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CBSE Class 12-science Answered

A series LCR circuit with L=0.12 H, C=480 nano farad , r=23 ohms is connected to a 230V variable frequency supply (a) What is the frequency for which current amplitude is maximum ? Obtain this maximum value (b) what is the source frequency for which average power absorbed by the circuit is maximum? obtain the value of this maximum power. (c) for which frequency of the source is the power transferred to the circuit half the power and resonance frequency. What is the current amplitude at these frequencies ? (d) What is the Q-Factor of the given circuit?
Asked by rutujasarangmehta | 15 Mar, 2016, 09:10: AM
answered-by-expert Expert Answer
(a) Given:
L=0.12 H
C = 480 nF = 480 × 10-9 F
R = 23 Ω
Ev = 230 volt
begin mathsize 14px style straight E subscript 0 space equals space square root of 2 straight E subscript straight v equals square root of 2 cross times 230 space volt end style
begin mathsize 14px style straight I subscript 0 equals fraction numerator straight E subscript 0 over denominator square root of straight R squared plus left parenthesis ωL minus 1 divided by ωC right parenthesis squared end root end fraction end style
I0 would be maximum, then
begin mathsize 14px style straight omega subscript straight r equals straight omega equals fraction numerator 1 over denominator square root of LC end fraction equals fraction numerator 1 over denominator square root of 0.12 cross times 480 cross times 10 to the power of negative 9 end exponent end root end fraction equals 4166.7 space rad space straight s to the power of negative 1 end exponent end style
begin mathsize 14px style straight I subscript 0 equals straight E subscript 0 over straight R equals fraction numerator square root of 2 cross times 230 over denominator 23 end fraction equals 14.14 space amp end style
(b) Average power absorbed by the circuit is maximum, when I = I0 at ω = ωr
begin mathsize 14px style straight P subscript av equals 1 half straight I subscript 0 superscript 2 space straight R equals 1 half cross times left parenthesis 14.14 right parenthesis squared cross times 23 equals 2299.3 space watt end style
(c) Power transferred to the circuit is half the power at resonance frequency, when
begin mathsize 14px style triangle straight omega equals fraction numerator straight R over denominator 2 straight L end fraction equals fraction numerator 23 over denominator 2 cross times 0.12 end fraction equals 95.83 space rad space straight s to the power of negative 1 end exponent end style
Therefore, frequencies at which power is transferred is half = begin mathsize 14px style straight omega subscript straight r plus-or-minus triangle straight omega equals 4166.7 plus-or-minus 95.83 equals 4262.53 space and space 4070.87 space rad space straight s to the power of negative 1 end exponent end style
Current amplitude at these frequencies is begin mathsize 14px style fraction numerator straight I subscript 0 over denominator square root of 2 end fraction equals fraction numerator 14.14 over denominator 1.414 end fraction equals 10 space amp end style
(d) begin mathsize 14px style straight Q equals fraction numerator straight omega subscript straight r space end subscript straight L over denominator straight R end fraction equals fraction numerator 4166.7 cross times 0.12 over denominator 23 end fraction equals 21.74 end style
Answered by Faiza Lambe | 15 Mar, 2016, 10:31: AM

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