You connect a 250-$\Omega$ resistor,a 1.20-mH inductor,and a 1.80-$\mu$F capacitor in series across a 60.0-Hz,120-V (peak)source.The approximate impedance of your circuit is
A)1.49 k$\Omega$
B)1.47 k$\Omega$
C)0.452 $\Omega$
D)250 $\Omega$
E)None of these is correct.

Question

Quizplus · Accepted Answer

The impedance of the circuit is given by:$$Z = \sqrt{R^2 + (XL - XC)^2}$$where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.In this case, R = 250 ohms, XL = 2πfL = 2π(60 Hz)(1.20 mH) = 452 ohms, and XC = 1/(2πfC) = 1/(2π(60 Hz)(1.80 μF)) = 147 ohms.Therefore, the impedance of the circuit is:$$Z = \sqrt{(250 ohms)^2 + (452 ohms - 147 ohms)^2} = 1.49 kΩ$$

You Connect a 250- Ω\OmegaΩ Resistor,a 1 μ\muμ F Capacitor in Series Across a 60

You Connect a 250- $\Omega$ Resistor,a 1 $\mu$ F Capacitor in Series Across a 60