MHT CET · Physics · Alternating Current
In the circuit shown the ratio of quality factor and the bandwidth is
- A \(10 \mathrm{~s}\)
- B \(8 \mathrm{~s}\)
- C \(6 \mathrm{~s}\)
- D \(4 \mathrm{~s}\)
Answer & Solution
Correct Answer
(A) \(10 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\(\text {Bandwidth }=2 \Delta \omega=\frac{\mathrm{R}}{\mathrm{L}}\)
\(\text {Q factor }=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}\)
\(\frac{\mathrm{Q} \text { factor }}{\text { Band width }}=\frac{\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}}{\frac{\mathrm{R}}{\mathrm{L}}}=\frac{1}{\mathrm{R}^2} \frac{\mathrm{L}^{3 / 2}}{\sqrt{\mathrm{C}}}\)
\(\frac{\mathrm{Q}_{\text {pacs; }}}{\text { Bandwidth }}=\frac{3 \sqrt{3}}{100 \sqrt{27 \times 10^{-6}}}=\) \(\frac{3 \sqrt{3}}{100 \times 3 \sqrt{3} \times 10^{-3}}=10 \mathrm{~s}\)
\(\text {Q factor }=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}\)
\(\frac{\mathrm{Q} \text { factor }}{\text { Band width }}=\frac{\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}}{\frac{\mathrm{R}}{\mathrm{L}}}=\frac{1}{\mathrm{R}^2} \frac{\mathrm{L}^{3 / 2}}{\sqrt{\mathrm{C}}}\)
\(\frac{\mathrm{Q}_{\text {pacs; }}}{\text { Bandwidth }}=\frac{3 \sqrt{3}}{100 \sqrt{27 \times 10^{-6}}}=\) \(\frac{3 \sqrt{3}}{100 \times 3 \sqrt{3} \times 10^{-3}}=10 \mathrm{~s}\)
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