JEE Mains · Physics · STD 12 -7. Alternating current
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 \(\left.\frac{22}{7}\right]\)
- A \(88\, {A}\) and \(\tan ^{-1}\left(\frac{11}{6}\right)\)
- B \(0.88\, {A}\) and \(\tan ^{-1}\left(\frac{11}{6}\right)\)
- C \(8.8 \,{A}\) and \(\tan ^{-1}\left(\frac{11}{6}\right)\)
- D \(8.8 \,{A}\) and \(\tan ^{-1}\left(\frac{6}{11}\right)\)
Answer & Solution
Correct Answer
(C) \(8.8 \,{A}\) and \(\tan ^{-1}\left(\frac{11}{6}\right)\)
Step-by-step Solution
Detailed explanation
Inductive reactance, \(X_{L}=\omega L\) \(=100 \pi \times 0.07\) \(=100 \times \frac{22}{7} \times \frac{7}{100}\) \(=22 \Omega\) \(\therefore \quad \text { Impedance, } Z =\sqrt{R^{2}+X_{L}^{2}}\) \(=\sqrt{12^{2}+22^{2}}\) \(=\sqrt{144+484}\) \(=\sqrt{628}\)…
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