Start Learning Free

Before You Begin

Question #43463

Difficulty: 8/10 | 🟢 Free

The Question

A series RLC circuit consists of a resistor R=100 ΩR = 100 \ \Omega, an inductor L=1 HL = 1 \text{ H}, and a capacitor C=100 μFC = 100 \ \mu\text{F}, driven by an AC voltage source V(t)=10sin⁡(ωt) VV(t) = 10 \sin(\omega t) \text{ V}. Two ideal peak detectors are connected across the inductor and the capacitor to measure their respective voltage amplitudes, VL,peakV_{L,peak} and VC,peakV_{C,peak}. The outputs of these detectors are fed into a digital logic circuit. Input A of a 2-input OR gate receives logic '1' if VL,peak≥11 VV_{L,peak} \ge 11 \text{ V}, and logic '0' otherwise. Input B of the same OR gate receives logic '1' if VC,peak≥11 VV_{C,peak} \ge 11 \text{ V}, and logic '0' otherwise. Let YY be the Boolean output of the OR gate. Determine the sequence of outputs (Y1,Y2,Y3)(Y_1, Y_2, Y_3) at three driving frequencies: ω1=502 rad/s\omega_1 = 50\sqrt{2} \text{ rad/s}, ω2=100 rad/s\omega_2 = 100 \text{ rad/s}, and ω3=1002 rad/s\omega_3 = 100\sqrt{2} \text{ rad/s}.

Options

  • A.

    (1,0,1)(1, 0, 1)

  • B.

    (0,1,0)(0, 1, 0)

  • C.

    (0,0,0)(0, 0, 0)

  • D.

    (1,1,1)(1, 1, 1)

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Logic Gate Boolean Algebra|Bandwidth in RLC|Voltage Magnification in Resonance

Why This Question Is Tricky

Common Mistakes

This question tests a critical misconception regarding voltage magnification in RLC circuits.

Students often memorize that the maximum voltage across the inductor and capacitor occurs exactly at the resonant frequency ω0=1/LC\omega_0 = 1/\sqrt{LC}, and that this maximum value is QV0Q V_0.

While this is a valid approximation for high-Q circuits, it fails completely for low-Q circuits.

In this problem, the quality factor is Q=1RLC=1Q = \frac{1}{R}\sqrt{\frac{L}{C}} = 1.

For low Q, the actual peak voltages for the capacitor and inductor occur at ωC=ω01−1/(2Q2)\omega_C = \omega_0 \sqrt{1 - 1/(2Q^2)} and ωL=ω0/1−1/(2Q2)\omega_L = \omega_0 / \sqrt{1 - 1/(2Q^2)} respectively, and their maximum values exceed QV0Q V_0.

Students who blindly apply the high-Q approximation will incorrectly conclude that the peak voltage never exceeds QV0=10 VQ V_0 = 10 \text{ V}, leading them to choose the trap option (0,0,0)(0, 0, 0) or assume resonance gives the only '1' output (0,1,0)(0, 1, 0).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterSemiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
ConceptLogic Gate Boolean Algebra
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree