Physics · Alternating Current · NEET
Q is a dimensionless number that measures how sharp the resonance is. For a series LCR circuit, NCERT gives Q = w0*L/R = 1/(w0*C*R), where w0 = 1/sqrt(L*C) is the resonant angular frequency. A large Q means the current peak at resonance is tall and narrow, so the circuit responds strongly to a very small band of frequencies near w0 and weakly to all others.
Sharpness = how narrow the current-vs-frequency peak is. It is measured by the bandwidth 2*(delta-w) = R/L, which is the frequency width where power drops to half its peak value. Smaller bandwidth = sharper peak. Since Q = w0*L/R = w0 / (bandwidth), a larger Q directly means a smaller bandwidth and a sharper resonance. So Q and sharpness say the same thing.
Better tuning means the circuit selects a very narrow band of frequencies - useful in radios and communication. Better tuning = higher Q = smaller bandwidth (R/L). To get high Q you want low R and high L (and the right C). The 2016 NEET question uses exactly this: the option with the smallest R and largest L wins because it gives the smallest R/L.
They are the two frequencies w0 - delta-w and w0 + delta-w on either side of resonance where the power falls to half the peak power (current falls to 1/sqrt(2) of its peak). The gap between them is the bandwidth = R/L. In the 2021 PYQ, w0 = 50 rad/s and bandwidth R/L = 8, so the half-power frequencies are 50 - 4 = 46 and 50 + 4 = 54 rad/s.
No. Q = w0*L/R depends only on the circuit components L, C and R, not on the applied voltage or current. Changing the 230 V source strength changes the current amplitude but not Q or the shape sharpness of the resonance curve.
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?
A series LCR circuit containing 5.0 H inductor, 80 uF capacitor and 40 ohm resistor is connected to a 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Q = w0*L/R has units (rad/s * H)/ohm which cancel out, so Q is a pure dimensionless number.
Bandwidth = 2*(delta-w) = R/L (in rad/s). It equals the gap between the two half-power frequencies, and Q = w0 / bandwidth.
Lower the resistance R, or increase the inductance L (keeping resonance fixed by adjusting C). Q = w0*L/R rises when R falls or L rises.
A high-Q tuning circuit has a very narrow bandwidth, so it responds only to the one station's frequency and rejects nearby stations, giving clear reception.
No. Q depends only on L, C and R. Changing the source voltage changes the current size but not the sharpness of resonance.
