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प्रश्न
(i) An a.c. source of emf ε = 200 sin omegat is connected to a resistor of 50 Ω . calculate :
(1) Average current (`"I"_("avg")`)
(2) Root mean square (rms) value of emf
(ii) State any two characteristics of resonance in an LCR series circuit.
उत्तर
(1) Given :
`ε = 200 sin omegat`
`R = 50 Omega`
`ε_0 = 200`
`ε_(av) = ∫_(omegat = 0)^(2pi) (200 sin omegat d (omegat))/(2pi)`
= `(200[-cos (omegat)]_(0)^(2pi))/(2pi)`
= `(-200[cos 2pi - cos 0])/(2pi)`
`ε_(a"v") = 0`
`I_(a"v") = ε_(a"v")/50 = 0/50 = 0 A`
(2) RMS value of EMF
`ε_(RMS) = ε_0/sqrt(2)`
= `200/sqrt(2)`
= `200/1.41 = 141.84 V`
(ii) Two characteristic of resonance in series LCR
(a)At resonance impedance of the current is minimum
Z = R
(b) The net reactance is zero, so the circuit behaves as purely resistive circuit. At resonance frequency peak current is maximum.
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