The rms value for the saw-tooth voltage of peak value $V_{0}$ from t = 0 to t = 2T as shown in figure is $\frac{V_{0}}{\sqrt{x}}$ Find the value of x.

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answer is 3.

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Detailed Solution

x = 3

general equation of V

$V = \frac{V_{0}}{T / 2} t - V_{0} = \frac{2 V_{0}}{T} t - V_{0}$

$V_{rms} = {[\frac{\int_{0}^{T} V^{2} dt}{T}]}^{\frac{1}{2}} = {[\frac{\int_{0}^{T} {(\frac{2 V_{0}}{T} t - V_{0})}^{2} dt}{T}]}^{\frac{1}{2}} = \frac{V_{0}}{\sqrt{3}}$

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