How Do You Calculate the Amplitude of Combined Alternating Currents?

[Enochfoul]

Homework Statement

Two alternating currents i₁ and i₂ flow into a circuit node,the output current i is given by i₁ + i₂.

Calculate the amplitude of i and the first time it occurs.

i1=5sin(50t +Pi/3)

i2=6cos50t

Homework Equations

The Attempt at a Solution

5 sin ( 50t +60˚)
= 5 ( sin 50t cos60˚ + sin60˚cos50t )
= 2.5 sin 50t + 4.33 cos 50t

2.5 sin 50t + 4.33 cos 50t + 6 cos 50t
i1+i2 = 2.5 sin 50t +10.33 cos 50t

R² = 2.5²+10.33²
R = √113
= 10.6

arctan 10.33 / 2.5 = 33.8˚

so i = 10.6 and occurs when 50t = 33.8˚

Does this look right?

 

[cnh1995]

arctan 10.33 / 2.5 = 33.8˚

This should be more than 45°. Peak current looks correct to me.

 

[Enochfoul]

This should be more than 45°. Peak current looks correct to me.

Hi Any ideas what I should do to correct the value?

 

[cnh1995]

Hi Any ideas what I should do to correct the value?

tan^-1 (10.33/2.5)= tan^-1 4.13 which is clearly more than 45^°(tan 45=1). Calculate the actual value using a calculator.

 

[Enochfoul]

tan^-1 (10.33/2.5)= tan^-1 4.13 which is clearly more than 45^°(tan 45=1). Calculate the actual value using a calculator.

76.38°?

 

[gneill]

Your current function isn't a phasor (no imaginary component), so your angle computation is not correct.

Call your current function f(θ). How would you normally go about finding a minimum or maximum for a function f(θ) with respect to θ?

 

[Enochfoul]

Your current function isn't a phasor (no imaginary component), so your angle computation is not correct.

Call your current function f(θ). How would you normally go about finding a minimum or maximum for a function f(θ) with respect to θ?

Ok so do I need to be looking into something like the following?

"The result of adding 2 phasors can be summarised below."
a∠α + b∠β = c∠g
c = √{ [a cos(a) + b cos(b) ]^2 + [a sin(a) + b sin(b) ]^2}
g = arctan{ [a sin(a) + b sin(b) ]/ [a cos(a) + b cos(b)] }

With this in mind would this solution work?

a angle a + b angle b = c angle g
c = sqrt{ [a cos(a) + b cos(b) ]2 + [a sin(a) + b sin(b) ]2}
g = atan{ [a sin(a) + b sin(b) ]/ [a cos(a) + b cos(b)] }

cos 50t = sin (50t + pi/2)

a = 5
b = 6
angle a = pi/3
angle b = pi/2

Thanks

 

[gneill]

No, your current is given as a function in the time domain. It's not a phasor.

You did fine reducing it to the expression:

I(t) = 2.5 sin 50t +10.33 cos 50t

How might you go about finding a maximum for that function?

 

[Enochfoul]

No, your current is given as a function in the time domain. It's not a phasor.

You did fine reducing it to the expression:

I(t) = 2.5 sin 50t +10.33 cos 50t

How might you go about finding a maximum for that function?

From my notes it states that:
"It is often useful to be able to quote the maximum value of a sinusoidal waveform and the phase angle at which this maximum occurs. This involves only a simple deduction provided the waveform is expressed in one of the now familiar forms:"
Rsin(x+α)
or
Rcos(x+α)

The first maximum on the sinθ curve occurs at θ=90°, therefore x+α=90°
The first maximum on the cosθ curve occurs at θ=0° therefore x+α=0°

Am I going down the right lines?

 

[gneill]

If you want to go that route then recognize that you have an expression of the form:

a⋅sin(θ) + b⋅cos(θ)

where θ = 50 t ; a = 5/2 ; b = 10.33

You need to find a way to write this as a function of the form you stated, either Rsin(θ+α) or Rcos(θ+α). That will require using an appropriate trig identity and a bit of a trick with the constants a and b to turn them into sines and cosines of another angle (the phase angle). Do you know that trick?

