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.
A 110 V, 60 Hz AC line is connected to a 6.55 mH coil. Find the current through the coil.
4 answers
Xl = 6.28*60*(6.55*10^-3h) = 2.47 Ohms.
= Reactance of coil.
I=V / Xl = 110 / 2.47 = 44.5Amps., rms.
= Reactance of coil.
I=V / Xl = 110 / 2.47 = 44.5Amps., rms.
My Imax result is equivalent to Henry's Irms result. The ratio of the two is sqrt2
Thank u! I was halfway there.