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Homework Help: Triac - Power dissipated in light bulb

  1. Jul 26, 2014 #1
    1. The problem statement, all variables and given/known data

    A triac is used in a dimmer switch to control a 100W light bulb. If the firing angle is set for α=∏/3 estimate the power dissipated in the bulb Rated at 100W and the voltage source 230V @ 50Hz.

    2. Relevant equations

    Vpeak = Vrms/0.707

    3. The attempt at a solution

    Vpeak = 230/0.707 = 325V

    Vload = 325 * √((2∏-(2*∏/3) + sin(2∏/3))/4∏)

    Vload = 206Vrms

    100W = 230 * I

    I = 100/230

    I = 0.43A

    power dissipated = 206 * 0.43 = 88.6 Watts

    Could someone please let me know if I'm on the right track? Or give me a nudge in the right direction? Thanks in advance
  2. jcsd
  3. Jul 26, 2014 #2


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    Staff: Mentor

    Hi steviespark; Welcome to Physics Forums.

    You've made an assumption that the current will be the same for the two cases (at full conduction versus the new conditions where the conduction angle is limited). That won't be true; the RMS current will change when the RMS voltage changes. So... what remains constant (at least by approximation or assumption!) between the scenarios? (Hint: what property of a lightbulb causes power to be dissipated when current flows though it?)
  4. Jul 27, 2014 #3
    Much appreciated Gneil,

    V/I = R

    230/0.43 = 534.884,

    Calculate I for new Voltage:

    V/R = I

    206/534.884 = 0.386

    Power Dissipated = 206 * 0.386 = 79.5 W

    Is this correct?
  5. Jul 27, 2014 #4
    Hello Steviespark,

    How did you get this?

    Vload = 325 * √((2∏-(2*∏/3) + sin(2∏/3))/4∏)?

    I'm doing the same course and feel stuck many times.
  6. Jul 27, 2014 #5


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    Staff: Mentor

    The method appears to be correct, but you should keep more decimal places in intermediate results. In particular, use a couple more decimal places for your current value.
  7. Jul 27, 2014 #6


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    Staff: Mentor

    Hi bizuputyi, Welcome to Physics Forums.

    One approach would be to perform the integration for the power for the given waveform.
  8. Jul 27, 2014 #7
    Thanks Gneill :)
  9. Jul 27, 2014 #8
    Yes, he is the best.
    Thank you Gneill.
  10. Aug 13, 2014 #9
    [itex] P = \frac{1}{T} \int_0^T \frac{V_{rms}^2}{R}dt = \frac{1}{RT} \int_0^T V_{peak}^2 \sin^2(\omega t) dt= \frac{325^2}{2*529*0.02} \int_0^{0.02} 1+\cos(2*2*π*50 t) dt = 100W[/itex]

    That seems to be correct for an ordinary sine wave. How do I do the same with the firing angle included?
  11. Aug 13, 2014 #10
    Is it becoming something like:

    [itex] P = \frac{1}{T} \int_0^{\frac{π}{3}} 0 dt + \frac{1}{T} \int_{\frac{π}{3}}^π \frac{V_{rms}^2}{R} dt + \frac{1}{T} \int_π^{\frac{4π}{3}} 0 dt + \frac{1}{T} \int_{\frac{4π}{3}}^{2π} \frac{V_{rms}^2}{R} dt[/itex]?
  12. Aug 13, 2014 #11


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    Yup, you set the integration limits to cover the conducting periods.

    Note: I didn't verify your formula in detail, but the overall form looks right to me.
  13. Aug 13, 2014 #12
    Calculating with 50Hz I've got:

    [itex] P = 2*\frac{V_{rms}^2}{2RT} \int_{0.00333}^{0.01} 1+\cos(2*2*π*50t) dt \approx 50W [/itex]

    Which is not correct I think as firing angle [itex] \frac{π}{2} [/itex] would result in half power. I must have done something wrong.
  14. Aug 13, 2014 #13


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    Okay, I've given in and taken a closer look at the setup.

    You know, you can work strictly with the angular information. No need to muck about with the actual frequency and time periods. In angular terms the total period of a cycle is ##2\pi## radians. You integrate from ##\pi / 3## to ##5 \pi /3## to account for your firing angle. Since the argument of the integral consists of squared terms, you don't have to worry about positive versus negative half cycles.

