Question : A series LCR circuit containing 5.0 H inductor, 80 µF capacitor and 40 Ω resistor is connected to 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 :
1) 25 rad/s and 75 rad/s
2) 50 rad/s and 25 rad/s
3) 46 rad/s and 54 rad/s
4) 42 rad/s and 58 rad/s
Doubt by Diya
Solution :
In LCR Series Circuit
L = 5 H
C = 80 µF = 80×10-6 F
R = 40 Ω
ɛrms = 230 V
Let P₀ = Power of the LCR series circuit at resonance
and P = Power of the LCR series circuit when P₀ becomes half.
and P = Power of the LCR series circuit when P₀ becomes half.
P = P₀ / 2 (Given)
We know
P = ɛI cosΦ
P = (Iz)I ×(R/Z)
P = I²R
Clearly Power is directly proportional to I²
So when
P = P₀ / 2
then
I² = I₀2/2
I = I₀/√2
I² = I₀2/2
I = I₀/√2
We know at I₀/√2 there are two values of Angular Frequencies ω1 and ω2
Now, expression to calculate ω1 and ω2
ω1 = ω₀ - (Δω/2)
ω2 = ω₀ + (Δω/2)
ω₀ = 1/√LC
= 1/√(5×80×10-6)
= 1/√(400×10-6)
= 1/(20×10-3)
= 1000/20
= 50 rad/s
Also,
Quality Factor (Q) = ω₀/Δω = ω₀L/R
So
So
Δω = R/L = 40/5 = 8 rad/s
Now,
ω1 = 50 - (8/2)
ω1 = 50 - 4
ω1 = 46 rad/s
ω1 = 50 - 4
ω1 = 46 rad/s
ω2 = 50 + (8/2)
ω2 = 50 + 4
ω2 = 54 rad/s
ω2 = 50 + 4
ω2 = 54 rad/s
ω1 = 46 rad/s & ω2 = 54 rad/s
Hence, "3) 46 rad/s and 54 rad/s" would be the correct option.
