Question

A series LCR circuit containing 5.0 H inductor, 80 µF capacitor and 400) 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:​

Answer 1

Answer:A series LCR circuit containing 5.0 H inductor, 80 µF capacitor and 400) 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:​

Explanation:

Answer 2

Given:

Inductance $=5H$

Capacitance $=80$ μ$F$

Resistance $=40$Ω

Potential difference $=230V$

To find:

The angular frequencies at which power transferred is half the power at resonant frequency.

Explanation:

We know, half input frequency is the frequency used in a circuit when the power dissipation in a circuit is half of the magnitude of power loss when the frequency is resonant frequency.

These frequencies are given by

ω₁ = ω₀ + Δω

ω₂ = ω₀ - Δω

Where,ω₀ is the resonant frequency and Δω is the bandwidth given by $\frac{R}{2L}$

Solution:

We know, resonant frequency (ω₀)  in an AC circuit is given by  $\frac{1}{\sqrt{LC} }$.

Substituting the given values in the equation, we get,

$f_{R} =\frac{1}{\sqrt{5(80*10^{-6} )} }$

$f_{R} =\frac{10^{3} }{\sqrt{400} }$

$f_{R} =50Hz$

ω₀ $=50Hz$

Now, for the bandwidth (Δω)

Δω = $\frac{R}{2L}$

Substituting the given values in the equation, we get

Δω $=\frac{40}{2(5)}$

Δω $=4$

Substituting the known values in the two frequencies, we get

ω₁ $=54Hz$   and  ω₂ $=46Hz$

Final answer:

Hence, the angular frequencies are 54 Hz and 46 Hz.