An inductor 200 μH, capacitor 500 μF, resistor 10 Ω are connected in series with a 100 V, variable frequency a.c. source. Calculate the
(i) frequency at which the power factor of the circuit is unity
(ii) current amplitude at this frequency
(iii) Q-factor

Solution

Given,
Inductor, L = 200 μH
Capacitor, C = 500 μF
Resistor, R = 10 Ω
Effective voltage, V = 100 V

(i) Power factor, $\cos θ = \frac{R}{Z} = 1$

So, $R = Z$

$\Rightarrow$ $R = \sqrt{R^{2} + {(ωL - \frac{1}{ωC})}^{2}}$
$R^{2} = R^{2} + {(ωL - \frac{1}{ωC})}^{2}$
   $ωL = \frac{1}{ωC}$

   $ω^{2} = \frac{1}{LC} or ω = \frac{1}{\sqrt{LC}}$
$2 πv = \frac{1}{\sqrt{LC}}$

$v = \frac{1}{2 π \sqrt{LC}}$

$ω_{0} = \frac{1}{2 π \sqrt{LC}} = \frac{1}{\sqrt{200 \times 10^{- 6} \times 500 \times 10^{- 6}}} = 3.16 \times 10^{- 3} rad s^{- 1}$

$ω_{0} = 3.16 \times 10^{- 3} rad / s$

(ii) The current amplitude at this frequency,

$I_{0} = \frac{V}{R} = \frac{100}{10} = 10 A$

(iii) The Q-factor,

$Q = \frac{X_{L}}{R} = \frac{ω_{0} L}{R} = \frac{3.16 \times 10^{- 3} \times 200 \times 10^{- 6}}{10} = 6.32 \times 10^{- 8}$

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An inductor 200 μH, capacitor 500 μF, resistor 10 Ω are connected in series with a 100 V, variable frequency a.c. source. Calculate the
(i) frequency at which the power factor of the circuit is unity
(ii) current amplitude at this frequency
(iii) Q-factor

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Electrostatic Potential And Capacitance

An inductor 200 μH, capacitor 500 μF, resistor 10 Ω are connected in series with a 100 V, variable frequency a.c. source. Calculate the(i) frequency at which the power factor of the circuit is unity(ii) current amplitude at this frequency(iii) Q-factor

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An inductor 200 μH, capacitor 500 μF, resistor 10 Ω are connected in series with a 100 V, variable frequency a.c. source. Calculate the
(i) frequency at which the power factor of the circuit is unity
(ii) current amplitude at this frequency
(iii) Q-factor