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GUJCET · Physics · Alternating Current
Resonance frequency of L－C－R AC circuit is \(\nu_{0}\). Now the capacitance is made 4 times，then the new resonance frequency will become ____________ ．
- A \(\frac{v_{0}}{4}\)
- B \(2 v_{0}\)
- C \(v_{0}\)
- D \(\frac{v_{0}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{v_{0}}{2}\)
Step-by-step Solution
Detailed explanation
(D) \(\frac{v_{0}}{2}\)
\(\quad v_{0}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
\(
v_{0} \propto \frac{2 \pi}{\sqrt{\mathrm{C}}}
\)
\(
\therefore \frac{v_{0}}{v_{0}{ }^{\prime}}=\sqrt{\frac{\mathrm{C}^{\prime}}{\mathrm{C}}}
\)
\(
\therefore \frac{v_{0}}{v_{0}{ }^{\prime}}=\sqrt{\frac{4 \mathrm{C}}{\mathrm{C}}}=2
\)
\(
\nu_{0}^{\prime}=\frac{\nu_{0}}{2}
\)
\(\quad v_{0}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
\(
v_{0} \propto \frac{2 \pi}{\sqrt{\mathrm{C}}}
\)
\(
\therefore \frac{v_{0}}{v_{0}{ }^{\prime}}=\sqrt{\frac{\mathrm{C}^{\prime}}{\mathrm{C}}}
\)
\(
\therefore \frac{v_{0}}{v_{0}{ }^{\prime}}=\sqrt{\frac{4 \mathrm{C}}{\mathrm{C}}}=2
\)
\(
\nu_{0}^{\prime}=\frac{\nu_{0}}{2}
\)
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