Question

A 2 µF capacitor, 100 Ω resistor and 8 H inductor are connected in series with an AC source.
(i) What should be the frequency of the source such that current drawn in the circuit is maximum? What is this frequency called?
(ii) If the peak value of e.m.f. of the source is 200 V, find the maximum current.
(iii) Draw a graph showing variation of amplitude of circuit current with changing frequency of applied voltage in a series LRC circuit for two different values of resistance R₁ and R₂ (R₁ > R₂).
(iv) Define the term 'Sharpness of Resonance'. Under what condition, does a circuit become more selective?

Answer 1

(i)  To draw maximum current from a series LCR circuit, the circuit must be at resonance. For resonance, we know at particular frequency 

\[X_L = X_C\] The frequency of the source will be

\[\nu = \frac{1}{2\pi\sqrt{LC}}\]

\[ = \frac{1}{2 \times 3 . 14 \times \sqrt{8 \times 2 \times {10}^{- 6}}}\]

\[ = 39 . 80 Hz\]

This frequency is known as the series resonance frequency.
(ii) The maximum current will be given by

\[I_0 = \frac{E_0}{R}\]

\[ = \frac{200}{100}\]

\[ = 2 A\]

(iii) 

(iv) 
Sharpness of Resonance : It is defined as the ratio of the voltage developed across the inductance (L) or capacitance (C) at resonance to the voltage developed across the resistance (R).

\[Q = \frac{1}{R}\sqrt{\frac{L}{C}}\]

It may also be defined as the ratio of resonance angular frequency to bandwidth of the circuit.

\[Q = \frac{\omega_r}{2 ∆ \omega}\]

Circuit become more selective if the resonance is more sharp, maximum current is more, the circuit is close to resonance for smaller range of (2

\[∆ \omega\] of frequencies. Thus, the tuning of the circuit will be good.