Question:
The fundamental frequency and peak voltage of an inverter are f and \(\text{V}_\text{m}\) respectively. When it contains all odd harmonics, then its frequency and rms voltage of \(5^{\text{th}}\) harmonic are
The fundamental frequency and peak voltage of an inverter are f and \(\text{V}_\text{m}\) respectively. When it contains all odd harmonics, then its frequency and rms voltage of \(5^{\text{th}}\) harmonic are
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Harmonic Rule of Thumb:
- Frequency always scales up: \( f_n = n \cdot f \)
- Amplitude always scales down: \( \text{RMS}_n = \frac{V_{\text{peak}, n}}{\sqrt{2}} = \frac{V_m}{n\sqrt{2}} \)
For \(n=5\), this gives \(5f\) and \(\frac{V_m}{5\sqrt{2}}\) instantly!
Updated On: Jun 25, 2026
- \( 5f \text{ and } \frac{5V_m}{\sqrt{2}} \text{ respectively} \)
- \( \frac{f}{5} \text{ and } \frac{V_m}{5} \text{ respectively} \)
- \( \frac{f}{5} \text{ and } \frac{V_m}{5\sqrt{2}} \text{ respectively} \)
- \( 5f \text{ and } \frac{V_m}{5\sqrt{2}} \text{ respectively} \)
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The Correct Option is D
Solution and Explanation
Concept:
According to Fourier series analysis, any non-sinusoidal periodic waveform can be broken down into a sum of pure sinusoidal components consisting of a fundamental frequency along with higher integer multiples called harmonics.
For an output voltage waveform from an inverter (such as a square wave):
• The frequency of the $n^{\text{th}}$ harmonic component is $n$ times the fundamental frequency: \[ f_n = n \cdot f \]
• The peak amplitude voltage of the $n^{\text{th}}$ harmonic component is inversely proportional to its harmonic order: \[ V_{n(\text{peak})} = \frac{V_m}{n} \]
• The corresponding Root-Mean-Square (RMS) voltage for any pure sine wave component is its peak amplitude divided by the square root of two: \[ V_{n(\text{rms})} = \frac{V_{n(\text{peak})}}{\sqrt{2}} \]
Step 1: Calculating the frequency of the 5th harmonic.
We are looking for the properties of the $5^{\text{th}}$ harmonic component ($n = 5$). Using the frequency relation: \[ f_5 = 5 \times f = 5f \]
Step 2: Finding the peak voltage of the 5th harmonic.
The peak voltage amplitude drops by a factor of 5 for the fifth harmonic: \[ V_{5(\text{peak})} = \frac{V_m}{5} \]
Step 3: Computing the RMS voltage of the 5th harmonic.
To find the RMS value, divide the peak voltage of this harmonic component by $\sqrt{2}$: \[ V_{5(\text{rms})} = \frac{V_{5(\text{peak})}}{\sqrt{2}} = \frac{\left(\frac{V_m}{5}\right)}{\sqrt{2}} = \frac{V_m}{5\sqrt{2}} \]
Step 4: Combining the answers.
The frequency is $5f$ and the RMS voltage is $\frac{V_m}{5\sqrt{2}}$, which matches option (4). Hence, the correct choice is option (4).
• The frequency of the $n^{\text{th}}$ harmonic component is $n$ times the fundamental frequency: \[ f_n = n \cdot f \]
• The peak amplitude voltage of the $n^{\text{th}}$ harmonic component is inversely proportional to its harmonic order: \[ V_{n(\text{peak})} = \frac{V_m}{n} \]
• The corresponding Root-Mean-Square (RMS) voltage for any pure sine wave component is its peak amplitude divided by the square root of two: \[ V_{n(\text{rms})} = \frac{V_{n(\text{peak})}}{\sqrt{2}} \]
Step 1: Calculating the frequency of the 5th harmonic.
We are looking for the properties of the $5^{\text{th}}$ harmonic component ($n = 5$). Using the frequency relation: \[ f_5 = 5 \times f = 5f \]
Step 2: Finding the peak voltage of the 5th harmonic.
The peak voltage amplitude drops by a factor of 5 for the fifth harmonic: \[ V_{5(\text{peak})} = \frac{V_m}{5} \]
Step 3: Computing the RMS voltage of the 5th harmonic.
To find the RMS value, divide the peak voltage of this harmonic component by $\sqrt{2}$: \[ V_{5(\text{rms})} = \frac{V_{5(\text{peak})}}{\sqrt{2}} = \frac{\left(\frac{V_m}{5}\right)}{\sqrt{2}} = \frac{V_m}{5\sqrt{2}} \]
Step 4: Combining the answers.
The frequency is $5f$ and the RMS voltage is $\frac{V_m}{5\sqrt{2}}$, which matches option (4). Hence, the correct choice is option (4).
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