# Triac firing angle question

1. Mar 22, 2017

### David J

1. The problem statement, all variables and given/known data
If the firing angle is set to $\alpha = \frac{\pi}{3}$ estimate the power dissipated in the bulb if it is rated at $100w$ and the voltage source is $230V$ @ $50Hz$

2. Relevant equations

$Vpeak = \frac{230}{0.707}$ = 325V

$Vload = 325\sqrt\frac{\left(2\pi -\left(\frac{2\pi} {3}\right)+\sin\left(\frac{2\pi}{3}\right)\right)}{4\pi}$

$Vload = 206Vrms$

The bulb is rated at 100w so $100=\frac{230}{I}$

$I rms=\frac{100}{230}=0.43A$

$R=\frac{230}{0.43}=534.884$

3. The attempt at a solution

I have just started looking at this question and picked up on an old post from 2015
(Triac - Power dissipated in light bulb)
In that post they had taken the given voltage of 230V (assuming it to be rms) and calculated the peak value by dividing by 0.707 to give a peak value of 325V. I get that part.
The next part is to calculate the "load" voltage using $Vload = 325\sqrt\frac{\left(2\pi -\left(\frac{2\pi}{3}\right)+\sin\left(\frac{2\pi}{3}\right)\right)}{4\pi}$.

I am a little confused by this as I have tried following this equation many times and I cannot end up with a value of 206 and I was wondering if someone could give a little advice.

I am guessing that $\alpha=\frac{\pi}{3}$ is the time / part of the wave when the firing takes place so it maybe at a lower part of the sine wave so to speak. This the value that everyone was getting of 206V rms

If I can understand how to arrive at the value of 206 I should be able to complete this and was wondering is anyone could point out why I cannot arrive at 206. Is the equation wrong or probably I am not carrying out the calculation correct ???

Tanks in advance for any help with this

2. Mar 22, 2017

### cnh1995

You can derive the expression for Vrms in terms of source voltage and firing angle.
Plot the output voltage and use the definition of rms voltage.

3. Mar 22, 2017

### Staff: Mentor

You should keep more digits during your calculation of the bulb resistance. In particular, you've kept only two digits in the intermediate value for the current, truncating after the second digit. As a result your bulb's resistance value will be a bit high.

Edit: FYI, the previous thread being referred to is here:
Triac - Power dissipated in light bulb

4. Mar 22, 2017

### David J

I was working in degrees. I changed it to radians and it worked out correctly

$v peak= \frac{230}{0.707}$

$Vload=325\sqrt\frac{\left(2\pi-\left(\frac{2\pi}{3}\right)+\sin\left(\frac{2\pi}{3}\right)\right)}{4\pi}$

$Vload=325\sqrt\frac{\left(6.283185307-2.094395102\right)+0.866025403}{12.56637061}$

$Vload=325\sqrt\frac{4.188790205+0.866025403}{12.56637061}$

$Vload=325\sqrt\frac{5.054815608}{12.56637061}$

$Vload=325\sqrt {0.402249445}$

$Vload=325\times0.634231381$

$Vload=206.125199 V$

I found the $I rms$ using $\frac{100}{230}=0.434782608 amps$

I found the $R$ using $\frac{230}{0.434782608}=529 \Omega$

So the current flowing through the filament at the firing time will be equal to $I=\frac{206.125199}{529}=0.389650659 amps$

So the Power dissipated in the filament during the firing time will be equal to $I=206.125199\times 0.389650659=80.3168\omega$

I think this is correct now. Thanks once again for the help, much appreciated