What I was hinting at earlier was that you could differentiate the current function, set it to zero and solve for the angle. This will yield the angle of the first maximum (or the first minimum, you should check with either the second derivative test or just plug the angle into the original function to see if the result is positive or negative).

 

[Enochfoul]

"What I was hinting at earlier was that you could differentiate the current function, set it to zero and solve for the angle"

Any further clues you could give me to start doing this?

Thanks

 

[gneill]

"What I was hinting at earlier was that you could differentiate the current function, set it to zero and solve for the angle"

Any further clues you could give me to start doing this?

Thanks

Do you know how to differentiate a function?

 

[Enochfoul]

Do you know how to differentiate a function?

I have limited experience differentiating but I've had a go.

Do you mean by the current function:

I(t) = 2.5 sin 50t +10.33 cos 50t

i.e

d/dt(2.5 sin (50t)+10.33cos (50t)

Answer

=125cos(50t)-1033 sin(50t)\2

I do not know what you mean by set it to zero and solve for the angle

Thanks

 

[gneill]

In post #10 I wrote your current function as:

I(t) = a⋅sin(θ) + b⋅cos(θ)

where θ = 50 t ; a = 5/2 ; b = 10.33

Differentiate a⋅sin(θ) + b⋅cos(θ). It's simple because the "50t" is removed from trig arguments and replaced with θ. This θ is the angle you're looking for, when the current reaches a maximum.

To find a maximum or minimum you find the value of θ that makes the derivative zero.

 

[Enochfoul]

I have limited experience differentiating but I've had a go.

Do you mean by the current function:

I(t) = 2.5 sin 50t +10.33 cos 50t

i.e

d/dt(2.5 sin (50t)+10.33cos (50t)

Answer

=125cos(50t)-1033 sin(50t)\2

I do not know what you mean by set it to zero and solve for the angle

Thanks

Is my calculation to find

In post #10 I wrote your current function as:

I(t) = a⋅sin(θ) + b⋅cos(θ)

where θ = 50 t ; a = 5/2 ; b = 10.33

Differentiate a⋅sin(θ) + b⋅cos(θ). It's simple because the "50t" is removed from trig arguments and replaced with θ. This θ is the angle you're looking for, when the current reaches a maximum.

To find a maximum or minimum you find the value of θ that makes the derivative zero.

Thanks I will have a crack at this tomorrow. Is my calculation for the amplitude correct?

 

[gneill]

Yes, your amplitude is fine. Your derivative is okay, too. But it's probably easier to leave it in symbolic form with a, b, and θ, solve for the angle in terms of those constants, then plug in the numbers at the end. Thus:

$\frac{d}{dθ}(a~ sin(θ) + b~ cos(θ)) = a~ cos(θ) - b~ sin(θ) = 0$

Solve for θ.

 

[Charles Link]

If I may add a constructive comment, there is a trick every engineer should really know when they encounter the form $y=A \cos(\omega t)+B \sin(\omega t)$ It looked like you were heading toward that route in your OP but never got there. And the trick is to factor out sqrt(A^2+B^2) and let A/sqrt(A^2+B^2)=$\cos(\phi)$ and B/sqrt(A^2+B^2)=$\sin(\phi)$ Then y=sqrt(A^2+B^2) $\cos(\omega t-\phi)$ I first saw this trick in a mechanics text in my sophomore year in college and it has come in handy countless times. (Note that $\tan(\phi)=B/A$)

 

[gneill]

If I may add a constructive comment, there is a trick every engineer should really know when they encounter the form $y=A \cos(\omega t)+B \sin(\omega t)$ It looked like you were heading toward that route in your OP but never got there. And the trick is to factor out sqrt(A^2+B^2) and let A/sqrt(A^2+B^2)=$\cos(\phi)$ and B/sqrt(A^2+B^2)=$\sin(\phi)$ Then y=sqrt(A^2+B^2) $\cos(\omega t-\phi)$ I first saw this trick in a mechanics text in my sophomore year in college and it has come in handy countless times. (Note that $\tan(\phi)=B/A$)

That is the trick that I was hinting at in post #10.

