Question

An inductor of 200 mh, capacitor of 400 f and a resistor of 10 are connected in series to ac source of 50 v of variable frequency. Calculate the i) angular frequency at which maximum power dissipation occurs in the circuit and the corresponding value of the effective current, and ii) value of q-factor in the circuit.

Answer 1

Given that,

Inductance, $L=200\ mH=0.2\ H$

Capacitance, $C=400\ \mu F=4\times 10^{-4}\ F$

Resistance, R = 10 ohms

To find,

Angular frequency at which maximum power dissipation and the value of q-factor in the circuit.

Solution,

(a) Maximum power dissipation occurs in the circuit at resonance i.e.

$\omega L=\dfrac{1}{\omega C}\\\\\omega=\dfrac{1}{\sqrt{LC} }\\\\\omega=\dfrac{1}{\sqrt{0.2\times 4\times 10^{-4}} }\\\\\omega=111.8\ rad/s$

(b) Q factor of the circuit is given by :

$Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}} \\\\Q=\dfrac{1}{10}\sqrt{\dfrac{0.2}{4\times 10^{-4}}} \\\\Q=2.23$

Therefore, this is the required explanation.

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LCR circuit

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