1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Power drawn from capacitor in series w bulb & voltage supply

  1. Mar 14, 2015 #1
    Hi again - partially stuck with a question and I'm not sure where to go. Hoping someone would be able to give me a hint :)

    A 75 W non-inductive light bulb is designed to run from an ac supply of 120 V rms to 50 Hz. If the only supply available is 240 V rms show that the bulb can be run at the correct power by placing a capacitor C in series with it. Show that the capacitor draws 75 W from the supply and calculate it's capacitance.

    So I've said that in order for the bulb to still run at the same power, it will need the same charge to flow across it (I'm not sure if this is the right argument though?) Therefore, the power the capacitor draws will be 0.5*Q*V, where V=240 V and Q=75/120 & this gives 75 W.

    For the capacitance, I said that 75 W will be equal to V^2/2Z, where Z=1/(omega*C); the impedance of the capacitor. This gives me 8.3 micro Farads and the answer is 9.6 micro Farads. What have I done wrong?
  2. jcsd
  3. Mar 14, 2015 #2


    User Avatar
    Homework Helper
    Gold Member

    The main thing you have ignored is the phase difference between the voltage across the capacitor and the current through the capacitor.
  4. Mar 14, 2015 #3
    Oh I think I see. There's a pi/2 phase difference between them? Not sure how to account for it though?

    Also - was my original argument about the charge being the same valid? I wasn't completely convinced that it was :s
  5. Mar 14, 2015 #4


    User Avatar
    Homework Helper
    Gold Member

    To be honest, when I saw the bit about charge, I didn't think too much about it because I thought that was the wrong track.

    I also wasn't happy about "the capacitor draws power" neither in your attempt, nor in the question. As far as I'm concerned, ideal capacitors don't draw power, and real ones don't draw much.

    But back to the plot, you know the voltage across the lamp, the power dissipated by the lamp and hence the current through it.

    For the capacitor you don't know the voltage across it (yet), but you do know the current. You would like to know V because then you could find Z, which as you said, will give you C.

    Now can you draw a diagram relating the voltages across the lamp Vr, the capacitor Vc and the supply Vs, taking account of phase?
    This will allow you to calculate Vc.
  6. Mar 14, 2015 #5
    I see - I've got it now!

    Think the "charge thing" I used was just a coincience that it worked
  7. Mar 15, 2015 #6
    Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook