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NCERT Exemplar · Q12

Q.In a series LCR circuit, the peak current ImaxI_{max} is plotted against the driving angular frequency ω\omega. The plot is a resonance (bell-shaped) curve: the vertical axis ImaxI_{max} is marked in amperes at 0.50.5 and 1.01.0, and the horizontal axis ω\omega is marked in rad/s at 0.5, 1.0, 1.50.5,\ 1.0,\ 1.5 and 2.02.0. Starting near zero, the current rises to a peak of about 1.01.0 A close to ω0≈1.2\omega_0\approx 1.2 rad/s and then falls back towards zero. Find the bandwidth of this resonance curve and indicate the half-power points on it.

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Bandwidth of a resonance curve is the frequency interval between the two half-power points, where the current drops to Imax/2I_{max}/\sqrt2. With Imax=1.0I_{max}=1.0 A the half-power level is ≈0.71\approx0.71 A; the two frequencies at that level are about 1.01.0 and 1.41.4 rad/s, so the bandwidth is roughly 0.40.4 rad/s.

Concept: bandwidth and half-power points

At resonance (ω0\omega_0) the series LCR current is maximum, ImaxI_{max}. The half-power points ω1\omega_1 and ω2\omega_2 are the frequencies on either side of ω0\omega_0 where the power delivered is half the maximum. Since power ∝I2\propto I^2, half power means

I=Imax2≈0.707 Imax.I=\frac{I_{max}}{\sqrt2}\approx0.707\,I_{max}.

The bandwidth is Δω=ω2−ω1\Delta\omega=\omega_2-\omega_1.

Reading the curve

  1. The peak height is Imax=1.0I_{max}=1.0 A, occurring near ω0≈1.2\omega_0\approx1.2 rad/s.
  2. The half-power current is 1.02≈0.71\dfrac{1.0}{\sqrt2}\approx0.71 A.
  3. Draw a horizontal line at I=0.71I=0.71 A. It cuts the rising side of the curve at ω1≈1.0\omega_1\approx1.0 rad/s (right at the printed 1.01.0 gridline) and the falling side at ω2≈1.4\omega_2\approx1.4 rad/s (a little short of the printed 1.51.5 gridline). These are the two points to mark on the graph. …

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