Triac - Power dissipated in light bulb

[steviespark]

Homework Statement

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.

Homework Equations

Vpeak = Vrms/0.707

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

 

[gneill]

Homework Statement

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.

Homework Equations

Vpeak = Vrms/0.707

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

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?)

 

[steviespark]

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?

 

[bizuputyi]

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.

 

[gneill]

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?

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.

 

[gneill]

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.

Hi bizuputyi, Welcome to Physics Forums.

One approach would be to perform the integration for the power for the given waveform.

 

[steviespark]

Thanks Gneill :)

 

[bizuputyi]

Yes, he is the best.
Thank you Gneill.

 

[bizuputyi]

$P = \frac{1}{T} \int_0^T \frac{V_{rms}^2}{R}dt = \frac{1}{RT} \int_0^T V_{peak}^2 \sin^2(\omega t) dt= \frac{325^2}{2*529*0.02} \int_0^{0.02} 1+\cos(2*2*π*50 t) dt = 100W$

That seems to be correct for an ordinary sine wave. How do I do the same with the firing angle included?

 

[bizuputyi]

Is it becoming something like:

$P = \frac{1}{T} \int_0^{\frac{π}{3}} 0 dt + \frac{1}{T} \int_{\frac{π}{3}}^π \frac{V_{rms}^2}{R} dt + \frac{1}{T} \int_π^{\frac{4π}{3}} 0 dt + \frac{1}{T} \int_{\frac{4π}{3}}^{2π} \frac{V_{rms}^2}{R} dt$?

 

[gneill]

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.

 

[bizuputyi]

Calculating with 50Hz I've got:

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

Which is not correct I think as firing angle $\frac{π}{2}$ would result in half power. I must have done something wrong.

 

[gneill]

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 $\sqrt{2}\;E_{RMS}$.

 

[bizuputyi]

I totally get your drift and ended up with steviespark's result. Thank your for your effort to help!

 

[The Electrician]

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.

 

[bizuputyi]

That looks even more accurate then, great. I want to gain that thinking you guys have, hopefully I get there some day.

 

[The_daddy_2012]

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:

325.269sin(x)V

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:

-325.269cos(x)

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?

 

[The_daddy_2012]

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.

 

[gneill]

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.

The integral is solving for power dissipated over a cycle. Given voltage and resistance, what's the equation for power?

 

[The_daddy_2012]

V^2R=P

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

i can't see how R is found in this situation though?

 

[The_daddy_2012]

sorry correction V^2/R = P

 

[gneill]

i can't see how R is found in this situation though?

What's the resistance of a 100W light bulb? You're given the working voltage (the mains source voltage).

 

[The_daddy_2012]

Seriously, how do you do that?

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

That method would never have occurred to me.

Thank you very much Gneill, a legend as always!

 

[Gremlin]

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

Using this as a starting point,
b= 5π/3
a= π/3
R = V²/P = 230² / 100 = 529Ω
E_peak = 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/3^5π/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.

 

[gneill]

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?

 

[Gremlin]

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?

0.5 * [x- (0.5 sin 2x)]

1/(2π R) * E2 * (0.5 [ θ-(0.5 sin 2θ)])

 

[gneill]

0.5 * [x- (0.5 sin 2x)]

1/(2π R) * E2 * (0.5 [ θ-(0.5 sin 2θ)])

Okay.

By the way, you can use the x² and x₂ icons on the edit panel to create superscripts and subscripts.

 

[Gremlin]

Okay.

By the way, you can use the x² and x₂ icons on the edit panel to create superscripts and subscripts.

Yes, scripts I've got the hang of, but i didn't need to use any in the post you quoted - and i say that as you phrased your comment like i should have used some.

That doesn't appear to solve the problem though as now I've got when θ = 5π/3:

3.0086 x 10^-4 * 325.3² * (0.5 (5π/3 - (0.5 sin (2* 5π/3)))) = 81.8W

Which is not far from the right answer, but isn't the right answer and i haven't used θ = π/3, which is 16.38W.

 

[gneill]

You need to plug in both end points. What you've derived is the indefinite integral (without the constant of integration). When you want to use it for a definite integral you need to plug both endpoint values into the indefinite integral and take the difference.

 

[Gremlin]

You need to plug in both end points. What you've derived is the indefinite integral (without the constant of integration). When you want to use it for a definite integral you need to plug both endpoint values into the indefinite integral and take the difference.

Thanks.

Isn't that what I've done with a = π/3 and b = 5π/3? Arent they the end points and difference between then 81.8 - 16.38 = 65.42W? But that would be the wrong answer.

 

[gneill]

Thanks.

Isn't that what I've done with a = π/3 and b = 5π/3? Arent they the end points and difference between then 81.8 - 16.38 = 65.42W? But that would be the wrong answer.

Sorry, I didn't pick up on that. Going back to your expression in post #28 where you evaluate the $5 \pi / 3$ endpoint, the expression looks fine but you're arriving at an incorrect result (81.8 W). Perhaps a finger problem on the calculator?

What do you get for just the sub-expression $\frac{1}{2} ( \theta - \frac{1}{2} sin(2 \theta))$ ?

 

[Gremlin]

2.572.

 

[Merlin3189]

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.

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

That looks even more accurate then, great.

May I just query these comments? While it is generally good advice to keep intermediate results more accurately than you need your result, it does seem very inappropriate here to be worrying about fractions of a watt. If the required accuracy of the answer is only about 5 or 10%, then I find it hard to understand why anyone would ask OP'er to work to greater accuracy. It only encourages people to give answers to spurious levels of precision - as witness the quote from bizuputyi.

As pointed out by The Electrician,

...
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.

so in light of that, I can't see how (s)he can suggest that a result like 84.45 watt is "better" To me a "better" answer would be 80 +10/-0 watt or something similar.On a different point

You integrate from $\pi / 3$ to $5 \pi /3$ to account for your firing angle. .

I think that arithmetically this is correct, but I feel it is rather misleading. For a triac firing at π/3, surely the correct range of conduction is from π/3 to π and from 4π/3 to 2π? Arithmetically this may be equal to π/3 to 5π/3 or to the more sensible, 2x integral from π/3 to π, but if you draw the graph and mark in π/3 and 5π/3 as the range to integrate, these points do not correspond to the triac firing points, so could lead a student to misunderstand what is actually happening.

 

[gneill]

2.572.

That seems to be low. I see something closer to 2.8.

 

[Gremlin]

That seems to be low. I see something closer to 2.8.

Hmm, I'm working degrees, that's not the issue is it?

2 sin (5π/3) = 0.182516164
0.182516164 * 0.5 = 0.091258082
5π/3 = 5.235987756 - 0.091258082 = 5.144729674
5.144729674 * 0.5 = 2.572364837