Answer :

Ans. 20

For the given Quasi square waveform,

$V_n=\frac{4V_s}{n\pi }\mathrm{c}\mathrm{o}\mathrm{s}\left(n\sigma \right)$

Given that, the magnitude of 3rd harmonic component from the output voltage is zero i.e.V3 = 0

$V_3=\frac{4V_s}{3\pi }\cos n\sigma =0$

⇒ cos (3σ) = 0

⇒ σ = π/6

The magnitude of the fifth harmonic component from the output voltage is,

$V_5=\frac{4V_s}{5\pi }\cos \frac{5\pi }{6}$

The magnitude of the fundamental harmonic component from the output voltage is,

$V_1=\frac{4V_s}{\pi }\cos \frac{\pi }{6}$

$\frac{V_5}{V_1}=\frac{\frac{4V_s}{5\pi }\cos \frac{5\pi }{6}}{\frac{4V_s}{\pi }\cos \frac{\pi }{6}}=-0.2$

Now, the value of the magnitude of the 5th harmonic component as a percentage of the magnitude of the fundamental component is

$\mathrm{\%}\left|\frac{V_5}{V_1}\right|=20\mathrm{\%}$