Q154 / 180 · KCET 2020 · Physics

The power factor of $R-L$ circuit is $\frac{1}{\sqrt{3}}$. If the inductive reactance is $2 \Omega$. The value of resistance is

  • A
    $2 \Omega$
  • Correct answer
    B
    $\sqrt{2} \Omega$
  • C
    $0.5 \Omega$
  • D
    $\frac{1}{\sqrt{2}} \Omega$

Solution

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 $\begin{aligned} \frac{1}{3} &=\frac{R^{2}}{R^{2}+4} \\ \Rightarrow \quad R^{2}+4 &=3 R^{2} \Rightarrow R=\sqrt{2} \Omega \end{aligned}$

Hence, the correct answer is option: 'B'.

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