fluctuates, average power is non-zero (due to the phase difference), and RMS power is greater than the average power.
C :x: Incorrect. Average power is non-zero due to the phase difference.
D :x: Incorrect. Instantaneous power fluctuates, and RMS power is greater than the average power.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Assuming constant instantaneous power in AC circuits with a phase difference.: Remember that instantaneous power depends on the product of instantaneous voltage and current, which fluctuate differently based on the phase difference.
  2.Ignoring the phase difference when calculating average power: The phase difference significantly affects the average power; it will only be zero if the phase difference is 90 degrees.
 **Quick Tips**:
  1.Remember the relationship between RMS, average, and peak values for sinusoidal waveforms.
  2.Use trigonometric identities to simplify calculations involving products of sinusoidal functions.

P.NARASIMHA · Answer

The instantaneous power is constant, the average power is zero, and the RMS power is equal to the peak power.

P.NARASIMHA · Answer

The correct Option is "The instantaneous power fluctuates, the average power is non-zero, and the RMS power is greater than the average power."

P.NARASIMHA · Answer

The instantaneous power fluctuates, the average power is zero, and the RMS power is equal to the average power.

P.NARASIMHA · Answer

The instantaneous power is constant, the average power is non-zero, and the RMS power is less than the average power.

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