Question

You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impedance of your circuit is approximately A) 3772 B) 30222 C) 2652 D) 1002 E) 1072

Answer 1

Impedance, Z = 107 ohms

Step-by-step explanation:

It is given that,

Resistance, R = 100 ohms

Inductance,
`L=800\ mH=800* 10^(-3)\ H=0.8\ H`

Capacitance,
`C=10\ \mu F=10* 10^(-6)\ F=10^(-5)\ F`

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

`Z=√(R^2+(X_C-X_L)^2)`...........(1)

Where

`X_C` is the capacitive reactance,
`X_C=(1)/(2\pi fC)`

`X_C=(1)/(2\pi * 60* 10^(-5))=265.65\ \Omega`

`X_L` is the inductive reactance,
`X_L={2\pi fL}`

`X_L={2\pi * 60* 0.8}=301.59\ \Omega`

So, equation (1) becomes :

`Z=√((100)^2+(265.65-301.59)^2)`

Z = 106.26 ohms

or

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.