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Question: The voltage over a cycle varies as \[V = V_{0}\sin\omega\text{t for 0 } \leq t \leq \ \frac{\pi}{\o...

The voltage over a cycle varies as

V=V0sin⁡ωt for 0 ≤t≤ πωV = V_{0}\sin\omega\text{t for 0 } \leq t \leq \ \frac{\pi}{\omega}

=−V0sin⁡ωt for πω ≤t≤ 2πω= - V_{0}\sin\omega\text{t for }\frac{\pi}{\omega}\ \leq t \leq \ \frac{2\pi}{\omega}

The average value of the voltage for one cycle is

A

V02\frac{V_{0}}{\sqrt{2}}

B

V02\frac{V_{0}}{2}

C

Zero

D

2V0π\frac{\text{2}\text{V}_{0}}{\pi}

Answer

2V0π\frac{\text{2}\text{V}_{0}}{\pi}

Explanation

Solution

: The average value of the voltage is

Vav(vkSlr)=∫02π/ωVdt∫02π/ωdt=∫0π/ωV0sin⁡ωtdt+∫π/ω2π/ω(−V0sin⁡ωt)dt2πω=ω2π[∣−V0cos⁡ωtω∣π/ω+∣V0cos⁡ωtω∣π/ω2π/ω]=V02π[−cos⁡π+cos⁡0+cos⁡2π−cos⁡π]=2V0πV_{av(vkSlr)} = \frac{\int_{0}^{2\pi/\omega}{Vdt}}{\int_{0}^{2\pi/\omega}{dt}} = \frac{\int_{0}^{\pi/\omega}{V_{0}\sin\omega tdt + \int_{\pi/\omega}^{2\pi/\omega}{( - V_{0}\sin\omega t)dt}}}{\frac{2\pi}{\omega}} = \frac{\omega}{2\pi}\left\lbrack {\left| \frac{- V_{0}\cos\omega t}{\omega} \right|^{\pi/\omega}}_{} + \left| \frac{V_{0}\cos\omega t}{\omega} \right|_{\pi/\omega}^{2\pi/\omega} \right\rbrack = \frac{V_{0}}{2\pi}\lbrack - \cos\pi + \cos 0 + \cos 2\pi - \cos\pi\rbrack = \frac{2V_{0}}{\pi}