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Anonymous Thu 07 Jan 13:34:27 2021 No.12554237 View Reply Original Report

Quoted By: >>12554249 >>12554316 >>12554341 >>12554699 >>12555189 >>12555283 >>12555673 >>12555833 >>12555956 >>12558219 >>12560382 >>12560393

I need help with the following question...

Two sinusoidal voltages of the same frequency have peak values of 8V and 6V, respectively. They have a phase difference of 90o. Determine the peak value of the sum of the two voltages.

I know the answer is 10. I just don't know how to get it. I added the RMS voltages of each and got ~10 but that doesn't take the 90 degree phase shift in to account so I don't think that's the correct way to solve this.

Anonymous

Anonymous Thu 07 Jan 2021 13:43:12 No.12554249 Report

Quoted By:

>>12554237
Coffee keeps me awake at night.

Anonymous

Anonymous Thu 07 Jan 2021 14:15:45 No.12554316 Report

Quoted By:

>>12554237
Don't worry about RMS
Both are the same frequency, so when wave1 is at peak, its 8 Vp.
Wave 2 being 90 deg out of phase means you need to find the peak voltage of wave 2 when it is 90deg off. That would be the 6V peak voltage at (?/2). Use some algebra to calculate it.
Then you add that (?/2) peak voltage to the 8Vp.

Anonymous

Anonymous Thu 07 Jan 2021 14:28:37 No.12554341 Report

Quoted By: >>12555189

>>12554237
first wave is $8sin(x)$ while the second is $6sin(x-\pi/2)$ . Add them up, find maxima by taking the derivative, voila, 10V.

Anonymous

Anonymous Thu 07 Jan 2021 14:36:44 No.12554354 Report

Quoted By: >>12555083 >>12555106 >>12559892 >>12560396

STOP POSTING FUCKING SMUT.

I WANT THIS SHIT OUT OF MY LIFE FOR GOOD. IT RUINS ALL MY MOTIVATION AND INTEREST. I FALL TIME AND AGAIN TO MASTURBATION, WHICH IS FOLLOWED BY MORE OF THE SAME THING.

THIS DOESN'T HELP ANYHING. KILL YOUR SELF.

Anonymous

Anonymous Thu 07 Jan 2021 17:03:12 No.12554699 Report

Quoted By: >>12560412

>>12554237
represent voltages as phasors. add phasors together. compute length of phasor. done

Anonymous

Anonymous Thu 07 Jan 2021 19:23:58 No.12555083 Report

Quoted By:

>>12554354
I feel you anon

Anonymous

Anonymous Thu 07 Jan 2021 19:35:32 No.12555106 Report

Quoted By:

>>12554354
This, but less autistic.

Anonymous

Anonymous Thu 07 Jan 2021 20:05:04 No.12555189 Report

Quoted By: >>12560104

>>12554237
>>12554341
Easier than that.
Take the vector form and draw a triangle by adding those vectors.

Anonymous

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Anonymous Thu 07 Jan 2021 20:32:24 No.12555283 Report

Quoted By: >>12558175 >>12560714

>>12554237
Fuck off cofeenigger

Anonymous

Anonymous Thu 07 Jan 2021 22:38:45 No.12555673 Report

Quoted By:

>>12554237
Isn't 90 degrees a hint to use the Pythagorean?
SQRT( 8*8 + 6*6) = 10.
Essentially this is the well known 3, 4, 5 triangle.

Anonymous

Anonymous Thu 07 Jan 2021 23:34:07 No.12555833 Report

Quoted By:

>>12554237
I bet these two dirty milfs would let me get in bed with them
*sniiiffffffffffffffff*
Oh yeah

Anonymous

Anonymous Fri 08 Jan 2021 00:26:31 No.12555956 Report

Quoted By: >>12559866

>>12554237

This is basic fucking calculus and trigonometry,

$0 = \frac{\mathrm{d} }{\mathrm{d} x} \left [ 8sin(x) + 6sin(x - \frac{\pi}{2}) \right ] = 8cos(x) + 6cos(x - \frac{\pi}{2}) = 8cos(x) + 6sin(x)$
So, $0 = 8cos(x) + 6sin(x)$ , hence, $arctan(-\frac{4}{3}) = x$ .

