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An inductor L, a capacitor of 20 \mu \mathrm{F} and a resistor of 10 \Omega are connected in series with an AC source of frequency 50 Hz. If the current is in phase with the voltage, then the inductance of the inductor is:

Option: 1

2.00 H


Option: 2

0.51 H


Option: 3

1.5 H


Option: 4

0.99 H


Answers (1)

best_answer

In an L-C-R circuit, the current and the voltage are in phase \mathrm{(\phi=0)} , when 

\mathrm{\tan \phi=\frac{\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}}{\mathrm{R}}=0 \quad}

or 
\mathrm{\quad \omega \mathrm{L}=\frac{1}{\omega \mathrm{C}} or \quad \mathrm{L}=\frac{1}{\omega^2 \mathrm{C}}}
Here, \mathrm{\omega=2 \pi \mathrm{f}=2 \times 3.14 \times 50 \mathrm{~s}^{-1}=314 \mathrm{~s}^{-1} and \mathrm{C}=20 \mu \mathrm{F}=20 \times 10^{-6} \mathrm{~F} }
\mathrm{\therefore \quad \mathrm{L}=\frac{1}{\left(314 \mathrm{~s}^{-1}\right)^2 \times\left(20 \times 10^{-6} \mathrm{~F}\right)}=0.51 \mathrm{H} }

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shivangi.bhatnagar

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