Engineering Ac circuit non inductive resistor and a coil

AI Thread Summary
A non-inductive resistor draws 8A at 100 volts, leading to a resistance of 12.5 ohms. To connect a coil with negligible resistance in series for a 220V, 60Hz supply, the inductance must be calculated using the impedance formula Z^2 = R^2 + XL^2. The voltage divider rule for complex impedance is applicable here to find the voltage across the resistor and the coil. Ultimately, any finite inductance can supply the load as long as the RMS current remains consistent.
herbgriffin
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Homework Statement


A non inductive resistor takes 8A at 100 volts. What inductance of a coil of negligible resistance must be connected in series in order that this load can be supplied from a 220 volt 60 Hz mains.



Homework Equations


Z^2 = R^2 + XL^2
R=V/I
XL = 2pi(frequency) L(inductance)

The Attempt at a Solution


i have solve the resistance before the coil is added.. R=12.5 ohms...
i am stuck on how to get the value of the coil since there is no given current..
 
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(Stipud educators. The correct answer to this question is that any finite inductance will "supply" the load!)

The implied question phrased in more than half-witted manner is, "What will supply the load at the same RMS current?

You know the new voltage. It's 220 VAC. So it's not the same one you used to find R.

Use the voltage divider rule for complex impedance in series. Remember the one for finding the voltage across one resistor that's in series with a second resistor? It looks the same.
 

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