MHT CET · Physics · Alternating Current
A series resonant circuit consists of inductor ' \(L\) ' of negligible resistance and a capacitor ' C ' which produces resonant frequency ' f '. If L is changed to 3 L and ' C ' is changed to 6 C , the resonant frequency will become.
- A \(\frac{\mathrm{f}}{6}\)
- B \(\frac{\mathrm{f}}{3}\)
- C \(\frac{\mathrm{f}}{2 \sqrt{2}}\)
- D \(\frac{\mathrm{f}}{3 \sqrt{2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{f}}{3 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Resonant frequency,
\(f=\frac{1}{2 \pi \sqrt{L C}}\)
If L becomes 3 L and C becomes 6 C , then the frequency will become
\(f^{\prime}=\frac{1}{2 \pi \sqrt{3 L} \cdot 6 \mathrm{C}}=\frac{1}{2 \pi \cdot 3 \sqrt{2} \sqrt{\mathrm{LC}}}=\frac{f}{3 \sqrt{2}}\)
\(f=\frac{1}{2 \pi \sqrt{L C}}\)
If L becomes 3 L and C becomes 6 C , then the frequency will become
\(f^{\prime}=\frac{1}{2 \pi \sqrt{3 L} \cdot 6 \mathrm{C}}=\frac{1}{2 \pi \cdot 3 \sqrt{2} \sqrt{\mathrm{LC}}}=\frac{f}{3 \sqrt{2}}\)
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