Respuesta :
Answer:
(a). (i). The reactants are [tex]X_{L} =31.66\ \Omega [/tex] .
(II). The inductance of the circuit is 0.1583 Henry.
(b). The resonant angular frequency is 229.4 rad/s.
Explanation:
Given that,
Capacitor = 120.0 μC
Frequency = 200.0 rad/s
Impedance = 100.0 -10j
(I). We need to calculate the [tex]X_{C}[/tex]
[tex]X_{C}=\dfrac{1}{C\times\omega}[/tex]
Put the value into the formula
[tex]X_{C}=\dfrac{1}{120\times10^{-6}\times200}[/tex]
[tex]X_{C}=41.66\ \Omega[/tex]
(II). We know that,
Formula of impedance is
[tex]Z=\sqrt{R^2+X_{L}^2+X_{C}^2}[/tex]...(I)
Given equation of impedance is
[tex]Z=(100-10j)[/tex]...(II)
On Comparing of equation (I) and (II)
[tex]R = 100[/tex]
[tex]X_{L}-X_{C}=-10[/tex]
Now, put the value of [tex]X_{C}[/tex]
[tex]X_{L=41.66-10[/tex]
[tex]X_{L}=31.66\ \Omega[/tex]
We need to calculate the inductance
Using formula of inductance
[tex]X_{L}=\omega\times L[/tex]
Put the value into the formula
[tex]L=\dfrac{X_{L}}{\omega}[/tex]
[tex]L=\dfrac{31.66}{200}[/tex]
[tex]L=0.1583\ Henry[/tex]
(b). We need to calculate the resonant angular frequency
Using formula of the resonant angular frequency
[tex]angular\ frequency =\dfrac{1}{\sqrt{L\times C}}[/tex]
[tex]angular\ frequency =\dfrac{1}{\sqrt{0.1583\times120\times10^{-6}}}[/tex]
[tex]angular\ frequency =229.4\ rad/s[/tex]
Hence, (a). (i). The reactants are [tex]X_{L} =31.66\ \Omega [/tex] .
(II). The inductance of the circuit is 0.1583 Henry.
(b). The resonant angular frequency is 229.4 rad/s.
