Single Phase A.C. Coil question help please

1. Jan 20, 2013

Physicist3

1. A coil connected to a 230V 50Hz supply takes a current of 10A at a phase angle of 30o Calculate the resistance and inductance of the coil

2. i = ImSinωt, R=V/I and XL=2∏fL

3. To get the resistance I used the value of 10A to get the value of Imby dividing 10 by sin30 which = 20A. I then multiplied this by 0.707 to get RMS value as I believe the Voltage to be RMS as well. I then divided voltage by current to get resistance of 16.26Ω but the answer in the book says 21.65Ω and I cant seem to get this value. For the inductance I divided voltage by current again to get XL which I get as 16.26 again. I then divided this by 2∏ x 50 to get a value of 0.052H but the book says 39.8mH and I cant seem to work out how to get to the answers in the book

Last edited by a moderator: Jan 20, 2013
2. Jan 20, 2013

Staff: Mentor

Model it by a series circuit: R and L. Of the 230V applied, use trig to determine how much voltage appears across the R and how much across Ѡ.L

You know the current in each series element is 10A.

Parameters are already in RMS, so no further conversion is needed.

3. Jan 20, 2013

Physicist3

Sorry but am I missing something because im really struggling as how to find how much voltage is in each element if ive been given a source voltage, frequency and current?

4. Jan 20, 2013

Staff: Mentor

You are also told the power-factor angle, so ....

You draw a right-angled triangle showing how the voltage across R adds to the voltage across L to produce 230V. Mark in the angle whose value you know.

5. Jan 21, 2013

Staff: Mentor

Explaining further: because the elements are in series, it is convenient that we take the current as reference, because that current is common to each element (it's identical in each element at all times). The voltage across the R is in phase with this current, (it always is for a resistor) so you can draw the voltage phasor as being parallel to the current phasor. The voltage across the L is 90 degrees ahead of the current, so you can draw the inductor's voltage phasor as 90 degrees ahead of the current.