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In A.C. circuit in which inductance and capacitance are joined in series. Current is found to be maximum when the value of inductance is \(0.5 \mathrm{H}\) and the value of capacitance is \(8 \mu \mathrm{F}\). The angular frequency of applied alternating voltage will be:
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Verified Answer
The correct answer is:
\(500 \mathrm{~Hz}\)
Current is max, when at resonance condition
So, \(\omega=\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{0.5 \times 8 \times 10^{-6}}}\)
\(\begin{aligned}
& \omega=\frac{1}{\sqrt{4 \times 10^{-6}}} \\
& \omega=\frac{1 \times 1000}{2} \\
& \omega=500 \mathrm{~Hz}
\end{aligned}\)
So, \(\omega=\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{0.5 \times 8 \times 10^{-6}}}\)
\(\begin{aligned}
& \omega=\frac{1}{\sqrt{4 \times 10^{-6}}} \\
& \omega=\frac{1 \times 1000}{2} \\
& \omega=500 \mathrm{~Hz}
\end{aligned}\)
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