Question

A 9.50-Ω resistor, 8.50-mH inductor, and 90.0-µF capacitor are connected in series to a 45.0-V (rms) source having variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.

Answer 1

The energy delivered to the circuit during one period is 177.87 mJ.

Step-by-step explanation:

Given that,

Resistance
`R=9.50\ \Omega`

Inductance L=8.50 mH

Capacitance
`C=90.0\ \mu F`

Voltage
`V_(rms) = 45.0\ V`

We need to calculate the resonance frequency

Using formula of resonance frequency

`f_(r)=(1)/(2\pi√(LC))`

`f_(r)=\frac{1}{2\pi*\sqrt{8.50*10^(-3)*90.0*10^(-6)}}`

`f_(r)=181.9\approximate\ 182\ Hz`

`2f_(r)=2*181.9=363.8=364\ Hz`

We need to calculate the period

`T=(1)/(f)`

`T=(1)/(364)`

`T=0.0028\ sec`

We need to calculate the inductive resistance

Using formula of inductive resistance

`X_(L)=2\pi Lf`

`X_(L)=2*\pi*8.50*10^(-3)*364`

`X_(L)=19.44`

We need to calculate the capacitive resistance

Using formula of capacitive resistance

`X_(C)=(1)/(2\pi fC)`

`X_(C)=(1)/(2\pi*364*90.0*10^(-6))`

`X_(C)=4.86\ \Omega`

We need to calculate the impedance

Using formula of impedance

`Z=\sqrt{R^2+(X_(L)-X_(C))^2}`

`Z=√(9.50^2+(19.44-4.86)^2)`

`Z=17.401\ \Omega`

We need to calculate the angle

`\phi=tan^(-1)((X_(L)-X_(C))/(R))`

`\phi=tan^(-1)((19.44-4.86)/(9.50))`

`\phi=56.91^(\circ)`

We need to calculate the energy

Using formula of energy

`Q=(V^2)/(Z)T\cos\phi`

Put the value into the formula

`Q=((45.0)^2)/(17.401)*0.0028*\cos(56.91^(\circ))`

`Q=177.87\ mJ`

Hence, The energy delivered to the circuit during one period is 177.87 mJ.