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 series LCR circuit has impedance

$Z=R+j(\omega L-\frac{1}{\omega C})$

The average power delivered from the source is

$P=V_{\mathrm{rms}}{2}^2\frac{R}{{\mathrm{|Z|}}^2}$

At resonance the reactive part cancels ($\omega L=\frac{1}{\mathrm{\omega C}}$), so

$\mathrm{|Z|}=R$ and the resonant power is

$P$₀=Vrms221R

We need the angular frequencies where the power is one‑half of this value:

$P=\frac{}{}2P$₀

Substituting the expressions for $P$ and $P$₀ gives

$\frac{R}{{\mathrm{|Z|}}^2}$=121R

Multiplying both sides by ${\mathrm{|Z|}}^2$ and by $R$ yields

${\mathrm{|Z|}}^2=2R^2$

Since

${\mathrm{|Z|}}^2=R^2+{\omega L-\frac{1}{\mathrm{\omega C}}}^2$

the condition becomes

${\omega L-\frac{1}{\mathrm{\omega C}}}^2=R^2$

or

$\left|\omega L-\frac{1}{\mathrm{\omega C}}\right|=R$

Insert the given component values: $L=5 H$, $C=80 \mu F=80\times 10-6 F$, $R=40 \Omega$. Then

$5\omega -\backslash frac\{12500\}\{\omega \}=\pm 40$

Multiplying by $\omega$ gives two quadratic equations.

1. $5\omega ^\{2\}-40\omega -12500=0$

2. $5\omega ^\{2\}+40\omega -12500=0$

Dividing each by 5:

1. $\omega ^\{2\}-8\omega -2500=0$

2. $\omega ^\{2\}+8\omega -2500=0$

Solving the first quadratic:

$\omega =\backslash frac\{8\pm \backslash sqrt\{8^\{2\}+4·2500\}\}\{2\}=\backslash frac\{8\pm \backslash sqrt\{10064\}\}\{2\}$

Only the positive root is physical, giving

$\omega \approx \backslash frac\{8+100.32\}\{2\}=54.16\backslash ;\{\backslash text\{rad/s\}\}$

Solving the second quadratic:

$\omega =\backslash frac\{-8\pm \backslash sqrt\{(-8)^\{2\}+4·2500\}\}\{2\}=\backslash frac\{-8\pm \backslash sqrt\{10064\}\}\{2\}$

The positive root yields

$\omega \approx \backslash frac\{-8+100.32\}\{2\}=46.16\backslash ;\{\backslash text\{rad/s\}\}$

Thus the two angular frequencies at which the transferred power drops to one‑half of its resonant value are

$\omega \approx 46\backslash ;\backslash text\{rad/s\}$ and $\omega \approx 54\backslash ;\backslash text\{rad/s\}$.

These match the provided answer.