A circuit containing resistance $R_{1}$ , Inductance $L_{1}$ and capacitance $C_{1}$ connected in series resonates at the same frequency $n$ as a second combination of $R_{2}$ , $L_{2}$ and $C_{2}$ . If the two are connected in series, then the circuit will resonate at:

n

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2 n

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\sqrt{\frac{L_{2} C_{2}}{L_{1} C_{1}}}

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\sqrt{\frac{L_{1} C_{1}}{L_{2} C_{2}}}

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Solution

The correct option is B $n$
Condition of resonance $ω = \frac{1}{\sqrt{L C}}$

So $ω_{1} = \frac{1}{\sqrt{L_{1} C_{1}}}$

$ω_{2} = \frac{1}{\sqrt{L_{2} C_{2}}}$

Since both frequency are same $L_{1} C_{1} = L_{2} C_{2}$

Equivalent capacitance of $C_{1} a n d C_{2}$ in series = $\frac{C_{1} C_{2}}{C_{1} + c_{2}}$

Equivalent inductance of $L_{1}$ and $L_{2} = L_{1} + L_{2}$

Resonance for combination = $\sqrt{\frac{(C_{1} + C_{2})}{C_{1} C_{2} (L_{1} + L_{2})}}$

Using $L_{1} C_{1} = L_{2} C_{2}$ , resonance of combination = $\frac{1}{\sqrt{L_{1} C_{1}}}$

So the frequency of resonance remains same = $n$

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Similar questions

L_{1} C_{1}

L_{2} C_{2}

ω

R = 100 Ω

L_{1} = 1.7 m L

L_{2} = 2.3 m H

C_{1} = 4 μ F, C_{2} = 2.5 μ F

C_{3} = 3.5 μ F

f_{o}

\sqrt{3}

μ

Ω

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