AC emf applied to capacitor and lamp in series

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SUMMARY

The discussion focuses on calculating the required capacitance to operate a lamp in series with an alternating emf of 200V at 50Hz. The solution reveals that the capacitive reactance is derived from the total voltage applied across the circuit rather than the voltage across the lamp alone. The calculated capacitance needed to run the 20V, 5W lamp is approximately 3.797 µF, based on the relationship between current, voltage, and capacitive reactance. The misunderstanding arises from the phase difference between the voltages across the lamp and capacitor, which necessitates using the total voltage for reactance calculations.

PREREQUISITES
  • Understanding of alternating current (AC) circuits
  • Familiarity with capacitive reactance (X = 1/wC)
  • Knowledge of Ohm's Law (U = I*R)
  • Basic principles of phase relationships in AC circuits
NEXT STEPS
  • Study the concept of phasors in AC circuit analysis
  • Learn about the different types of capacitors and their applications
  • Explore the use of complex numbers in electrical engineering
  • Investigate the effects of reactance and resistance in series circuits
USEFUL FOR

Physics students, electrical engineering students, and professionals involved in circuit design and analysis will benefit from this discussion, particularly those interested in AC circuit behavior and capacitor applications.

mja78
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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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