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 answer is 46 rad/s and 54 rad/s.

To find the angular frequencies where the power transferred to the circuit is half the power at the resonant angular frequency, we need to determine the half-power frequencies. First, let us list the given values for the series LCR circuit:

Inductance, $L=5.0\text{H}$

Capacitance, $C=80\text{μF}=80\times {10}^{-6}\text{F}$

Resistance, $R=40\text{Ω}$

Step 1: Calculate the Resonant Angular Frequency

The resonant angular frequency (${\omega }_0$) occurs when the inductive reactance equals the capacitive reactance. It is given by the formula:

${\omega }_0=\frac{1}{\sqrt{LC}}$

Substituting the given values into the formula:

${\omega }_0=\frac{1}{\sqrt{5.0\times 80\times {10}^{-6}}}$

${\omega }_0=\frac{1}{\sqrt{400\times {10}^{-6}}}$

${\omega }_0=\frac{1}{20\times {10}^{-3}}=\frac{1}{0.02}=50\text{rad/s}$

Step 2: Calculate the Bandwidth and Half-Power Deviation

The power transferred to the circuit is half of the maximum resonant power at the frequencies where the current drops to $\frac{1}{\sqrt{2}}$ of its maximum value. These frequencies are found symmetrically on either side of the resonant frequency. The frequency difference from resonance is $\frac{\Delta \omega }{2}$, where $\Delta \omega$ is the bandwidth given by $\frac{R}{L}$.

Thus, the deviation from the resonant frequency is:

$\frac{\Delta \omega }{2}=\frac{R}{2L}$

Substituting the values of R and L:

$\frac{R}{2L}=\frac{40}{2\times 5.0}=\frac{40}{10}=4\text{rad/s}$

Step 3: Determine the Half-Power Angular Frequencies

The half-power frequencies (${\omega }_1$ and ${\omega }_2$) are calculated by subtracting and adding this deviation to the resonant frequency:

${\omega }_1={\omega }_0-\frac{R}{2L}$

${\omega }_1=50-4=46\text{rad/s}$

And for the upper frequency limit:

${\omega }_2={\omega }_0+\frac{R}{2L}$

${\omega }_2=50+4=54\text{rad/s}$

Therefore, the angular frequencies at which the power is halved are exactly 46 rad/s and 54 rad/s.