NEET · Physics · STD 12 -7. Alternating current
An ac circuit contains a resistance of \(1 k \Omega\), a capacitor of \(0.1 \mu F\) and an inductor of 1 mH connected in series. The resonance frequency of the circuit is approximately :
- A 13.5 kHz
- B 10.1 kHz
- C 20.7 kHz
- D 15.9 kHz
Answer & Solution
Correct Answer
(D) 15.9 kHz
Step-by-step Solution
Detailed explanation
(D) 15.9 kHz
Given \(L=1 mH =10^{-3} H\) and \(C=0.1 \mu F=10^{-7} F\).
The resonance frequency of a series RLC circuit is given by \(f_r=\frac{1}{2 \pi \sqrt{L C}}\).
Substituting the values:
\(f_r=\frac{1}{2 \pi \sqrt{10^{-3} \times 10^{-7}}}\)
\(f_r=\frac{1}{2 \pi \sqrt{10^{-10}}}\)
\(f_r=\frac{10^5}{2 \pi} Hz\)
\(f_r=\frac{50}{\pi} kHz \approx 15.9 kHz\)
Given \(L=1 mH =10^{-3} H\) and \(C=0.1 \mu F=10^{-7} F\).
The resonance frequency of a series RLC circuit is given by \(f_r=\frac{1}{2 \pi \sqrt{L C}}\).
Substituting the values:
\(f_r=\frac{1}{2 \pi \sqrt{10^{-3} \times 10^{-7}}}\)
\(f_r=\frac{1}{2 \pi \sqrt{10^{-10}}}\)
\(f_r=\frac{10^5}{2 \pi} Hz\)
\(f_r=\frac{50}{\pi} kHz \approx 15.9 kHz\)
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