    Use the peak voltage in the calculation. The RMS value makes the assumption that the curve is a pure sinewave. With the firing angle active, it's not, and the RMS value of the new curve will be different. Using the peak value will sum the actual power dissipated.

    Your starting integral should look something like:

    $$\frac{1}{2\pi} \int_a^b \frac{E^2 sin(\theta)^2}{R} d\theta$$

    Where a and b are the angular integration limits as set by the firing angle, and ##E## is ##\sqrt{2}\;E_{RMS}##.
  15. Aug 14, 2014 #14
    I totally get your drift and ended up with steviespark's result. Thank your for your effort to help!
  16. Aug 14, 2014 #15

    The Electrician

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    Gold Member

    steviespark didn't carry enough digits in his calculations. A better result is 80.45 watts.

    Also, as gneill said, you don't need to muck around with actual frequency and time periods; even more you don't need to involve the actual voltages and resistances. The result is simply the ratio of a couple of integrals times 100 watts:


    As an aside, in the real world, incandescent light bulbs don't have a constant resistance independent of applied voltage. Their resistance varies with temperature of the filament, which in turn changes quite a bit with dissipated power.

    Attached Files:

    Last edited: Aug 14, 2014
  17. Aug 14, 2014 #16
    That looks even more accurate then, great. I want to gain that thinking you guys have, hopefully I get there some day.
  18. Mar 18, 2015 #17
    I'm really stumped on this exact question.

    What i have so far is.

    I have found the peak voltage which makes sense as this gives the amplitude of the wave.

    Vpeak= 325.269V

    from this i find the equation for the voltage sine wave is:


    I now need to find the average voltage of the circuit between pi/3 and pi

    so intergrating

    1/pi (integral) 325.269sin(x).dx gives:


    inputting values for x of pi/3 and pi

    [-325.269cos(pi)] - [-325.269cos(pi/3)]

    [103.536] - [-51.725]

    155.261V average between

    returning this from, average value to peak value

    155.261*rt2 =219.57V

    Which is way higher than anyone elses values.

    What have i done wrong?
  19. Mar 18, 2015 #18
    Just tried something different looking at other peoples solutions, but i have no idea why everyone is using sin^2? it makes no sense. the voltage wave is a sin wave. not a sin^2 wave.
  20. Mar 18, 2015 #19


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    Staff: Mentor

    The integral is solving for power dissipated over a cycle. Given voltage and resistance, what's the equation for power?
  21. Mar 18, 2015 #20

    so V does equal 325.269sin(x) but it needs to be squared?

    i cant see how R is found in this situation though?
  22. Mar 18, 2015 #21
    sorry correction V^2/R = P
  23. Mar 18, 2015 #22


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    Staff: Mentor

    What's the resistance of a 100W light bulb? You're given the working voltage (the mains source voltage).
  24. Mar 18, 2015 #23
    Seriously, how do you do that?

    how can you look at an equation and say, hey, im going to do it this way.

    That method would never have occured to me.

    Thank you very much Gneill, a legend as always!
  25. Jun 12, 2015 #24
    Using this as a starting point,
    b= 5π/3
    a= π/3
    R = V2/P = 2302 / 100 = 529Ω
    Epeak = 325.269

    $$\frac{1}{2\pi * R} \int_{π/3}^{5π/3} E^2 sin(\theta)^2 d\theta$$

    $$\frac{1}{2\pi R} E^2 \int_{π/3}^{5π/3} sin(\theta)^2 d\theta$$

    sin2 A = ½ (1 - cos 2θ)

    $$\frac{1}{2\pi R} E^2 \int_{π/3}^{5π/3} 1/2 (1 - cos 2θ) d\theta$$

    1/(2π R) * E2 [ ½ (θ-sin 2θ)] π/35π/3

    1/(2π 529) * 325.2692 [ ½ (5π/3 -sin (2 5π/3))] = 77.56W

    1/(2π 529) * 325.2692 [ ½ (π/3 -sin (2 π/3))] = 15.5W

    Any advice as to where it's starts going awry? Apologies for the formatting i don't know all the code.
    Last edited: Jun 12, 2015
  26. Jun 12, 2015 #25


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    Staff: Mentor

    I think something's gone awry in your integration of cos2θ. The "2" in the argument makes a difference. Try performing the integration alone symbolically. What method will you use?
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