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A coil, a capacitor, and an ac source of rms voltage 24 V are connected in series. By varying the source frequency, a maximum current of 6 A is observed (resonance). If the same coil is then connected to a battery of emf 12 V with internal resistance 4 ohm, find the steady current through the coil.
- 1.5 A
- 2.0 A
- 1.0 A
- 3.0 A
Correct answer: 1.5 A
Solution
At resonance in a series LCR circuit, XL = XC, so impedance = R (the coil's resistance). Hence R = V_rms/I_max = 24/6 = 4 ohm. When the coil (resistance 4 ohm) is connected to a 12 V battery with 4 ohm internal resistance, the dc steady current is I = emf/(R_coil + r) = 12/(4 + 4) = 1.5 A. (For dc steady state the inductor offers no opposition.)
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