In a parallel RLC circuit, if the resistance is 10Ω, inductance is 0.2H, and cap

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Question: In a parallel RLC circuit, if the resistance is 10Ω, inductance is 0.2H, and capacitance is 50μF, what is the quality factor Q? (2021)

Options:

Correct Answer: 3

Exam Year: 2021

Solution:

Quality factor Q = (1/R) * √(L/C) = (1/10) * √(0.2/50e-6) ≈ 3.

In a parallel RLC circuit, if the resistance is 10Ω, inductance is 0.2H, and capacitance is 50μF, what is the quality factor Q? (2021)

In a parallel RLC circuit, if the resistance is 10Ω, inductance is 0.2H, and capacitance is 50μF, what is the quality factor Q? (2021)

Step 1: Identify the values given in the problem. We have resistance (R) = 10Ω, inductance (L) = 0.2H, and capacitance (C) = 50μF.
Step 2: Convert the capacitance from microfarads to farads. 50μF = 50 * 10^-6 F = 0.000050 F.
Step 3: Use the formula for the quality factor Q in a parallel RLC circuit: Q = (1/R) * √(L/C).
Step 4: Substitute the values into the formula: Q = (1/10) * √(0.2/0.000050).
Step 5: Calculate the value inside the square root: 0.2 / 0.000050 = 4000.
Step 6: Find the square root of 4000: √4000 ≈ 63.2456.
Step 7: Multiply by (1/10): Q = (1/10) * 63.2456 ≈ 6.32456.
Step 8: Round the answer to a reasonable number of significant figures: Q ≈ 6.32.

Quality Factor (Q) – The quality factor Q of a parallel RLC circuit is a measure of the sharpness of the resonance peak and is calculated using the formula Q = (1/R) * √(L/C), where R is resistance, L is inductance, and C is capacitance.
Parallel RLC Circuit – A parallel RLC circuit consists of a resistor (R), inductor (L), and capacitor (C) connected in parallel, and its behavior is characterized by its resonance and quality factor.