Question

You connect a 250-2 resistor, a 1.20-mH inductor, and a 1.80-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The approximate impedance of your circuit A) 1.49 k22 B) 1.47 kO2 C) 0.4522 D) 250 22 E) None of these is correct.

Answer 1

Answer:

The  impedance is 1.49 kΩ.

(A) is correct option.

Explanation:

Given that,

Resistor = 250

Inductor $L= 1.20\ mH$

Capacitor $C= 1.80\ \muF$

Frequency f = 60.0 Hz

Voltage = 120 V

We need to calculate the $\omega$

Using formula of  $\omega$

$\omega = 2\pif$

Put the value into the formula

$\omega=2\times3.14\times60.0$

$\omega=376.8\ rad/s$

We need to calculate the $X_{L}$

Using formula of  $X_{L}$

$X_{L}=\omega L$

Put the value into the formula

$X_{L}=376.8\times1.20\times10^{-3}$

$X_{L}=0.4522\ \Omega$

We need to calculate the $X_{C}$

Using formula of  $X_{C}$

$X_{C}=\dfrac{1}{\omega C}$

Put the value into the formula

$X_{C}=\dfrac{1}{376.8\times1.80\times10^{-6}}$

$X_{C}=1474.404\ \Omega$

We need to calculate the impedance

Using formula of impedance

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

$Z=\sqrt{250^2+(0.4522-1474.404)^2}$

$Z=1.49\ k\Omega$

Hence, The  impedance is 1.49 kΩ.