- #1
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.
> 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.