 

[NascentOxygen]

so i = 10.6 and occurs when 50t = 33.8˚

Does this look right?

Your very first move when faced with this problem should have been to sketch the curves so you can see exactly what you are dealing with.

➫ Get to know wolframalpha and it will turn drudgery into fun!
http://m.wolframalpha.com/input/?i=plot+5sin(50t+Pi/3)+6cos(50t)+for+0<t<0.01&incCompTime=true

You can see their sum starts (i.e., when t=0) at an angle that's a little less than the peak, so a little less than 90°.

 

[Enochfoul]

Your very first move when faced with this problem should have been to sketch the curves so you can see exactly what you are dealing with.

➫ Get to know wolframalpha and it will turn drudgery into fun!
http://m.wolframalpha.com/input/?i=plot+5sin(50t+Pi/3)+6cos(50t)+for+0<t<0.01&incCompTime=true

You can see their sum starts (i.e., when t=0) at an angle that's a little less than the peak, so a little less than 90°.

Could the problem be looking for an actual time value rather than an angle? i.e in the graph the amplitude 10.6 occurs around 0.005. On top of this using the calculation 2.5sin(50) + 10.33cos(50) I got 8.55. Arctan8.55 = 83.32° which is a little less than the 90° you stated or I am swinging and missing again?

 

[NascentOxygen]

Could the problem be looking for an actual time value rather than an angle? i.e in the graph the amplitude 10.6 occurs around 0.005.

Once you determine the angle (and remember, you'll need it in radians), you can easily calculate the time.

On top of this using the calculation 2.5sin(50) + 10.33cos(50) I got 8.55.

As far as I can see, that expression does not arise anywhere in the solution to this exercise.

 

[NascentOxygen]

If you replace "plot" with "maximum of" I think you'll find wolframalpha will tell you the ordinates of the maximum in that range of time...so you can check your calculations.

 

[Enochfoul]

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?

 

[NascentOxygen]

That looks right.

 

[NascentOxygen]

So to make sure I understand why you solved the exercise this way, you haven't yet learned to add as phasors, e.g.,

I₁ = 5∠Pi/3
I₂ = 6∠0

then I₁ + I₂ = ...

 

[Enochfoul]

So to make sure I understand why you solved the exercise this way, you haven't yet learned to add as phasors, e.g.,

I₁ = 5∠Pi/3
I₂ = 6∠0

then I₁ + I₂ = ...

No I haven't. Tbh the notes given by uni are very limited so I'm trying to pick up bits and bobs from other sources.

 

[NascentOxygen]

Homework Statement

Two alternating currents i₁ and i₂ flow into a circuit node,the output current i is given by i₁ + i₂.

Calculate the amplitude of i and the first time it occurs.

i1=5sin(50t +Pi/3)

i2=6cos50t

Homework Equations

The Attempt at a Solution

5 sin ( 50t +60˚)
= 5 ( sin 50t cos60˚ + sin60˚cos50t )
= 2.5 sin 50t + 4.33 cos 50t

2.5 sin 50t + 4.33 cos 50t + 6 cos 50t
i1+i2 = 2.5 sin 50t +10.33 cos 50t

R² = 2.5²+10.33²
R = √113
= 10.6

arctan 10.33 / 2.5 = 33.8˚

when t=0 you can easily calculate the phase angle, the angle by which this sum leads a reference sinewave:
arctan 10.330/10.628 = 76.39° (= 1.333^c)
note: superscript c is the symbol for radians

so i1 + i2 = 10.628 sin(50t + 76.39°)

so i = 10.6 and occurs when 50t = 33.8˚

and this sinusoid peaks when (50t + 1.333^c) = Pi/2

It seems you were almost finished back in post #1. Now, from this last line solve for t.

 

[Enochfoul]

It seems you were almost finished back in post #1. Now, from this last line solve for t.

So my post #23 is not correct?

 

[NascentOxygen]

So my post #23 is not correct?

It's right.

Either way, the answer should be the same.

 

[Enochfoul]

It's right.

Either way, the answer should be the same.

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

 

[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