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A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Omega is connected to a 230 V variable frequency supply. What is the source frequency for which current amplitude is maximum. Obtain this maximum vlaue. |
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Answer» SOLUTION :(a) `omega_(0)=4167 "rad s"^(-1), v_(0)=663Hz` `I_(0)^("max")=14.1A` (b) `barP=(1//2)I_(0)^(2)R` which is maximum at the same frequency (663 Hz) for which `I_(0)` is maximum `barP_("max")=(1//2)(I_("max"))^(2)R=2300W`. (c ) At `omega=omega_(0)pm Deltaw` [Approximation good if `(R//2L) lt lt omega_(0)`]. `Delta omega=R//2L=95.8"rad s"^(-1), Deltav=Delta omega//2pi =15.2Hz.` POWER ABSORBED is half the peak power at n = 648 Hz and 678 Hz. At these frequencies, current AMPLITUDE is`(1//sqrt(2))` times `I_(0)^("max")`, i.e., current amplitude(at half the peak power points) is 10 A. (d) Q = 21.7 |
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