An inductor and a resistor are connected in series to an ac source of variable frequency. When the frequency…
An inductor and a resistor are connected in series to an ac source of variable frequency. When the frequency of the applied ac is $50 \mathrm{~Hz}$, the power factor of the circuit is $\frac{\sqrt{3}}{2}$. If the frequency of the ac is increased by $200 \%$, the power factor of the circuit is ____.
$0.8$
$0.9$
$0.7$
$0.5$
Solution
Step 1: Calculate the initial inductive reactance
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The power factor of a series RL circuit is given by \(\cos \phi =\frac{R}{Z}\), where \(Z\) is the impedance.
The impedance of a series RL circuit is \(Z=\sqrt{R^{2}+X_{L}^{2}}\), where \(X_{L}\) is the inductive reactance.
Given the initial frequency \(f_{1}=50\mathrm{~Hz}\) and the initial power factor \(\cos \phi _{1}=\frac{\sqrt{3}}{2}\), we can write:\(\cos \phi _{1}=\frac{R}{Z_{1}}=\frac{R}{\sqrt{R^{2}+X_{L1}^{2}}}=\frac{\sqrt{3}}{2}\)
Squaring both sides and simplifying:
\(\frac{R^{2}}{R^{2}+X_{L1}^{2}}=\frac{3}{4}\)\(4R^{2}=3R^{2}+3X_{L1}^{2}\)\(R^{2}=3X_{L1}^{2}\)\(R=\sqrt{3}X_{L1}\)
Since inductive reactance is \(X_{L}=2\pi fL\), we have \(X_{L1}=2\pi (50)L=100\pi L\). Therefore, the relationship between R and L is \(R=\sqrt{3}(100\pi L)\).
Step 2: Calculate the new frequency and inductive reactance
The frequency of the AC source is increased by \(200\%\).
The new frequency, \(f_{2}\), is:\(f_{2}=f_{1}+200\%\text{\ of\ }f_{1}=f_{1}+2f_{1}=3f_{1}\)\(f_{2}=3(50\mathrm{~Hz})=150\mathrm{~Hz}\)
The new inductive reactance, \(X_{L2}\), is proportional to the new frequency:
\(X_{L2}=2\pi f_{2}L=2\pi (150)L=300\pi L\)
Using the relationship from Step 1, \(100\pi L=\frac{R}{\sqrt{3}}\), we can express the new reactance in terms of R:\(X_{L2}=3(100\pi L)=3\left(\frac{R}{\sqrt{3}}\right)=\sqrt{3}R\)
Step 3: Calculate the new power factor The new power factor, \(\cos \phi _{2}\), is given by:\(\cos \phi _{2}=\frac{R}{Z_{2}}=\frac{R}{\sqrt{R^{2}+X_{L2}^{2}}}\)Substituting the expression for \(X_{L2}\) from Step 2:\(\cos \phi _{2}=\frac{R}{\sqrt{R^{2}+(\sqrt{3}R)^{2}}}=\frac{R}{\sqrt{R^{2}+3R^{2}}}=\frac{R}{\sqrt{4R^{2}}}=\frac{R}{2R}=\frac{1}{2}\)
Answer: The power factor of the circuit is \(\frac{\mathbf{1}}{\mathbf{2}}\).