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An alternating voltage given by V= 140 sin 314 t is connected across a pure resistor of 50 2. Find (i) the frequency of the source. (ii) the rms current through the resistor. |
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Answer» Solution :(i) As alternating voltage is GIVEN by `V = 140 sin 314 t` and `R = 50 Omega` `therefore` PEAK voltage `V_(m) =140 V` and angular FREQUENCY `omega = 314 rad s^(-1)` `therefore` Frequency `v= omega/(2pi) = 314/(2 xx 3.14) = 50 Hz` (ii) rms voltage `V_(rms) = V_(m)/sqrt(2) = 140/sqrt(2) V` `therefore` rms voltage `I_(rms) =V_(rms)/R = 140/(sqrt(2) xx 50) = 1.98 A` |
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