MHT CET · Physics · Alternating Current
In LCR circuit the inductance is changed from L to \(9 \mathrm{~L}\). For same resonant
frequency the capacitance should be changed from \(\mathrm{C}\) to
- A \(9 \mathrm{C}\)
- B \(3 \mathrm{C}\)
- C \(\frac{C}{9}\)
- D \(\frac{C}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{C}{9}\)
Step-by-step Solution
Detailed explanation
\(\omega=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}\)
\(\therefore \quad \mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}\)
\(\quad \mathrm{C}_{2}=\frac{\mathrm{L}_{1} \mathrm{C}_{1}}{\mathrm{~L}_{2}}=\frac{\mathrm{L}}{9 \mathrm{~L}} \times \mathrm{C}=\frac{\mathrm{C}}{9}\)
\(\therefore \quad \mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}\)
\(\quad \mathrm{C}_{2}=\frac{\mathrm{L}_{1} \mathrm{C}_{1}}{\mathrm{~L}_{2}}=\frac{\mathrm{L}}{9 \mathrm{~L}} \times \mathrm{C}=\frac{\mathrm{C}}{9}\)
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