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 maximum current in your circuit is approximately
A)56.9 mA
B)80.5 mA
C)0.480 mA
D)37.8 mA
E)165 mA

Question

Quizplus · Accepted Answer

The impedance of the circuit is given by:

Z = R + j(XL - XC)

Where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

R = 250 Ω
XL = 2πfL = 2π(60)(1.20 x 10^-3) = 0.452 Ω
XC = 1/(2πfC) = 1/(2π(60)(1.80 x 10^-6)) = 147.5 Ω

Z = 250 + j(-147.05) = 250 - j147.05

The magnitude of the impedance is given by:

|Z| = √(250^2 + 147.05^2) = 290.96 Ω

The maximum current in the circuit is given by:

Imax = Vmax / |Z| = (120 V)/(290.96 Ω) = 0.412 A = 412 mA

Therefore, the closest answer choice is B) 80.5 mA.

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