Sol. Current amplitude or peak value of the current in series LR circuit is given by

I M = …(1)

While current at resonance is given by

I R = …(2)

Given that, at f 1 and f 2 , the current amplitude falls to of the current at resonance.

Hence = …(3)

This gives ω 1 L – = – R …(4)

and ω 2 L – = + R …(5)

Adding eqns. (4) and (5), we get

L( ω 1 + ω 2 ) = or ω 1 ω 2 = …(6)

Subtracting eqns. (4) and (5), we get

( ω 2 – ω 1 )L + = 2R

Putting the value of from eqn. (6),

( ω 2 – ω 1 )L + × ω 1 ω 2 L = 2R

or ( ω 2 – ω 1 ) = or 2 π (f 2 –f 1 ) =

∴ (f 2 – f 1 ) =