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The power factor of $R-L$ circuit is $\frac{1}{\sqrt{3}}$. If the inductive reactance is $2 \Omega$. The value of resistance is
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Verified Answer
The correct answer is:
$\sqrt{2} \Omega$
Given, power factor $=\frac{1}{\sqrt{3}}$
Inductive reactance, $X_{L}=2 \Omega$
The power factor in a $R-L$ circuit is given by
Power factor $=\frac{\text { Resistance }}{\text { Impedance }}=\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}$ $\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{R}{\sqrt{R^{2}+4}}$
Squaring both sides, we get
$\begin{aligned}
\frac{1}{3} &=\frac{R^{2}}{R^{2}+4} \\
\Rightarrow \quad R^{2}+4 &=3 R^{2} \Rightarrow R=\sqrt{2} \Omega
\end{aligned}$
Inductive reactance, $X_{L}=2 \Omega$
The power factor in a $R-L$ circuit is given by
Power factor $=\frac{\text { Resistance }}{\text { Impedance }}=\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}$ $\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{R}{\sqrt{R^{2}+4}}$
Squaring both sides, we get
$\begin{aligned}
\frac{1}{3} &=\frac{R^{2}}{R^{2}+4} \\
\Rightarrow \quad R^{2}+4 &=3 R^{2} \Rightarrow R=\sqrt{2} \Omega
\end{aligned}$
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