A series RLC circuit is driven by a 1.0-kHz oscillator.The circuit parameters are L = 5.0 mH,C = 4.0 $\mu$F,R = 10 $\Omega$.What is the impedance of this circuit?
A)8.4 $\Omega$
B)10 $\Omega$
C)13 $\Omega$
D)0.24 k$\Omega$
E)81 $\Omega$

Question

Quizplus · Accepted Answer

The impedance of a series RLC circuit can be found using the formula:

Z = √(R^2 + (ωL - 1/ωC)^2)

where R is the resistance, L is the inductance, C is the capacitance, and ω is the angular frequency.

In this case, R = 10 Ω, L = 5.0 mH = 0.005 H, C = 4.0 μF = 0.000004 F, and ω = 2πf = 2π(1000) = 6283.2 rad/s.

Plugging these values into the formula, we get:

Z = √(10^2 + (6283.2*0.005 - 1/(6283.2*0.000004))^2) ≈ 13,000 Ω ≈ 13 kΩ

So the answer is choice C, 13 kΩ.

A Series RLC Circuit Is Driven by a 1 μ\muμ F,R = 10

A Series RLC Circuit Is Driven by a 1 $\mu$ F,R = 10