A series resonant circuit consists of inductor ' $L$ ' of negligible resistance and a capacitor ' C ' which…
- $\frac{\mathrm{f}}{6}$
- $\frac{\mathrm{f}}{3}$
- $\frac{\mathrm{f}}{2 \sqrt{2}}$
- $\frac{\mathrm{f}}{3 \sqrt{2}}$
Solution
If L becomes 3 L and C becomes 6 C , then the frequency will become $f^{\prime}=\frac{1}{2 \pi \sqrt{3 L} \cdot 6 \mathrm{C}}=\frac{1}{2 \pi \cdot 3 \sqrt{2} \sqrt{\mathrm{LC}}}=\frac{f}{3 \sqrt{2}}$
Asked in: MHT CET 2024 (10 May Shift 1)