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

The impedance is 1.49 kΩ.

(A) is correct option.

Step-by-step 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*3.14*60.0`

`\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*1.20*10^(-3)`

`X_(L)=0.4522\ \Omega`

We need to calculate the
`X_(C)`

Using formula of
`X_(C)`

`X_(C)=(1)/(\omega C)`

Put the value into the formula

`X_(C)=(1)/(376.8*1.80*10^(-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=√(250^2+(0.4522-1474.404)^2)`

`Z=1.49\ k\Omega`

Hence, The impedance is 1.49 kΩ.