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An alternating voltage given by `V=70 sin 100 pi t` is connected across a pure resistor of `25 Omega.` Find (i) the frequency of the source. (ii) the rms current through the resistor. |
Answer» `V_(0)=70, omega =100 pi` (i) `2piV=100 pi therefore V=50 Hz` (ii) `I_(rms)=(V_(rms))/(R ) =(V_(0)//sqrt(2))/(R )=(70)/(sqrt(2xx25))` `=(14)/(5xx1.414)=1.98` ampere. |
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