# Series RL Circuit - Description for Circuit Current not clear

#### flowtech

Joined Mar 23, 2018
6
I have been tasked with calculating VR and VL given only this information:-

1/ R=10 Ohm
2/ L = 20 mH
3/ I = 5.0 Cos (500t + 10°) (A)

I cant decipher the precise meaning of the current statement in this format though 3/?

Can anybody describe whet this means precisely please to give me a heads up? Thanks

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#### Papabravo

Joined Feb 24, 2006
14,678
1/ and 2/ represent constant values of resistance and inductance according to the units given

3/ is a description of a sinusoidal waveform. The peak amplitude is 5.0 Amperes. Cos is a shorthand notation for the Cosine function. The frequency of the cosine function is 500 Hz. When you multiply a frequency by time you get a dimensionless quantity which is required for the argument of a transcendental function. 10° is a phase offset and is also a dimensionless quantity. For your purposes you need the RMS (Root Mean Square) current. Do you know how to compute that from the Peak current?

Is that enough "heads up" for you?

#### WBahn

Joined Mar 31, 2012
26,300
3/ is a description of a sinusoidal waveform. The peak amplitude is 5.0 Amperes. Cos is a shorthand notation for the Cosine function. The frequency of the cosine function is 500 Hz. When you multiply a frequency by time you get a dimensionless quantity which is required for the argument of a transcendental function.
How does multiplying (500 cycles per second) by (seconds) yield a dimensionless quantity?

This is why units matter and why leaving them off is not a good idea.

The only units that work in that equation are 500 radians/second, which is about 79.6 Hz since there are 2π radians/cycle.

#### flowtech

Joined Mar 23, 2018
6
1/ and 2/ represent constant values of resistance and inductance according to the units given

3/ is a description of a sinusoidal waveform. The peak amplitude is 5.0 Amperes. Cos is a shorthand notation for the Cosine function. The frequency of the cosine function is 500 Hz. When you multiply a frequency by time you get a dimensionless quantity which is required for the argument of a transcendental function. 10° is a phase offset and is also a dimensionless quantity. For your purposes you need the RMS (Root Mean Square) current. Do you know how to compute that from the Peak current?

Is that enough "heads up" for you?
Thanks Papabravo but I am not seeing how you concluded that 500t relates to Hz.
What do you see that indicates Hz? I interpreted 500t as Radians/Sec. WBahn thought this related to Radians/Sec also?
I am interested in and would appreciate your further comments on this one please Papabravo. I do need to sign off on how to move these supposedly simple calculations forward with the correct values from what has been supplied to me in the question.

#### Papabravo

Joined Feb 24, 2006
14,678
How does multiplying (500 cycles per second) by (seconds) yield a dimensionless quantity?

This is why units matter and why leaving them off is not a good idea.

The only units that work in that equation are 500 radians/second, which is about 79.6 Hz since there are 2π radians/cycle.
Ya got me. The 500 is radians/sec, which multiplied by time is radians which is dimensionless. You also can convert radians to degrees to add the phase offset, or you can convert 10° to radians and add that to 500t.

The TS has not yet said if he knows how to get the RMS value of a sinusoidal function.

#### flowtech

Joined Mar 23, 2018
6
How does multiplying (500 cycles per second) by (seconds) yield a dimensionless quantity?

This is why units matter and why leaving them off is not a good idea.

The only units that work in that equation are 500 radians/second, which is about 79.6 Hz since there are 2π radians/cycle.
I am still unsure about the next step though. Haven't firmed up on key values to move forward yet with the calcs.

#### flowtech

Joined Mar 23, 2018
6
Thanks Papabravo but I am not seeing how you concluded that 500t relates to Hz.
What do you see that indicates Hz? I interpreted 500t as Radians/Sec. WBahn thought this related to Radians/Sec also?
I am interested in and would appreciate your further comments on this one please Papabravo. I do need to sign off on how to move these supposedly simple calculations forward with the correct values from what has been supplied to me in the question.
I Peak * 1/SQRT(2) to convert to I RMS

#### WBahn

Joined Mar 31, 2012
26,300
I am still unsure about the next step though. Haven't firmed up on key values to move forward yet with the calcs.
There are a few approaches to finding the voltages across the components in the circuit. Which one you choose to use depends on which analysis techniques you know and that, in turn, largely on your math background. It seems odd that you would have been assigned a problem like this out of the blue without any exposure to any kind of analysis techniques, so how have you been shown (or what do your texts or reference material have to say) about how to work with voltages and currents in series circuits and about how voltage and current relate to each other in resistors, capacitors, and inductors?

#### WBahn

Joined Mar 31, 2012
26,300
Ya got me. The 500 is radians/sec, which multiplied by time is radians which is dimensionless. You also can convert radians to degrees to add the phase offset, or you can convert 10° to radians and add that to 500t.

The TS has not yet said if he knows how to get the RMS value of a sinusoidal function.
Depending on exactly what they are trying to find, they may not need to find the RMS value -- I don't see anything in the information provided that indicates that that is expected. They can express the voltage across each component as a function of time in the same format that the current is specified in.

#### Papabravo

Joined Feb 24, 2006
14,678
OK. The TS did say that the goal was to calculate VR and VL. It did not say to express the answer in a functional form from which instantaneous values could be calculated nor did it ask for voltage as a function of time which could be inferred from the use of lower case letters. The use of capital letters, to me at least, implies a single number is expected, and using RMS values would be my first choice. If there is additional information from the original problem statement that the TS has not shared with us, I might change my view.

