A \(0.07\,{H}\) inductor and a \(12 \,\Omega\) resistor are connected in series to a \(220 \,{V}, 50 \,{Hz}\) ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take \(\pi\) as \(\left.\frac{22}{7}\right]\)

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Answer & Solution

Correct Answer

(C) \(8.8 \,{A}\) and \(\tan ^{-1}\left(\frac{11}{6}\right)\)

Detailed explanation

Inductive reactance, \(X_{L}=\omega L\)

\(=100 \pi \times 0.07\)

\(=100 \times \frac{22}{7} \times \frac{7}{100}\)

\(=22 \Omega\)

\(\therefore \quad \text { Impedance, } Z =\sqrt{R^{2}+X_{L}^{2}}\)

\(=\sqrt{12^{2}+22^{2}}\)

\(=\sqrt{144+484}\)

\(=\sqrt{628}\)

\( \approx 25 \Omega\)

\(\therefore \text { Current, } I=\frac{V}{Z}=\frac{220}{25} =8.8 A\)

\(\Rightarrow \quad I=8.8 \Lambda\)

Also, phase angle between current and source voltage can be given as

\(\tan \phi =\left(\frac{X_{L}}{R}\right) \Rightarrow \tan \phi=\frac{22}{12}\)

\(\Rightarrow \quad \phi =\tan ^{-1}\left(\frac{11}{6}\right)\)

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