Consider the symmetric about the time axis voltage waveform shown in the figure below.The mean and root mean square values of the waveform are

\frac{V_{a}}{5}, V_{a}

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V_{a}, \sqrt{V_{a}}

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V_{a}, V_{a}^{2}

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\frac{.1}{V_{a}}, V_{a}

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Solution

The correct option is A $\frac{V_{a}}{5}, V_{a}$
For average value, $V_{m} = \frac{A r e a u n d e r c u r v e}{T o t a l t i m e}$

$\frac{V_{a} \times T - V_{a} \times (2 T - T) + V_{a} \times (3 T - 2 T) - V_{a} (4 T - 3 T) + V_{a} (5 T - 4 T)}{5 T}$
$V_{m} = \frac{V_{a}}{5}$

For r.m.s value, $V_{r m s} = \sqrt{\frac{V_{a}^{2} + V_{a}^{2} + V_{a}^{2} + V_{a}^{2} + V_{a}^{2}}{5}}$
$V_{r m s} = V_{a}$

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Consider the symmetric about the time axis voltage waveform shown in the figure below.The mean and root mean square values of the waveform are

The correct option is A Va5,VaFor average value, Vm=Area under curveTotal timeVa×T−Va×(2T−T)+Va×(3T−2T)−Va(4T−3T)+Va(5T−4T)5TVm=Va5For r.m.s value, Vrms=√V2a+V2a+V2a+V2a+V2a5 Vrms=Va