JEE Mains · Physics · STD 12 -7. Alternating current
A series combination of resistor of resistance \(100\,\Omega\), inductor of inductance \(1\,H\) and capacitor of capacitance \(6.25\,\mu F\) is connected to an ac source. The quality factor of the circuit will be \(.............\).
- A \(4\)
- B \(3\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(\text { Quality factor }=\frac{X_L}{R}=\frac{\omega L}{R}\) \(\omega=\frac{1}{\sqrt{ LC }}=\frac{1}{\sqrt{1 \times 6.25 \times 10^{-6}}}=\frac{10^3}{2.5}=400 / sec\) \(\text { Q-factor }=\frac{400 \times 1}{100}=4\)
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