50 Hz waveform

Thread Starter

liam

Joined Apr 22, 2008
12
could any one help us with this i have no idea where to start Find the instantaneous value of a 50 Hz waveform with a maximum value of 200v. 0.001 sec after it has passed through a voltage zero. many thanks . liam
 

mik3

Joined Feb 4, 2008
4,843
could any one help us with this i have no idea where to start Find the instantaneous value of a 50 Hz waveform with a maximum value of 200v. 0.001 sec after it has passed through a voltage zero. many thanks . liam
v=Asin(2πft)
where v is the instantaneous voltage
A is the peak (maximum) voltage
f is the frequency
t is the time

so v=200sin(2πx50x0.001) =1.097 Volts

You said when it has passed through zero: so it may be positive like + 1.098 or negative like - 1.097
 

Caveman

Joined Apr 15, 2008
471
The dimensional analysis is:

t in seconds.
2*PI = radians/cycle
f = cycles/sec.
2*PI*f*t = radians/cycle * cycles/sec * sec = radians.

So, take the sine in radians. The 61.8 magnitude is correct.
 

Thread Starter

liam

Joined Apr 22, 2008
12
Thanks for the replies guys. but im still none the wiser as were you got the answer from . i understand that the answer is 61.8v. but where did this come from . when i tried working out the equation v=200sin(2πx50x0.001) =1.097 Volts i got - 0.107 ????
 

Caveman

Joined Apr 15, 2008
471
Okay, step by step
1. 2*PI radians/cycle * 50 cycles/second * 0.001 seconds = 0.314 radians

This is just stating that 0.001 seconds into a 50Hz sine wave is 0.314 radians in.

2. Take the sine of this, but remember to set up your calculator to calculate your trigonometric functions in radians. So, sin(0.314 radians) = 0.309

Just so you understand, what this says is that at 0.001 seconds into a 50Hz sine wave, the voltage is 0.309 times the peak.

3. Multiply by the peak voltage, 200V. 200V*0.309 = 61.8V.

Got it?
 
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