# Physics

A 110 V, 60 Hz AC line is connected to a 6.55 mH coil. Find the current through the coil.

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1. The impedance is
Z = 2 pi f L

where f = 60 Hz and L = 6.55*10^-3 H
= 2.47 Ohms

The peak voltage is
Vmax = Vsqrt2 = 155.6 V

Current will lag voltage by 90 degrees,
and the peak current will be
Vmax/Z = 63.0 Amperes

The resistance of the coil has been neglected.

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posted by drwls
2. Xl = 6.28*60*(6.55*10^-3h) = 2.47 Ohms.
= Reactance of coil.

I=V / Xl = 110 / 2.47 = 44.5Amps., rms.

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posted by Henry
3. My Imax result is equivalent to Henry's Irms result. The ratio of the two is sqrt2

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posted by drwls
4. Thank u! I was halfway there.

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posted by Noorah

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