How to analyze AC using radians and degrees

[yosimba2000]

I just found out V = Acos(wt+phi) is made up of both radians and degrees, where w is rad/s and phi is degree.
You have to convert either one to the other to calculate it.

Assume phi is 0 degrees. I(t) = Icos(wt), and w is radian/s.
So then capacitor impedance is -j/wC.
V(t) = (I<0) * (1/wC<-90)
V(t) = 1/(wC) < -90 <-- change 90 degree to pi/2 radian

V(tr) = I/wC*cos(wt-pi/2) <--- this is solved using radians!

Ok, then assume we change from radians to degrees. So I(t) = Icos(57.3wt). We use 57.3 since there are about 57.3 degrees per radian.

Capacitor impedance then becomes -j/(57.3wC).
V(t) = (I<0) * (1/(57.3wC)) < -90
V(t) = I/(57.3wC)<-90

V(td) = I/(57.3wC)*cos(57.3wt-90) <----this is solved using degrees!

The resulting V(t) values do not match!
Assume all values are 1.

So V(tr) = 1/(1*1)cos(1-pi/2) using radians, comes out to be 0.84147.

And V(td) = 1/(57.3)*cos(57.3-90) using degrees, comes out to be 0.0146.

How come they don't match? They should be equivalent.

 

[Svein]

So I(t) = Icos(57.3wt). We use 57.3 since there are about 57.3 degrees per radian.

And what is the dimension of "w" then? Mixing dimensions is a major headache and is best avoided.

 

[Simon Bridge]

I just found out V = Acos(wt+phi) is made up of both radians and degrees, where w is rad/s and phi is degree...

Where have you seen this?

It is very bad practise - there is nothing wrong with having $\phi$ in radians. The equation you quote does not mix the units.

It is best practise to use mutually compatable units.
i.e. It is also possible for $\omega$ to be in RPM ... consistent units for the other variables are REVs (the number of rotations) for angle and mins for time.

When you need to specify a conversion factor in an equation, you should use the full version of the factor ... so it's $180/\pi$deg/rad rather than $57.3$deg/rad.

 

[yosimba2000]

And what is the dimension of "w" then? Mixing dimensions is a major headache and is best avoided.

w is in radians per second. After multiplying my 57.3 degrees/radian, 57.3*w has units of degree/second.
Yes it's not very fun to convert units, but I couldn't wrap my head around the fact that we could use both degrees and radians in the same equation. I just wanted to show myself that it is equivalent even if I use degrees instead of radians.

Where have you seen this?

It is very bad practise - there is nothing wrong with having $\phi$ in radians. The equation you quote does not mix the units.

It is best practise to use mutually compatable units.
i.e. It is also possible for $\omega$ to be in RPM ... consistent units for the other variables are REVs (the number of rotations) for angle and mins for time.

When you need to specify a conversion factor in an equation, you should use the full version of the factor ... so it's $180/\pi$deg/rad rather than $57.3$deg/rad.

Our homework provides solutions where phi is in degrees, and that's also how we learned it in class. So our professor told us we must take care to convert phi to radian before analyzing the final equation.

What I just noticed the discrepancy arises from the coefficient in front. cos(wt-pi/2) solved in radians is equivalent to cos(57.3wt-90) solved in degrees. But the coefficients in front of cos(wt-pi/2) is 1/wC, and the coefficient in front of cos(57.3wt-90) is I/(57.3wC). When all values are 1, 1/wC = 1, but 1/(57.3wC) = 1/57.3

So cos(wt-pi/2) in radians is equal to cos(57.3wt-90) in degrees
But 1*cos(wt-pi/2) in radians is not equal to (1/57.3) *cos(57.3wt-90) in degrees.

So the coefficients, which are the impedance factor are causing the problem...
So something went wrong with the calculation of the impedance, but I cannot spot it. I followed the derivation of the capacitor impedance directly as it was done for when the frequency was in radians/second.

Here's my capacitor imepdance derivation:
I=C dv/dt

For radians:
V = Ae^j(wt+phi)
dv/dt = jwAe^jwt+phi
dv/dt = jw(V)
I = C dv/dt = Cjw(V)
V/I = Capacitor Impedance = 1/jwC

For degrees:
V = Ae^{j((180/pi)*wt+phi)}
dv/dt = jw(180/pi)*Ae^{j((180/pi))*wt+phi)}
I = C dv/dt = Cjw(180/pi)(V)
V/I = Impedance = 1/[jwC(180/pi)]

 

[Simon Bridge]

Our homework provides solutions where phi is in degrees, and that's also how we learned it in class. So our professor told us we must take care to convert phi to radian before analyzing the final equation.

... that is different from what you said: you are just over-thinking what your prof said.

Of course in real life you will get measurements in any units, and you will be asked to provide answers in many different units.
That does not mean the equation has different units in it. You just convert whichever units are inconvenient so they are consistent inside the equation.
To avoid the kind of confusion you are experiencing, you should always do the conversion, as an extra step, outside of the equations.

