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
- A \(2 \Omega\)
- B \(\sqrt{2} \Omega\)
- C \(0.5 \Omega\)
- D \(\frac{1}{\sqrt{2}} \Omega\)
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
\(\frac{1}{3} =\frac{R^{2}}{R^{2}+4} \)
\( \Rightarrow \quad R^{2}+4 =3 R^{2} \Rightarrow R=\sqrt{2} \Omega\)
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
\(\frac{1}{3} =\frac{R^{2}}{R^{2}+4} \)
\( \Rightarrow \quad R^{2}+4 =3 R^{2} \Rightarrow R=\sqrt{2} \Omega\)
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