Calculating Resistance and Inductance of Coil in AC Circuit

[Squeeky21]

1. Homework Statement .
A wire coil has both resistance and inductance, and is connected in series to an AC power supply with 80Hz and 180V, drawing 0.4A current and 18W power. The circuit consists of a single loop.
1. Calculate the inductance and resistance of coil
2. Plot vector of impedance of coil at 50Hz in the impedance place
3. A capacitor is added to the circuit. What capacitance would maximize the current and power delivered to the coil?
4. What will the new current and power be once the capacitance of part 3 is added?


2. Homework Equations
P/I^2=R
XL=V/I
L=XL/2*pi*f
Z=sqrt(R^2+XL^2)
XC=V/I
C=1/(2*∏*f*XC)

3. The Attempt at a Solution .
I found P/I^2=R--> 18W/.4A^2=112.5ohms.
Then I found XL=V/I-->170V/.4A=425ohms.
Then I found L=XL/2*pi*f-->425ohms/(2*pi*60Hz)=1.23H.
Next, using Z=sqrt(R^2+XL^2), I got sqrt(112.5^2+425^2)=439.637 oms.
Next, I have V/I=XC-->170V/.4A=425Ω.
Next, C=1/(2*∏*f*XC) -->1/(2*∏*60Hz*425Ω)=6.24*10^-6F.
I feel I am off somewhere, but I am not sure where.
I also am not sure how to recalculate for power and current because I keep getting my original values.
I feel I made an error somewhere, but I am not sure where.
I am also still confused as to what happens when I add the capacitor.

 

[gneill]

1. Homework Statement .
A wire coil has both resistance and inductance, and is connected in series to an AC power supply with 80Hz and 180V, drawing 0.4A current and 18W power. The circuit consists of a single loop.
1. Calculate the inductance and resistance of coil
2. Plot vector of impedance of coil at 50Hz in the impedance place
3. A capacitor is added to the circuit. What capacitance would maximize the current and power delivered to the coil?
4. What will the new current and power be once the capacitance of part 3 is added?


2. Homework Equations
P/I^2=R
XL=V/I
L=XL/2*pi*f
Z=sqrt(R^2+XL^2)
XC=V/I
C=1/(2*∏*f*XC)

3. The Attempt at a Solution .
I found P/I^2=R--> 18W/.4A^2=112.5ohms.
Then I found XL=V/I-->170V/.4A=425ohms.

Where did the 170V come from?

If the supply voltage is E = 180V, then you would expect the magnitude of the current to be given by:
$$ |I| = \frac{E}{\sqrt{R^2 + XL^2}} $$
Go from there.

I am also still confused as to what happens when I add the capacitor.

The capacitor reactance acts to cancel the inductive reactance (bringing the power factor towards unity). How do reactances XL and XC combine arithmetically?

 

[Squeeky21]

Thank you. I finally got this one, I think. Z=sqrt(R^2+(XL-XC)^2) correct?

 

[gneill]

Thank you. I finally got this one, I think. Z=sqrt(R^2+(XL-XC)^2) correct?

Yup. Carry on