ANS:A - 1,616 

@All.

Here is my explanation.

To calculate the impedance magnitude at 1,200 Hz below the resonance frequency (fr), we need to consider the components in the series circuit.

The impedance of a resistor (R) in an AC circuit is simply its resistance, which is 10 Ω in this case.

The impedance of an inductor (L) in an AC circuit is given by the formula ZL = jωL, where j is the imaginary unit (√(-1)), ω is the angular frequency, and L is the inductance. Since we are looking for the impedance magnitude, we need to calculate |ZL|.

Substituting the values into the formula, we have:
|ZL| = ωL = 2πfL,
= 2π(1,200 Hz)(90 mH).
≈ 678 Ω.

The impedance of a capacitor (C) in an AC circuit is given by the formula ZC = 1/(jωC). Again, we need to calculate |ZC|.

Substituting the values into the formula, we have:
|ZC| = 1/(ωC) = 1/(2πfC),
= 1/(2π(1,200 Hz)(0.015 F)),
≈ 1.11 Ω.

Now, to find the total impedance, we add the impedances of the three components since they are in series:
Z total = R + ZL + ZC.
= 10 Ω + 678 Ω + 1.11 Ω,
≈ 689.11 Ω.

Therefore, the impedance magnitude at 1,200 Hz below fr is approximately 689.11 Ω.