Resonance curve of an LCR circuit

Question

The resonance curve of a series LCR circuit is shown below. What happens to the curve when the resistance is halved and capacitance is doubled?

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a. Shifts to the left and peak sharpens
b. Shifts to right and peak broadens
c. Remains same
d. None

Solution

The resonant frequency of a series LCR circuit is given by:
!!ω = 1/sqrt(LC)!! ...eq.1

Since the Capacitance increases but the Inductance is constant, the frequency decreases. Hence the curve shifts to the left, towards lower frequencies.

The sharpness of the peak is determined by the quantity called quality factor, !!Q!!, which is given by:
!!Q = ωL/R!!

!!= 1/ωCR!! (using eq.1)

Greater the value of !!Q!!, sharper will be the peak.

When !!C!! is doubled and !!R!! is halved,

!!Q\^' = 1/ω\^'2CR/2!!

!!= 1/ω\^'!!

Since !!ω\^'!! is less than !!ω!!, !!Q\^'!! will be greater than !!Q!!. Therefore we will get a sharper peak.

Hence the answer is (a), ie., the curve shifts to left and the peak sharpens.

Application

This concept is used while tuning the radio. When we tune the radio, we are actually changing the capacitance of the circuit. This changes the resonant frequency, and hence we get to listen to a different frequency radio station which is at resonance with our circuit.

This principle is also used in metal detectors and security checks in the airports.

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