AC emf applied to capacitor and lamp in series


by mja78
Tags: capacitor, electrical circuits, lamp, reactance
mja78
mja78 is offline
#1
Oct4-09, 01:15 AM
P: 1
1. The problem statement, all variables and given/known data
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?


2. Relevant 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)


3. 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.
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
willem2
willem2 is offline
#2
Oct4-09, 05:33 AM
P: 1,351
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 [tex] I = \frac {200}{\sqrt{X^2 + R^2}} [/tex].

since X is about 10 times as large as R, this doesn't differ much from 200/X
bendavy
bendavy is offline
#3
Mar12-11, 07:50 AM
P: 1
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?


Register to reply

Related Discussions
Capacitor and Inductor in Series Engineering, Comp Sci, & Technology Homework 9
Phase shift using a series capacitor Engineering, Comp Sci, & Technology Homework 2
Capacitor in Series and Parallel Introductory Physics Homework 6
Capacitor in Series/Parallel Introductory Physics Homework 1
International Series in Pure & Applied Maths? Science & Math Textbook Listings 0