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Homework Help: AC emf applied to capacitor and lamp in series

  1. Oct 4, 2009 #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."

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

    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.
  2. jcsd
  3. Oct 4, 2009 #2
    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
  4. Mar 12, 2011 #3
    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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