A sinusoidal voltage of 200V ,50Hz is applied to a series LCR circuit

1.	A sinusoidal voltage of 200V ,50Hz is applied to a series LCR circuit . If R= 60ohm ,L=50.96mH , C= 398 microF , then(a) what is the frequency of source at resonance.(b) calculate impedance, current and dissipated power for resonance situation.
Answer» Answer: Part i) Resonance frequency is F = 35.33 Hz Part ii) At resonance condition net impedence is purely resistive in nature Z = R = 60 ohm current in the circuit is given as i = 3.33 A Power consumed in resonance condition is given as P = 666.67 W Explanation: Part i) When resonance OCCURS in series LCR circuit then power dissipated ACROSS the resistor is maximum so it is purely resistive circuit so we have $f = \frac{1}{2\pi}\sqrt{\frac{1}{LC}}$ now we know that L = 50.96 mH $C = 398 \mu F$ now we have $f = \frac{1}{2\pi}\sqrt{\frac{1}{(50.96 \times 10^{-3})(398\times 10^{-6})}$ $f = 35.33 Hz$ Part ii) At resonance condition net impedence is purely resistive in nature so we have $Z = R = 60 ohm$ current in the circuit is given as $i = \frac{V}{Z}$ $i = \frac{200}{60}$ $i = 3.33 A$ Power consumed in resonance condition is given as $P = i V$ $P = 3.33(200)$ $P = 666.67 W$ #Learn Topic : AC Resonance circuit brainly.in/question/6878333