KCET · Physics · Alternating Current
A resistor of \(500 \Omega\), an inductance of \(0.5 \mathrm{H}\) are in series with an AC which is given by \(V=100 \sqrt{2} \sin (1000 t)\). The power factor of the combination is
- A \(0.6\)
- B \(\frac{1}{\sqrt{2}}\)
- C \(\frac{1}{\sqrt{3}}\)
- D \(0.5\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Voltage \(V=100 \sqrt{2} \sin (1000 t)\)
and \(\quad \mathrm{V}=\mathrm{V}_{0} \sin \omega t\)
On comparing \(\omega=1000\)
Inductive reactance
\(X_{\mathrm{L}} =\omega \mathrm{L}=1000 \times 0.5=500 \Omega \)
\(\therefore \cos \phi =\frac{500}{\sqrt{(500)^{2}+(500)^{2}}}=\frac{1}{\sqrt{2}}\)
and \(\quad \mathrm{V}=\mathrm{V}_{0} \sin \omega t\)
On comparing \(\omega=1000\)
Inductive reactance
\(X_{\mathrm{L}} =\omega \mathrm{L}=1000 \times 0.5=500 \Omega \)
\(\therefore \cos \phi =\frac{500}{\sqrt{(500)^{2}+(500)^{2}}}=\frac{1}{\sqrt{2}}\)
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