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

 

[rezeew]

I would first commend the student for their attempt at solving the problem and using the appropriate equations. However, there are a few errors that need to be addressed.

To calculate the inductance, the formula should be L=XL/(2*pi*f), not XL/2*pi*f.

Also, the value of frequency given in the problem is 80Hz, not 60Hz. So the calculations for inductance and reactance should be based on 80Hz, not 60Hz.

For part 3, the question asks for the capacitance that would maximize the current and power delivered to the coil. In order to maximize the power, the capacitor should be chosen such that the reactance (XC) and resistance (R) are equal in magnitude. This can be achieved by setting XC=XL, which gives XC=XL=425Ω. Using the formula C=1/(2*pi*f*XC), we can solve for the capacitance which comes out to be 3.73*10^-6F.

For part 4, once the capacitance is added, the new impedance of the circuit can be calculated using the formula Z=sqrt(R^2+(XL-XC)^2). This will give a new value of impedance which can be used to calculate the new current and power using the formulas P=I^2*R and I=V/Z.

Overall, the approach taken by the student is correct but there are some mistakes in the calculations and interpretation of the problem. It is important to carefully read and understand the given information in order to solve the problem accurately.