The reactance of an inductor at $50 \mathrm{~Hz}$ is $10 \Omega$. The reactance of it at $200 \mathrm{~Hz}$ is
The reactance of an inductor at $50 \mathrm{~Hz}$ is $10 \Omega$. The reactance of it at $200 \mathrm{~Hz}$ is
- $10 \Omega$
- $40 \Omega$
- $2.5 \Omega$
- $20 \Omega$
Solution
Given, initial frequency, $f_1=50 \mathrm{~Hz}$
Initial reactance, $X_1=10 \Omega$
Final frequency, $f_2=200 \mathrm{~Hz}$
Final reactance, $X_2=$ ?
Since, reactance, $X=\omega L$
$
=2 \pi f L
$
$
\begin{array}{rlrl}
\Rightarrow & & X & \propto f \\
\Rightarrow & & \frac{X_2}{X_1} & =\frac{f_2}{f_1}=\frac{200}{50}=4 \\
\Rightarrow & & X_2 & =4 X_1 \\
& & =4 \times 10=40 \Omega
\end{array}
$
Asked in: AP EAMCET 2021 (23 Aug Shift 1)
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