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  • A LCR circuit has L = 10 mH, R = 3 ohm and connected in series to a source of&nbs

A LCR circuit has L = 10 mH, R = 3 ohm and \mathrm{C=1\mu F} connected in series to a source of \mathrm{15 \cos \omega t \text { volt }}. Calculate the current amplitude and the average power dissipated per cycle at a frequency that is 10% lower than the resonant frequency.

Option: 1

0.50 W


Option: 2

0.60 W


Option: 3

0.70 W


Option: 4

0.75 W


Answers (1)

best_answer

\text { Resonant frequency } \omega_{\mathrm{R}}=\frac{1}{\sqrt{L C}}

\mathrm{Here \; L=10 \mathrm{mH}=10 \times 10^{-3} \mathrm{H} \: and\: \mathrm{C}=1 \mu \mathrm{F}=1 \times 10^{-6} \mathrm{~F}.}

 

\mathrm{\text { and } \mathrm{C}=1 \mu \mathrm{F}=1 \times 10^{-6} \mathrm{~F} \text {. }}

\mathrm{\therefore \quad \omega_R=\sqrt{\frac{1}{\left(10 \times 10^{-3}\right)\left(1 \times 10^{-6}\right)}}=10^4 / \mathrm{s}}

Now 10% less frequency will be

\mathrm{\omega=10^4-10^4 \times \frac{10}{100}=9 \times 10^3 / \mathrm{s}}

At this frequency,

\mathrm{\begin{aligned} & X_L=\omega L=9 \times 10^3 \times\left(10 \times 10^{-3}\right)=90 \mathrm{ohm} \\ & X_C=\frac{1}{\omega C}=\frac{1}{\left(9 \times 10^3\right)\left(1 \times 10^{-6}\right)}=111.11 \mathrm{ohm} \\ & Z=\sqrt{\left[R^2+\left(X_L-X_C\right)^2\right]}=\sqrt{(3)^2+(90-111.11)^2}=21.32 \mathrm{ohm} \end{aligned}}

Current amplitude,

\mathrm{I_0=\frac{E_0}{Z}=\frac{15}{21.23}=0.704 \mathrm{amp}}

\mathrm{Average\; power, P=\frac{1}{2} E_0 I_0\; \cos \phi\\\\\; where\; \; \cos \phi=\frac{R}{Z}=\frac{3}{21.32}=0.141}

 

\mathrm{P=\frac{1}{2} \times 15 \times 0.704 \times 0.141=0.75 \text { watt. }}

 

Posted by

Shailly goel

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