An inductance of 1 mH a condenser of $10 μ F$ and a resistance of $50 ω$ are connected in series. The reactance of inductor and condensers are same. The reactance of either of them will be

100 ω

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30 ω

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3.2 ω

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10 ω

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Solution

The correct option is D $10 ω$
Given $ω L = \frac{1}{ω C} \Rightarrow ω^{2} = \frac{1}{L C}$
or $ω = \frac{1}{\sqrt{10^{- 3} \times 10 \times 10^{- 6}}} = \frac{1}{\sqrt{10^{- 8}}} = 10^{4}$
$X_{L} = ω L = 10^{4} \times 10^{- 3} = 10 ω$

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An inductance of 1 mH a condenser of 10μF and a resistance of 50ω are connected in series. The reactance of inductor and condensers are same. The reactance of either of them will be

The correct option is D 10ωGiven ωL=1ωC⇒ω2=1LC or ω=1√10−3×10×10−6=1√10−8=104 XL=ωL=104×10−3=10ω

An inductance of 1 mH a condenser of $10 μ F$ and a resistance of $50 ω$ are connected in series. The reactance of inductor and condensers are same. The reactance of either of them will be

The correct option is D $10 ω$
Given $ω L = \frac{1}{ω C} \Rightarrow ω^{2} = \frac{1}{L C}$
or $ω = \frac{1}{\sqrt{10^{- 3} \times 10 \times 10^{- 6}}} = \frac{1}{\sqrt{10^{- 8}}} = 10^{4}$
$X_{L} = ω L = 10^{4} \times 10^{- 3} = 10 ω$