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An alternating voltage given by V =140 sin 314 t isconnected across a pure resistor of 50 Omega find : (i)the frequency of the source. (ii) The rms current through the resistor. |
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Answer» Solution :(i) `2pi v = omega` `rArr 2pi v = 314 rad s^(-1)` `rArr v = 50 Hz` (II) `I_(rms) = (V_(rms))/(R )` Where `V_(rms) = (V_(0))/(sqrt(2))` `I_(rms) = (140)/(sqrt(2) xx 50) = 1.98 ~~ 2A` |
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