Why is the dc voltage less than the ac?

impedance of coil 0 ohms @ dc , 151 ohms @ 60 cycle ac. why expect difference in required supply voltage maintain .4 amps of current.. voltage difference measured greater expected difference. expected ac source voltage @ .4 amps 61 volts measured 80 volts.


@ 60 hz impedance of coil (xl) = 2pi fl = (6.28)*(60hz)*(.4h) = 151 ohm = (0 +j151)ohm
@ 60 hz impedance of resistor = 25 ohm = (25 + j0)ohm
total impedance (zt) = (r + jxl) = [(25 + j0) + (0 + j151)] = sq rt [(r^2) + (xl^2) = sq rt [(25^2) + (151^2)] = 153 ohm
voltage = (current)*(zt) = [(.4a)(153 ohm)] = 61 volts (but measured 80 volts - heavens amiss)

@ dc frequency 0, impedance of coil = 2pi fl = (6.28)*(0 hz)*(.4h) = 0 ohms = (0 + j0)
@ dc impedance of resistor remains = 25 ohm = {25 + j0) ohm
total impedance(zt) = (r + jxl) = [(25 + j0) + (0 + j0) ]= (25 + j0) ohm = sq rt [(25^2 + 0^2) = 25 ohm
voltage = (current)*((zt) = (,4a)*(25 ohm) = 10 volts (you measured 10 volts- well)

conclusion: 2 possibilities;

1. made mistake when measured 80 volts or current @ (80v/153 ohm) = .523 amps instead of .4 amps when 80 volts measured.

or;

2. inductance of coil .527h instead of .4h. make zt = 200 ohms , voltage = (.4a)*(200 ohm) = 80 volts measured.

in experament there 0.4h coil , 25 ohm resister conected in serise. when 0.4 amps flowing through circuit dc supply voltage 10v , ac 80v. why this?

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