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In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle ? Explain your answer |
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Answer» <P> Solution : Net POWER absorbed by an a.c. circuit is given by `P =V_(rms) I_(rms) COS phi`where `phi`is the phase difference between current and voltage.In EXERCISE 7.3, circuit is a purely inductive circuit where current LAGS behind the voltage by `pi/2`and circuit of Exercise 7.4 is a purely capacitive circuit in which current leads the voltage by `pi/2` Hence, in both circuits Power `=V_(rms) .I_(rms) .cos pi/2 = V_(rms).I_(rms)(0)= 0` (zero). |
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