The absolute value of the plug and chug is the answer.

Anonymous

Anonymous Fri 08 Jan 2021 15:34:36 No.12558175 Report

Quoted By:

>>12555283
>mad micropenis coping

Anonymous

Anonymous Fri 08 Jan 2021 15:36:41 No.12558183 Report

Quoted By:

Anonymous

Anonymous Fri 08 Jan 2021 15:48:28 No.12558219 Report

Quoted By:

>>12554237
i have the solution, but i dont talk to simps

Anonymous

Anonymous Fri 08 Jan 2021 23:36:00 No.12559866 Report

Quoted By: >>12560408

>>12555956
i guess I'm kind of retarded. Where do I plug arctan(?4/3) = x in to, to get 10?

Anonymous

Anonymous Fri 08 Jan 2021 23:42:37 No.12559892 Report

Quoted By:

>>12554354
This, but more autistic.

Anonymous

Anonymous Sat 09 Jan 2021 00:43:41 No.12560104 Report

Quoted By: >>12560903

>>12555189
I always hated the guys who did it this way, not because it's wrong, just because it is actually easier than the derivative method, but less obvious

Anonymous

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Anonymous Sat 09 Jan 2021 02:04:53 No.12560382 Report

Quoted By: >>12560714

>>12554237
kill yourself simp cuck

Anonymous

Anonymous Sat 09 Jan 2021 02:08:17 No.12560393 Report

Quoted By:

>>12554237
you should probably lay off coffee

Anonymous

Anonymous Sat 09 Jan 2021 02:08:57 No.12560396 Report

Quoted By:

>>12554354
So stop going on an internet forum that contains smut. Here's an idea: If you don't want to see smut, shut off your computer and do something useful. Read books. Learn guitar. Practice drawing. Lie in your bed and meditate on cute anime waifus WITHOUT fapping to improve your self-control. Stop trying to change other people's behaviour to pander to your own needs.

Anonymous

Anonymous Sat 09 Jan 2021 02:14:24 No.12560408 Report

Quoted By: >>12560680

>>12559866
I would've said back into the $8cos(x) + 6sin(x)$ expression, but I'm only getting -2.8.

I can't brain right now.

Anonymous

Anonymous Sat 09 Jan 2021 02:15:25 No.12560412 Report

Quoted By:

>>12554699
Based.
For retards: Take the phase angle and amplitude of each sinusoidal. Mark the amplitude on the horizontal axis, then rotate that point about the origin by the phase angle.
So if sinusoidal A has an angle of 0° and an amplitude of 8V, that's just the point at x=8, y=0. Sinusoidal B has an angle of 90° and an amplitude of 6V, that's x=0, y=6.
Now add the two phasors. Graphically, you can shift one of them so that its origin is where the other ends. Or just add the x and y values, respectively. The end point of the resulting sum phasor is x=8, y=6. The amplitude of the summed voltage is the length of this phasor (from the origin), in other words, the distance of the end point (8, 6) from the origin. That distance can be calculated with pythagoras: d = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10.

Anonymous

Anonymous Sat 09 Jan 2021 04:08:55 No.12560680 Report

Quoted By: >>12560683

>>12560408
the minima of the superposition of the two waves is not the same as the sum of the individual waves minima, so it doesn’t make sense what you’re doing

Anonymous

Anonymous Sat 09 Jan 2021 04:09:59 No.12560683 Report

Quoted By:

>>12560680
maxima* excuse me

Anonymous

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Anonymous Sat 09 Jan 2021 04:28:39 No.12560714 Report

Quoted By:

>>12555283
>>12560382

Anonymous

Anonymous Sat 09 Jan 2021 06:08:08 No.12560903 Report

Quoted By: >>12561158

>>12560104
but that method only works if it's 90 degrees out of phase, right?

Anonymous

Anonymous Sat 09 Jan 2021 08:10:12 No.12561158 Report

Quoted By:

>>12560903
nope. the key is that the two signals are periodic and have the same frequency. learn phasors

Anonymous

Anonymous ## Mod Sat 09 Jan 2021 09:45:17 No.12561299 Report

Quoted By:

This thread was moved to >>>/wsr/954397

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