AC emf applied to capacitor and lamp in series

AI Thread Summary
The discussion revolves around the calculation of capacitive reactance in a circuit with an alternating emf applied to a capacitor and a lamp in series. The confusion arises from the expectation that the lamp should take a portion of the voltage, leading to a different reactance calculation. However, the correct approach considers the total voltage of 200V, as the voltages across the capacitor and lamp are out of phase, necessitating the use of complex numbers for accurate calculations. The final capacitance required is determined to be approximately 3.797 µF, based on the total circuit parameters. Additionally, there is a query about the recommended type of capacitor for this application.
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Homework Statement


Hi, I found the following example in a high school physics book about electrical circuits. There's something about the example that I don't get, so I hope you can help me out. I give you the example text and the solution, and then I explain what I don't understand.

Example text:
"An alternating emf. of 200V and 50Hz is applied to a capacitor in series with 20V, 5W lamp. Find the capacitance required to run the lamp."

Solution:
Lamp:
V = 20V, P = I*V, i.e. I = P/V = 5W/20V = 0.25A

Circuit:
I = 200V/X (where X is the capacitive reactance)
X = 200V/I = 200V/0.25A = 800 ohm
X = 1/wC (where w is the angular frequency)
X = 1/wC = 800 ohm, i.e. C = 1/(w*800 ohm) = 1/(2*pi*50Hz*800 ohm) = 3.797 uF

What I don't understand:
Why is the reactance calculated as 200V/0.25A? I would expect the lamp to "take" 20V from the circuit, thus leaving 180V for the capacitor, i.e. X = 180V/0.25 = 720 ohm. Can someone please tell me why this is not the case?


Homework Equations


U = I*R (ohm's law)
P = I*V (power = current * voltage)
X = 1/wC (capacitive reactance)
w = 2*pi*f (angular frequency and frequency)


The Attempt at a Solution


I looked at an exercise and its solution in the book and it gave the same answer as the example from above. I looked for an answer on the internet but couldn't find a similar case. I read about lamps in electrical circuits, but what I found supported me in my expectation that the lamp should "take up" some voltage.
 
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They forgot to include the resistance of the lamp, altough the error made is much smaller than you think. The rms voltages across the capacitor and the lamp can add up to more than 200V, because the voltages are out of phase.

the correct equation with complex numbers is I = 200/(R+jX).

If you calculate the magnitudes you get I = \frac {200}{\sqrt{X^2 + R^2}}.

since X is about 10 times as large as R, this doesn't differ much from 200/X
 
The voltage across the capacitor and the voltage across the resistor are out of phase by ninety degrees so to add them use phythagoras theorem. The supply voltage is the hypotenuse of the triangle.
What type of capacitor ( mica,ceramic etc ) is recommended for such an application?
 
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