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Pls show the integration for Iavg for half cycle

Dear Student,

Let dq = i dt q = \int_{0}^{\frac{T}{2}} i dt q = \int_{0}^{\frac{T}{2}} i_{0} \sin ωt dt q = i_{0} {[- \frac{\cos ωt}{ω}]}_{0}^{\frac{T}{2}} q = - \frac{i_{0}}{ω} [\cos ω \times \frac{T}{2} - \cos 0 °] q = - \frac{i_{0}}{ω} [\cos \frac{2 π}{T} \times \frac{T}{2} - 1] . . . . . . . . . . . . . . . . . . [as ω = \frac{2 π}{T}] q = - \frac{i_{0}}{ω} [cosπ - 1] q = - \frac{i_{0}}{ω} [- 1 - 1] = \frac{2 i_{0}}{ω} . . . . . . . . . . . . . . . . . (1) Also, q = i_{avg} \times \frac{T}{2} . . . . . . . . . . . . . . (2) from (1) and (2), i_{avg} \times \frac{T}{2} = \frac{2 i_{0}}{ω} i_{avg} \times \frac{T}{2} = \frac{2 i_{0}}{\frac{2 π}{T}} i_{avg} = \frac{2 i_{0}}{π}

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