Calculating capacitance from an AC circuit

[imas145]

This question is from my physics book and neither I nor my teacher can find the mistake in my method. (The original question is not in English so I'll do my best translating it)

> When a coffee maker is connected to the wall (230 V, 50 Hz) its power consumption is 180 W. The power consumption is wanted to cut in half by connecting a capacitor in series. How large should the capacitor's capacitance be?$$U=230 V, f=50 Hz, P_1=180W, P_2=90W, X_L=0$$
$$P_1=UI=RI^2=R(\frac{U}{R})^2=\frac{U^2}{R} \leftrightarrow PZ^2=RU^2 \leftrightarrow P=\frac{RU^2}{Z^2}$$
$$Z_1=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2}=R=\frac{U^2}{P_1}$$
$$Z_2=\sqrt{R^2+(X_L-X_C)^2} = \sqrt{R^2+X_C^2}=\sqrt{R^2+\frac{1}{4\pi^2f^2C^2}}$$
$$Z=\frac{U}{I} \leftrightarrow ZI=U \leftrightarrow I=\frac{U}{Z}$$

My method

$$P_1=2P_2$$
$$UI_1=2UI_2$$
$$I_1=2I_2$$
$$\frac{U}{Z_1}=\frac{2U}{Z_2}$$
$$\frac{1}{Z_1}=\frac{2}{Z_2}$$
$$2Z_1=Z_2$$
$$2Z_1=\sqrt{R^2+\frac{1}{4\pi^2f^2C^2}}$$
$$R^2+\frac{1}{4\pi^2f^2C^2}=4Z_1^2$$
$$\frac{1}{4\pi^2f^2C^2}=4Z_1^2-R^2$$
$$4\pi^2f^2C^2(4Z_1^2-R^2)=1$$
$$C^2=\frac{1}{4\pi^2f^2(4Z_1^2-R^2)}$$
$$C=\frac{1}{2\pi f\sqrt{4Z_1^2-R^2}}$$

Correct method

$$P_1=2P_2$$
$$\frac{U^2}{Z_1}=\frac{2RU^2}{Z_2^2}$$
$$2Z_1^2=Z_2^2$$
$$2R^2=R^2+\frac{1}{4\pi^2f^2C^2}$$
$$\frac{1}{4\pi^2f^2C^2}=R^2$$
$$C=\frac{1}{2\pi fR}$$

As you can see, my answer differs by that I have $$\sqrt{4Z_1^2-R^2}$$ in the denominator, whereas the correct solution only has $$R$$. I can follow the correct method and I'm fairly sure all the algebra in my solution is also correct, so the mistake is most likely a theoretical one. My question is then: why did my method fail? My guess is that it has something to do with the phase difference because the second circuit has a capacitor as well as a resistor while the first one only has the resistor.

 

[cnh1995]

why did my method fail?

This question is from my physics book and neither I nor my teacher can find the mistake in my method. (The original question is not in English so I'll do my best translating it)

> When a coffee maker is connected to the wall (230 V, 50 Hz) its power consumption is 180 W. The power consumption is wanted to cut in half by connecting a capacitor in series. How large should the capacitor's capacitance be?$$U=230 V, f=50 Hz, P_1=180W, P_2=90W, X_L=0$$
$$P_1=UI=RI^2=R(\frac{U}{R})^2=\frac{U^2}{R} \leftrightarrow PZ^2=RU^2 \leftrightarrow P=\frac{RU^2}{Z^2}$$
$$Z_1=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2}=R=\frac{U^2}{P_1}$$
$$Z_2=\sqrt{R^2+(X_L-X_C)^2} = \sqrt{R^2+X_C^2}=\sqrt{R^2+\frac{1}{4\pi^2f^2C^2}}$$
$$Z=\frac{U}{I} \leftrightarrow ZI=U \leftrightarrow I=\frac{U}{Z}$$

