An inductor-coil, a capacitor and an AC source of rms voltage 24V are connected in series. When the frequency of the source is varied, a maximum rms current of 6.0A is observed. If this inductor coil is connected to a battery of emf 12V and internal resistance 4.0Ω, what will be the current?

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1 Answers

Erms = 24V

r = 4Ω, Irms = 6A

R = E/I = 24/6 = 4Ω

Internal Resistance = 4Ω

Hence net resistance = 4 + 4 = 8Ω

∴ Current = 12/8 = 1.5A

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