Respuesta :
Answer:
The energy delivered to the circuit during one period is 177.87 mJ.
Explanation:
Given that,
Resistance [tex]R=9.50\ \Omega[/tex]
Inductance L=8.50 mH
Capacitance [tex]C=90.0\ \mu F[/tex]
Voltage [tex]V_{rms} = 45.0\ V[/tex]
We need to calculate the resonance frequency
Using formula of resonance frequency
[tex]f_{r}=\dfrac{1}{2\pi\sqrt{LC}}[/tex]
[tex]f_{r}=\dfrac{1}{2\pi\times\sqrt{8.50\times10^{-3}\times90.0\times10^{-6}}}[/tex]
[tex]f_{r}=181.9\approximate\ 182\ Hz[/tex]
[tex]2f_{r}=2\times181.9=363.8=364\ Hz[/tex]
We need to calculate the period
[tex]T=\dfrac{1}{f}[/tex]
[tex]T=\dfrac{1}{364}[/tex]
[tex]T=0.0028\ sec[/tex]
We need to calculate the inductive resistance
Using formula of inductive resistance
[tex]X_{L}=2\pi Lf[/tex]
[tex]X_{L}=2\times\pi\times8.50\times10^{-3}\times364[/tex]
[tex]X_{L}=19.44[/tex]
We need to calculate the capacitive resistance
Using formula of capacitive resistance
[tex]X_{C}=\dfrac{1}{2\pi fC}[/tex]
[tex]X_{C}=\dfrac{1}{2\pi\times364\times90.0\times10^{-6}}[/tex]
[tex]X_{C}=4.86\ \Omega[/tex]
We need to calculate the impedance
Using formula of impedance
[tex]Z=\sqrt{R^2+(X_{L}-X_{C})^2}[/tex]
[tex]Z=\sqrt{9.50^2+(19.44-4.86)^2}[/tex]
[tex]Z=17.401\ \Omega[/tex]
We need to calculate the angle
[tex]\phi=tan^{-1}(\dfrac{X_{L}-X_{C}}{R})[/tex]
[tex]\phi=tan^{-1}(\dfrac{19.44-4.86}{9.50})[/tex]
[tex]\phi=56.91^{\circ}[/tex]
We need to calculate the energy
Using formula of energy
[tex]Q=\dfrac{V^2}{Z}T\cos\phi[/tex]
Put the value into the formula
[tex]Q=\dfrac{(45.0)^2}{17.401}\times0.0028\times\cos(56.91^{\circ})[/tex]
[tex]Q=177.87\ mJ[/tex]
Hence, The energy delivered to the circuit during one period is 177.87 mJ.
