Answer:
E = 0.13 J
Explanation:
At resonance condition we have
[tex]\omega = \sqrt{\frac{1}{LC}}[/tex]
[tex]\omega = \sqrt{\frac{1}{(13.5 \times 10^{-3}}{50\times 10^{-6})}}[/tex]
[tex]\omega = 1217.2 rad/s[/tex]
now if the frequency is double that of resonance condition then we have
[tex]x_L = 2\omega L[/tex]
[tex]x_L = 2(1217.2)(13.5\times 10^{-3})[/tex]
[tex]x_L = 32.86 ohm[/tex]
now we have
[tex]x_c = \frac{1}{2(1217)(50\times 10^{-6})}[/tex]
[tex]x_c = 8.22 ohm[/tex]
now average power is given as
[tex]P = i_{rms}V_{rms}\times \frac{R}{z}[/tex]
[tex]P = \frac{55}{\sqrt{(32.86 - 8.22)^2 + 13^2)}}(55)\times \frac{13}{\sqrt{(32.86 - 8.22)^2 + 13^2}}[/tex]
[tex]P_{avg} = 50.67 Watt[/tex]
Now time period is given as
[tex]T = \frac{2\pi}{\omega}[/tex]
so total energy consumed is given as
[tex]E_{avg} = 50.67(\frac{2\pi}{2(1217.2)})[/tex]
[tex]E = 0.13 J[/tex]
