Triac - Power dissipated in light bulb

[gneill]

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

Biiig issue! The angles we've been using are specified in radians.

 

[Gremlin]

Biiig issue! The angles we've been using are specified in radians.

In radians for $\frac{1}{2} ( \theta - \frac{1}{2} sin(2 \theta))$ i get 3.051 which is still a bit off from your 2.8.

For the whole thing for 5π/3 i get 97.13W,

And for π/3 i get 0.09058 and 2.88W

 

[gneill]

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.

Those are good points for practical calculations. The original problem statement did not make any explicit statement of accuracy required, but all values given were in nice exact numbers leading one to presume that the problem is more of a theoretical exercise than a practical one. Whether it is or not, keeping extra digits is a good habit to get into anyways, and at the very least can help flag where someone's calculation method is a problem.

On a different point

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.

I admit to having taken advantage of the symmetry of the sinewave curve in order to simplify the domain of integration. A "practical habit" of mine that I should have made explicit. Thanks for pointing it out.

 

[gneill]

In radians for $\frac{1}{2} ( \theta - \frac{1}{2} sin(2 \theta))$ i get 3.051 which is still a bit off from your 2.8.

I'm certain that something must still be going wrong in your calculation. Break down the pieces and post their values.

 

[Gremlin]

I'm certain that something must still be going wrong in your calculation. Break down the pieces and post their values.

2 sin (5π/3) = -1.732050808
-1.732050808 * 0.5 = -0.866025403
5π/3 = 5.235987756 - -0.866025403 = 6.102013159
6.102013159 * 0.5 = 3.05100658

3.05100658 * 325.3² = 322857.7908
1/(2π*529) = 3.008 * 10^-4 * 322857.7908 = 97.1156

 

[gneill]

2 sin (5π/3) = -1.732050808

Where does 2 sin (5π/3) come from ? Isn't it $\frac{1}{2} sin(5 \pi / 3)$ ?

 

[Gremlin]

Where does 2 sin (5π/3) come from ? Isn't it $\frac{1}{2} sin(5 \pi / 3)$ ?

In post #24

∫ sin (θ)² dθ became ∫ 0.5 (1-cos (2θ)) using the relationship sin² A = ½ (1 - cos 2θ).

Integrating 0.5 (1-cos (2θ)) i arrived at [ ½ (θ-sin 2θ)], which following your post #25 i amended to [ ½ (θ- (½ (sin 2θ)))].

Looking at 0.5 (1-cos(2x)) on this site http://www.integral-calculator.com/ that would appear correct.

 

[gneill]

But haven't you multiplied by 2 rather than 1/2?

The sub-expression we're looking at is:

½ (θ - ½ (sin 2θ))

So the piece in question is ½ (sin 2θ), not 2(sin 2θ)

 

[Gremlin]

But haven't you multiplied by 2 rather than 1/2?

The sub-expression we're looking at is:

½ (θ - ½ (sin 2θ))

So the piece in question is ½ (sin 2θ), not 2(sin 2θ)

You are correct. Like a doughnut I was operating under the delusion that 2 sin (5π/3) = sin (2*(5π/3)).

I now have the correct answer. You've the patience of a saint. Nice one.

 

[MrBondx]

Hi guys

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

I'm getting same as this but if I multiply with 31.83 that I am getting from (3.0086 x 10^-4) * 325.32^2

My answer is 89.9. Just wondering why my answer is off.

Please help

Thanks

 

[gneill]

Hi guysI'm getting same as this but if I multiply with 31.83 that I am getting from (3.0086 x 10^-4) * 325.32^2

My answer is 89.9. Just wondering why my answer is off.

Please help

Thanks

You'll have to show your work if we're to find a problem in it.

 

[MrBondx]

You'll have to show your work if we're to find a problem in it.

Thanks gneill

Here it is:

[V² / (2πR)] x [ ½ (θ - (½ sin 2(θ)))]

[325.269² / (2π x 529)] x [ ½ (5π/3 - (½ sin 2(5π/3)))]

31.83 x [ ½ (5.236 -( - 0.433))]

31.83 x 2.8345

90.2

 

[gneill]

Thanks gneill

Here it is:

[V² / (2πR)] x [ ½ (θ - (½ sin 2(θ)))]

[325.269² / (2π x 529)] x [ ½ (5π/3 - (½ sin 2(5π/3)))]

I think your basic form of the integral is okay but there may be a problem forming the definite integration (handling the integration limits).

Can you explain how the limits of integration that are involved here lead you to the above substitutions?

 

[MrBondx]

I think your basic form of the integral is okay but there may be a problem forming the definite integration (handling the integration limits).

Can you explain how the limits of integration that are involved here lead you to the above substitutions?

Thanks for the pointer. I guess i didn't subtract the lower limit from the the upper limit

[325.2692 / (2π x 529)] x [ intergral with upper limit] - [ intergral with lower limit]

31.83 x (2.8345 - 0.3071)

= 80.45

Thats it, isn't it?

Thanx gneill

 

[gneill]

Yup. That's it.

 

[Jerremy_S]

Hi All,

Having given a= π/3 Can you explain why b= 5π/3 ?

And Also, where this integral formula comes from? What is it called?

I am going back and forth through my textbooks and I cannot find anything.
A push in the right direction would be appreciated:)

Thanks

 

[gneill]

Hi All,

Having given a= π/3 Can you explain why b= 5π/3 ?

If you look at the actual conduction periods for the firing angle you'll see that conduction takes place in two separate periods from π/3 to π, and 4π/3 to 2π. Rather than write separate integrals for each, it is convenient to make use of the symmetry of the sine function whereby the area under the curve from π to 5π/3 is the same as the area from 4π/3 to 2π. Thus a single integral over the angles from π/3 to 5π/3 is equivalent to the two separate integrals.

And Also, where this integral formula comes from? What is it called?
View attachment 110276

That's the power over one period. It stems from the power relationship P = V²/R.