A LCR circuit has L = 10 mH, R = 3 ohm and C = 1 $π$ F connected in series to an a.c. source of the voltage 15 V. Calculate current amplitude and the average power dissipated per cycle at a frequency that is 10% lower than the resonant frequency.

Solution

Given, an LCR series circuit.
Inductor,L = 10 mH= 10 x 10^-3H
Resistor, R = 3 ohm
Capacitor,

C = 1 μF = 1 \times 10^{- 6} F

Voltage source, V = 15 V

Resonant frequency,

ω_{r} = \frac{1}{\sqrt{LC}}

Therefore,

ω_{r} = \sqrt{[\frac{1}{(10 \times 10^{- 3}) (1 \times 10^{- 6})}]} = 10^{4} / second .

Now, 10% less frequency will be

ω = 10^{4} - 10^{4} \times \frac{10}{100} = 9 \times 10^{3} / second .

At this frequency,

X_{L} = ωL = 9 \times 10^{3} \times (10 \times 10^{- 3}) = 90 ohm X_{C} = \frac{1}{ωC} = \frac{1}{(9 \times 10^{3}) (1 \times 10^{- 6})} = 111.11 ohm

∴

Impedence,

Z = \sqrt{[R^{2} + (X_{L} - X_{C})^{2}]}

= \sqrt{[(3)^{2} + (90 - 111.11)^{2}]} = 21.32 ohm

Current amplitude,

I_{0} = \frac{E_{0}}{Z} = \frac{15}{21.32} = 0.704 amp .

Average power,

P = \frac{1}{2} E_{0} I_{0} cosϕ

where ,

\cos ϕ = \frac{R}{Z} = \frac{3}{21.32} = 0.141

is the power factor.

Hence,
Average power dissipated per cycle is,

P = \frac{1}{2} \times 15 \times 0.704 \times 0.141 = 0.744 watt .

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