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

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, $L=800\ mH=800\times 10^{-3}\ H=0.8\ H$

Capacitance, $C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F$

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

$Z=\sqrt{R^2+(X_C-X_L)^2}$...........(1)

Where

$X_C$ is the capacitive reactance, $X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega$

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

$X_L={2\pi \times 60\times 0.8}=301.59\ \Omega$

So, equation (1) becomes :

$Z=\sqrt{(100)^2+(265.65-301.59)^2}$

Z = 106.26 ohms

or

Z = 107 ohms

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