Find (a) the average and (b) rms value for the Saw-tooth voltage of peak value $V_{0}$ as shown in the figure.

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Solution

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As the equation of the Saw-tooth wave shown in the figure will be $V=T2V_{0} t−V_{0}=V_{0}(T2ty −1)$ So (a) $V_{av}=∫_{0}dt∫_{0}Vdt =T2 ∫_{0}V_{0}(T2ty −1)dt=∣∣∣∣∣∣ 2V_{0} ∣∣∣∣∣∣ $ and (b) $V_{rms}=[∫_{0}dt∫_{0}Vdt ]_{1/2}=3 V_{0} $