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 :

The correct option is 46 rad/s and 54 rad/s.

To find the angular frequencies at which the power transferred to the series LCR circuit is half the power at resonance, we need to calculate the resonant angular frequency (${\omega }_0$) and the half-bandwidth ($\mathrm{\Delta \omega }$).

1. Identify the given parameters:
Inductance, L = 5.0 H
Capacitance, C = 80 µF = 80 × 10^-6 F
Resistance, R = 40 Ω

2. Calculate the resonant angular frequency (${\omega }_0$):
The resonant frequency of a series LCR circuit is given by:
${\omega }_0=\frac{1}{\sqrt{LC}}$

Substituting the given values:
${\omega }_0=\frac{1}{\sqrt{5.0\times 80\times {10}^{-6}}}$
${\omega }_0=\frac{1}{\sqrt{400\times {10}^{-6}}}$
${\omega }_0=\frac{1}{\sqrt{4\times {10}^{-4}}}$
${\omega }_0=\frac{1}{2\times {10}^{-2}}=50\text{rad/s}$

3. Calculate the half-bandwidth ($\mathrm{\Delta \omega }$):
The half-power frequencies lie at ${\omega }_0\pm \mathrm{\Delta \omega }$, where the deviation $\mathrm{\Delta \omega }$ is given by:
$\mathrm{\Delta \omega }=\frac{R}{2L}$

Substituting the values:
$\mathrm{\Delta \omega }=\frac{40}{2\times 5.0}$
$\mathrm{\Delta \omega }=\frac{40}{10}=4\text{rad/s}$

4. Determine the half-power angular frequencies:
The lower half-power frequency is:
${\omega }_1={\omega }_0-\mathrm{\Delta \omega }=50-4=46\text{rad/s}$

The upper half-power frequency is:
${\omega }_2={\omega }_0+\mathrm{\Delta \omega }=50+4=54\text{rad/s}$

Therefore, the angular frequencies of the source at which the power transferred is half of the power at resonant frequency are 46 rad/s and 54 rad/s.