Answer:
The impedance is 1.49 kΩ.
(A) is correct option.
Explanation:
Given that,
Resistor = 250
Inductor [tex]L= 1.20\ mH[/tex]
Capacitor [tex]C= 1.80\ \muF[/tex]
Frequency f = 60.0 Hz
Voltage = 120 V
We need to calculate the [tex]\omega[/tex]
Using formula of [tex]\omega[/tex]
[tex]\omega = 2\pif[/tex]
Put the value into the formula
[tex]\omega=2\times3.14\times60.0[/tex]
[tex]\omega=376.8\ rad/s[/tex]
We need to calculate the [tex]X_{L}[/tex]
Using formula of [tex]X_{L}[/tex]
[tex]X_{L}=\omega L[/tex]
Put the value into the formula
[tex]X_{L}=376.8\times1.20\times10^{-3}[/tex]
[tex]X_{L}=0.4522\ \Omega[/tex]
We need to calculate the [tex]X_{C}[/tex]
Using formula of [tex]X_{C}[/tex]
[tex]X_{C}=\dfrac{1}{\omega C}[/tex]
Put the value into the formula
[tex]X_{C}=\dfrac{1}{376.8\times1.80\times10^{-6}}[/tex]
[tex]X_{C}=1474.404\ \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{250^2+(0.4522-1474.404)^2}[/tex]
[tex]Z=1.49\ k\Omega[/tex]
Hence, The impedance is 1.49 kΩ.