My method

$$P_1=2P_2$$
$$UI_1=2UI_2$$ Is this equation correct?
$$I_1=2I_2$$
$$\frac{U}{Z_1}=\frac{2U}{Z_2}$$
$$\frac{1}{Z_1}=\frac{2}{Z_2}$$
$$2Z_1=Z_2$$
$$2Z_1=\sqrt{R^2+\frac{1}{4\pi^2f^2C^2}}$$
$$R^2+\frac{1}{4\pi^2f^2C^2}=4Z_1^2$$
$$\frac{1}{4\pi^2f^2C^2}=4Z_1^2-R^2$$
$$4\pi^2f^2C^2(4Z_1^2-R^2)=1$$
$$C^2=\frac{1}{4\pi^2f^2(4Z_1^2-R^2)}$$
$$C=\frac{1}{2\pi f\sqrt{4Z_1^2-R^2}}$$

Correct method

$$P_1=2P_2$$
$$\frac{U^2}{Z_1}=\frac{2RU^2}{Z_2^2}$$
$$2Z_1^2=Z_2^2$$
$$2R^2=R^2+\frac{1}{4\pi^2f^2C^2}$$
$$\frac{1}{4\pi^2f^2C^2}=R^2$$
$$C=\frac{1}{2\pi fR}$$

As you can see, my answer differs by that I have $$\sqrt{4Z_1^2-R^2}$$ in the denominator, whereas the correct solution only has $$R$$. I can follow the correct method and I'm fairly sure all the algebra in my solution is also correct, so the mistake is most likely a theoretical one. My question is then: why did my method fail? My guess is that it has something to do with the phase difference because the second circuit has a capacitor as well as a resistor while the first one only has the resistor.

 

[Asymptotic]

The math may have become unnecessarily complex.

Another approach:

1. Calculate heater resistance from the given voltage and wattage rating.
2. Power is cut in half when X_c equals heater resistance. X_c=R_heater
   
3. X_c=1/(2πfC). Rearrange to solve for C.

C=1/(2πfX_c)

 

[imas145]

Would the correct equation then be $P=UIcos\theta$ where $\theta$ is the phase difference, hence making my approach unusable since I don't know the phase difference?

 

[cnh1995]

Would the correct equation then be $P=UIcos\theta$ where $\theta$ is the phase difference, hence making my approach unusable since I don't know the phase difference?

No. You have assumed the voltage to be same in both the cases. Is that correct?

 

[imas145]

No. You have assumed the voltage to be same in both the cases. Is that correct?

I think so, since it’s plugged into the wall which supplies 230 volts

 

[cnh1995]

Ok, I was assuming U to be the voltage across the coffee-maker and not the supply voltage. If you are assuming U to be 230V, then you're right you should include cosθ in your equation.

I think so, since it’s plugged into the wall which supplies 230 volts

The active power in the circuit i.e. the power dissipated in the coffee maker is halved. You have assumed U=230V, means you wrote the equation for the "apparent power" in the circuit. Add power factor in the equation.

 

[Asymptotic]

Would the correct equation then be $P=UIcos\theta$ where $\theta$ is the phase difference, hence making my approach unusable since I don't know the phase difference?

No, that wasn't it. I believe the formula @cnh1995 highlighted in red is where things went wrong.

As you can see, my answer differs by that I have

$\sqrt{4Z_1^2-R^2}$

in the denominator, whereas the correct solution only has R

Power is cut in half when Xc equals heater resistance. Xc=R_heater

This load is an impedance consisting of the series connected heater resistance, and capacitive reactance. To reduce heater power in two, Z must be twice the heater resistance (that is, heater resistance plus capacitive reactance = Z). Half of the supply voltage is dropped across the heater, and the other half is dropped across the capacitor.

In a circuit built with an actual (non-ideal) resistance and capacitance it may be necessary to consider the minor role inductance and capacitance have in the heater, inductive and resistive properties of the capacitor, and factor in their resulting phase shifts, but in this ideal case Xc=R_heater, and is why the correct solution only had R in the formula $C=\frac{1}{2\pi fR}$

Edit:I'm forgetting the basics in my dotage; what I said above is wrong - It is NOT half of the supply voltage.
In a capacitor, current leads voltage by 90°. It is the vectoral sum of resistor and capacitor voltage that adds up to the supply voltage. Working back from calculated element resistance (180W at 230V) unless I've boogered it up again, Vr and Vc are both 162.7 volts with the capacitor added.