Question

The peak voltage of the ac source is equal to:

• A. the value of voltage supplied to the circuit
• B. the rms value of the ac source
• C. √2 times the rms value of the ac source
• D. \(\frac{1}{√2}\) times the rms value of the ac source

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Alternating Current (AC) changes its magnitude and direction periodically over time, typically following a sinusoidal wave pattern.
Because the value is constantly changing, we use different metrics to describe it:
1. Peak Voltage (\(V_0\)): The maximum absolute value of voltage in a cycle.
2. RMS Voltage (\(V_{rms}\)): The "Root Mean Square" value, which represents the effective DC-equivalent voltage that would produce the same heating effect in a resistor.
Key Formula or Approach:
For a purely sinusoidal alternating voltage \(V(t) = V_0 \sin(\omega t)\), the relationship between the peak value and the RMS value is derived by integrating the square of the function over a period.
The standard relationship is:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]
Step 2: Detailed Explanation:
The question asks for the peak voltage (\(V_0\)) in terms of the RMS value.
Starting from the definition:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]
We can rearrange this equation to solve for \(V_0\) by multiplying both sides by \(\sqrt{2}\):
\[ V_0 = \sqrt{2} \times V_{rms} \]
This means that at its maximum point in a cycle, the voltage reaches a value that is approximately 1.414 times higher than the rated RMS value.
Step 3: Final Answer:
The peak voltage of an AC source is \(\sqrt{2}\) times its RMS value.