If you really insist on putting the (non SI) conversion factor inside the equation, use notation that reflects this.
i.e. the length of circumference subtended by an angle is given by $S = R\theta$ ... if angles are in radians, and $S=\pi R \theta / 180$ when angles are in degrees. If you need to use both in the same document, then you need your notation to make it clear what you mean: i.e.
$S = R[\theta]_{rad} = \pi R [\theta]_{deg} / 180$ ... see how that's clear?

What I just noticed the discrepancy arises from the coefficient in front. cos(wt-pi/2) solved in radians is equivalent to cos(57.3wt-90) solved in degrees. But the coefficients in front of cos(wt-pi/2) is 1/wC, and the coefficient in front of cos(57.3wt-90) is I/(57.3wC). When all values are 1, 1/wC = 1, but 1/(57.3wC) = 1/57.3

$\cos(\omega t + \phi)$ assumes the quantity inside the cosine is in radians - this is the default for science, so you should try to get used to thinking in radians.
If you want to include the conversion in the equation then you can do:
$\cos([\omega]_{rad/s} [t]_{s} + [\phi]_{rad}) = \cos([\omega]_{rad/s} [t]_{s} + \pi [\phi]_{deg}/180)$

Note: Your calculator gives the option to work trig in degrees (the "deg-rad" mode), and most default to deg mode.
If you have $\omega$ in rad/s but you want to work in your calculator's deg mode, you need to use $\cos(180[\omega]_{rad/s}/\pi + [\phi]_{deg})$

It is common to quote the phase angle in degrees - so solving for the phase angle, in degrees, using the standard formula, gives:
\phi = 180[\arccos(\psi) - \omega t]/pi

So cos(wt-pi/2) in radians is equal to cos(57.3wt-90) in degrees
But 1*cos(wt-pi/2) in radians is not equal to (1/57.3) *cos(57.3wt-90) in degrees.

You used the wrong conversion factor - but there is another problem.

The trig function is defined as the ratio of lengths of a triangle. That means it is a length divided by a length - which is just a number: no units.
The cosine is not measured in degrees or radians, it is just itself.
All the cosine is doing is taking an angle and turning it into a number.

Than means that cos(wt-pi/2) evaluated in radians gives the same number as cos(57.3wt-90) evaluated in degrees ... the answer is not in radians or degrees, it's just a number.

When you write: 1*cos(wt-pi/2) then the answer has whatever units that the "1" at the start had.
If that is 1rad, then the answer is in radians.
This means that 1*cos(57.3wt-90) (evaluated in degrees) will give the same answer in radians.
In order to convert this radian answer to degress you use the correct conversion (57.3) *cos(57.3wt-90) ... (there are 57.3 degrees per radian) which will give the same angle in different units. So it is correct that they should not give the same answer.

So:
1*cos(wt-pi/2) = (57.3) *cos(57.3wt-90)
... using notation to make the units explicit gives:
$$1(\text{rad})\times\cos\left( [\omega]_{rad/s} [t]_{s} - \frac{\pi}{2}(\text{rad})\right) =
57.3(\text{deg/rad})\times 1(\text{rad})\times \cos \left(57.3(\text{deg/rad})[\omega]_{rad/s}[t]_{s} - 90(\text{deg}) \right)$$ ... which is more precise but harder to read.

 

[yosimba2000]

Oh wait I think I see it now.

If I just say the impedance is 1/wC, then the Voltage is Current*Impedance.
But impedance, being in 1/wC form, is like saying 1/(radians)(capacitance).

So the voltage is like Amps/(radians)(Capacitance) so it becomes Amps/(Rad*Farad)

And if degrees was used in impedance, voltage would be like Amps/(degrees*Capacitance) = Amps/(Degree*Farad)

So therefore, they give different numbers as they are in radian and degree domain respectively, but are still equivalent when one is converted to the other?

 

[Simon Bridge]

Well done.
Some people never make that breakthrough.
You are free to choose whatever units you like as long as you are consistent. You end up using whatever units make the maths easiest.

Just a niggle: $V = I/(\omega C)$ so in SI units that is [volts] = [amps][rad/s]^-1[farads]^-1.
So 1V = 1A.s.F^-1.rad^-1... in words that is 1 volt is the same as 1 Amp-second per Farad, per radian.

You could measure voltage in a different unit to the SI "volt" ... i.e. in units of Amp-seconds per Farad, per degree say ... maybe call this the Yosimba:

i.e. [yosimba] = [amps][deg/s]^-1[farads]^-1

You can use the conversion factor 1rad = 57.3deg to figure out the relative sizes of the yosimba and the volt.

 

[Jeff Rosenbury]

BTW, it is often better to convert to radians because they are technically unitless. This is because a radian is the radius subtended by the arc length of an angle. Since both the radius and arclenth are in meters, we get m/m. Hence this is a real number and unitless. This can save lots of time and headaches when manipulating equations and doing calculus.