KCET · Physics · Capacitance
A series LCR circuit contains inductance \( 5 \mathrm{mH} \), capacitance \( 2 \mu \mathrm{F} \) and resistance \( 10 \Omega \). If a frequency A.C. source is varied, what is the frequency at which maximum power is dissipated \( ? \)
- A \( \frac{10^{5}}{\Pi z} H z \)
- B \( \frac{10^{-5}}{\Pi} H \)
- C \( \frac{2}{\Pi} \times 10^{5} \mathrm{~Hz} \)
- D \( \frac{5}{\Pi} \times 10^{3} H z \)
Answer & Solution
Correct Answer
(D) \( \frac{5}{\Pi} \times 10^{3} H z \)
Step-by-step Solution
Detailed explanation
Maximum power is dissipated at resonance
\(L=5 m H\) and \(C=2 \mu F\)
The resonant frequency
\(\begin{aligned} f_{R} &=\frac{1}{2 \pi \sqrt{L C}} \\ &=\frac{1}{2 \pi \sqrt{5 \times 10^{-3} \times 2 \times 10^{-6}}} \\ &=\frac{1}{2 \pi \sqrt{10^{-8}}}=\frac{1}{2 \pi \times 10^{-4}} \\ &=F_{R}=\frac{10 \times 10^{3}}{2 \pi}=\frac{5 \times 10^{3}}{\pi} H z \end{aligned}\)
\(L=5 m H\) and \(C=2 \mu F\)
The resonant frequency
\(\begin{aligned} f_{R} &=\frac{1}{2 \pi \sqrt{L C}} \\ &=\frac{1}{2 \pi \sqrt{5 \times 10^{-3} \times 2 \times 10^{-6}}} \\ &=\frac{1}{2 \pi \sqrt{10^{-8}}}=\frac{1}{2 \pi \times 10^{-4}} \\ &=F_{R}=\frac{10 \times 10^{3}}{2 \pi}=\frac{5 \times 10^{3}}{\pi} H z \end{aligned}\)
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