A capacitor and a coil having a resistance R are in series are connected to a 6 Volt AC source. By varying the frequency of the source, maximum current of 600 mA is observed. If the same coil is now connected to a cell of emf 6 volt and internal resistance of 2 ohm, the current through it will be :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.5 A
The current is maximum at resonance, where \( X_L = X_C \) and the impedance equals the resistance \( R \) of the coil.
\( I_{\max} = 600\,\mathrm{mA} = 0.6\,\mathrm{A} \)
\( \displaystyle R = \frac{V}{I_{\max}} = \frac{6\,\mathrm{V}}{0.6\,\mathrm{A}} = 10\,\Omega \)
With a steady (DC) current the inductance of the coil has no effect, so only its resistance and the internal resistance of the cell oppose the current.
\( \displaystyle I = \frac{E}{R + r} = \frac{6}{10 + 2} = 0.5\,\mathrm{A} \)