An alternating e.m.f. of frequency f=12πLC is applied to a series LCR circuit. For this frequency of the applied e.m.f.

For f=12πLC ,ω=2πf=1LC=ω0This is the condition of resonance of LCR circuitAt resonance, Z=R, ans so V=iZ=iRQuality factor = Voltage across capacitor at resonanceVoltage across resistor at resonance =Voltage across inducyor at resonanceVoltage across resistor at resonance

An alternating e.m.f. of frequency f=12πLC is applied to a series LCR circuit. For this frequency of the applied e.m.f.

For f=12πLC ,ω=2πf=1LC=ω0This is the condition of resonance of LCR circuitAt resonance, Z=R, ans so V=iZ=iRQuality factor = Voltage across capacitor at resonanceVoltage across resistor at resonance =Voltage across inducyor at resonanceVoltage across resistor at resonance

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An alternating e.m.f. of frequency $f (= \frac{1}{2 π \sqrt{LC}})$ is applied to a series LCR circuit. For this frequency of the applied e.m.f.

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The circuit is at resonance and its impedance is made up only of a reactive part

The quality factor of the circuit is $ω L / R$ or $1 / ω C R$ and this is a measure of the voltage magnification (produced by the circuit at resonance) as well as the sharpness of resonance of the circuit

The sum of the p.d.'s across the inductance and capacitance equals the applied e.m.f. which is 180° ahead of phase of the current in the circuit

The current in the circuit is in phase with the applied e.m.f. and the voltage across $R$ equals this applied emf

answer is B, D.

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Detailed Solution

For $f (= \frac{1}{2 π \sqrt{LC}})$ , $ω = 2 π f = \frac{1}{\sqrt{L C}} = ω_{0}$

This is the condition of resonance of $L C R$ circuit

At resonance, $Z = R, a n s s o V = i Z = i R$

Quality factor = $\frac{V o l t a g e a c r o s s c a p a c i t o r a t r e s o n a n c e}{V o l t a g e a c r o s s r e s i s t o r a t r e s o n a n c e} = \frac{V o l t a g e a c r o s s i n d u c y o r a t r e s o n a n c e}{V o l t a g e a c r o s s r e s i s t o r a t r e s o n a n c e}$