How Do You Calculate the Amplitude of Combined Alternating Currents?

[Charles Link]

Thanks for all of your help and for the other guys help too. This question has been doing my head in.

You almost had the correct answer in post #1. The trick I mentioned in post #17 is probably your simplest solution. B/A=2.5/10.33. It should be a simple matter to find $\phi$ and write out the answer. And in the form $y=C\cos(\omega t-\phi)$ the shift is to the right for positive $\phi$

 

[Enochfoul]

It's right.

Either way, the answer should be the same.

Any chance you could have a look at the impedance question I posted. I can't decide if the final answer should be positive or negative.

 

[joe forrester]

Ok I've had another bash.

So If C = 2.5 sin 50t + 10.33 cos 50t
the dC/dt = 125 cos 50t - 516.5 sin 50t (diff of sin 50t is 50 cos 50t and diff of cos 50t is -50 sin 50t)
At a maximum dC/dt is 0
0 = 125 cos 50t - 516.5 sin 50t
rearrange and divide by cos 50t this give tan 50t = 2.5/10.33 giving 50t as 13.6 deg.

So 13.6 deg converted to radians would be 0.237.
0.237 ÷ 50 = 0.00474 which would be the time the amplitude of 10.6 would occur?

Using you tip of changing the graph to "Maximum of" it comes out as http://www4f.wolframalpha.com/Calculate/MSP/MSP72722a107agdc0he8ga0000405a140d63h2172g?MSPStoreType=image/gif&s=42&w=496.&h=33.

Im guessing this is correct then?

I have the same question myself and have followed it up until this point, but i am confused as to wear the value 125 and -516.5 come from?

 

[gneill]

I have the same question myself and have followed it up until this point, but i am confused as to wear the value 125 and -516.5 come from?

The function of time was differentiated. The values are a result of that operation.

Differentiating trig functions sin and cos:

$\frac{d}{dt} Asin(ω t) = Aωcos(ω t)$

$\frac{d}{dt} Bcos(ω t) = -Bωsin(ω t)$

 

[joe forrester]

The function of time was differentiated. The values are a result of that operation.

Differentiating trig functions sin and cos:

$\frac{d}{dt} Asin(ω t) = Aωcos(ω t)$

$\frac{d}{dt} Bcos(ω t) = -Bωsin(ω t)$

Thanks