(a) What do you understand by 'sharpness of resonance'for a series LCR
resonant circuit? How is it related with the quality factor 'Q'of the circuit?
using the graphs given in the diagram, explain the factors which affect it.
For which graph is the resistance (R) minimum?

(b) A2 \muF capacitor,100 \Omega resistor and 8 H inductor are connected in series
with an ac source. Find the frequency of the ac source for which the current drawn in the circuit is maximum.

If the peak value of emf of the source is 200 V, calculate the
(i)maximum current,and(ii) inductive and capacitive reactance of
the circuit at resonance.

 

 
 
 
 
 

Answers (1)
S safeer

(a) The circuit would be set to have a high sharpness of Resonance if the current in the circuit drops rapidly as the frequency of the applied AC Source shifts from its resonant value.
The sharpness of Resonance is measured by the quality factor(Q).
\therefore Q= \frac{1}{R}\sqrt{\frac{L}{C}}

R is minimum for C
(b) we have given; L= 8H \, and\, C= 2\mu F = 2\times 10^{-6}F
The frequency

 \left ( \nu\right )= \frac{1}{2\pi\times \sqrt{LC}}
where, L is inductor and 
           C is Capacitor
   \nu= \frac{1}{2\times 3.14\sqrt{8\times 2\times 10^{-6}}}
= 39\cdot 81Hz
(i)calculating the maximum current;
we have
V_{0}= 200V

At resonance Z=R
i_{max}= \frac{V_{0}}{Z}= \frac{V_{0}}{R}
i_{max}= \frac{200}{100}
(ii)To determine the inductive and Capacitance reactance;
we know:
at resonance
X_{L}= X_{C}
X_{L}= wL= 2\pi vL
on putting the values we have,
X_{L}= X_{C}= 2\times\pi\times 39\cdot 81\times 8= 2000\Omega

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