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KCET · Physics · Alternating Current

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

  1. A \(2 \Omega\)
  2. B \(\sqrt{2} \Omega\)
  3. C \(0.5 \Omega\)
  4. D \(\frac{1}{\sqrt{2}} \Omega\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{2} \Omega\)

Step-by-step Solution

Detailed explanation

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
\(\frac{1}{3} =\frac{R^{2}}{R^{2}+4} \)
\( \Rightarrow \quad R^{2}+4 =3 R^{2} \Rightarrow R=\sqrt{2} \Omega\)