#### WBahn

Joined Mar 31, 2012
26,300
OK. The TS did say that the goal was to calculate VR and VL. It did not say to express the answer in a functional form from which instantaneous values could be calculated nor did it ask for voltage as a function of time which could be inferred from the use of lower case letters. The use of capital letters, to me at least, implies a single number is expected, and using RMS values would be my first choice. If there is additional information from the original problem statement that the TS has not shared with us, I might change my view.
While convention does, indeed, prefer the use of lower case for instantaneous functions of time and upper case for aggregate values, it would seem that consistency within the same problem by the same writer should be given greater weight -- the original post used a capital I for the current expressed as an instantaneous time-domain function, so why wouldn't it seem reasonable that the same writer in the same paragraph wouldn't use capital letters for voltage in the same way, especially since nothing else to the contrary is in evidence? Like you, any further clarification and information from the TS on what is being sought would be most welcome.

#### flowtech

Joined Mar 23, 2018
6
It is my understanding that AC Current and Voltages values should always be given as RMS values unless clearly denoted otherwise. I elected to go with 5 Amp as the RMS current and 79.58 as the Frequency, and the calculations were straight forward. Both VR and VL calculated at 50 V and VS 70.71 V. Φ = 45°. My original question was more to do with interpreting this term I = 5.0 Cos (500t + 10°) (A) accurately. I don't think it qualifies as concise, seeing that it attracted two different interpretations from Papabravo and WBahn when first posted?

#### WBahn

Joined Mar 31, 2012
26,300
It is my understanding that AC Current and Voltages values should always be given as RMS values unless clearly denoted otherwise.
Not an unreasonable default position, but what qualifies as being "clearly denoted".

If you were in a field in which distances should always be given in meters and someone asks you what the perimeter was of a square that was 50 feet on a side, would you insist that the answer needs to be given in meters, or would you consider that the fact that the question was posed in units of feet qualifies as clearly denoting that the answer is most probably expected to be provided in feet?

The same applies here -- the problem is posed in instantaneous terms, so isn't reasonable to surmise that the answer is possible expected to be given in instantaneous terms?

If you can't ask what form the answer is expected in, the best bet would be to be safe and provide the answer both ways, perhaps with a comment indicating that it why it wasn't clear which was being requested. That will most likely score you some brownie points for being astute enough to recognize the issue and address it. It might also result in the question being rephrased to make it clearer in the future or, if the desired form was already considered sufficiently clear, some feedback on why it was considered so.

I elected to go with 5 Amp as the RMS current and 79.58 as the Frequency, and the calculations were straight forward.
If you only want to use a value that makes the calculations easy instead of the provided values, why not just go with 1 Arms? The current given is very clearly NOT equal to 5 Arms and you've already shown that you know how to convert from the peak value of a sinusoidal waveform to it's RMS equivalent.

Doesn't it seem like it would be a good idea to use the value of current that was given in the problem?

Both VR and VL calculated at 50 V and VS 70.71 V. Φ = 45°.
What is VS represent? What does Φ represent?

Don't introduce quantities without defining them? It forces people to have to guess what you mean and engineering is not about guessing (most of the time). By forcing people to reverse engineer your work in order to figure out what your terms mean you run the very real risk of calculating A when you mean to find B but since you calculated B correctly and didn't make it clear that you were trying to find A, people will work backward from your result, conclude that you meant to calculate B, and then tell you that you are correct when you are actually completely wrong.

This is likely a case in point. Many people care going to look at your undefined Φ and, because of your answer of 45°, conclude that you meant that to be the impedance angle of the RL combination. But I'm guessing that you actually meant it to be the angle of the voltage across the RL combination and that is NOT 45°. Beyond that, you were asked to find the voltages across each component and each voltage has its own magnitude and its own phase angle. Just because their magnitudes are equal does not mean that their phase angles are.

My original question was more to do with interpreting this term I = 5.0 Cos (500t + 10°) (A) accurately. I don't think it qualifies as concise, seeing that it attracted two different interpretations from Papabravo and WBahn when first posted?
There's nothing wrong with its conciseness. The problem is that it is dimensionally incorrect and forces the reader to assume. Since the argument of the sine and cosine functions are so commonly provided in this particular dimensionally incorrect format, it seldom causes serious problems because most people will make the correct assumption. In this case Papabravo simply made a simple goof by looking a bit too quickly, though I will contend that if either the author of the problem has been careful to make the problem dimensionally sounds OR if Papabravo was in the habit of religiously inspecting every equation for dimensional soundness, the goof would either not have gotten made or would have gotten caught immediately.

So it took two people being a bit sloppy to cause the mistake -- the primary burden is on the author to not be sloppy since some fraction of their audience WILL be a bit sloppy, either habitually or just on that instance because their morning coffee hasn't hit their brain in full force yet. But there's still a lesson for you as the reader of many problems over your entire career. If YOU develop the habit of inspecting units to the point that you have a hard time even looking at an equation that is not dimensionally sound and find that you have to correct it before you can use it, then you have largely sloppy-proofed yourself against the all-too-prevalent sloppy authors that are out there.

The dimensionally correct form would have been

I = 5.0 cos(500 r/s · t + 10°) A

Putting the A in parens at the end (A) is almost always an indication that the writer completely ignores units and therefore thinks that they need to tack on side notes to tell the user what units he wants the expression to have. Instead, they should recognize that if equations are made dimensionally sound from the beginning and if the units are tracked properly through the work, the final result will be dimensionally sound and will intrinsically have the correct units.

Although the units could have been tied to the coefficient (I actually prefer that, but it is technically not proper) and given as

I = 5.0 A cos(500 r/s · t + 10°)

though this particular example opens the possibility for someone to interpret A cos as Acos or the inverse cosine function. This can be resolved as follows

I = (5.0 A) cos(500 r/s · t + 10°)

This is my personal preference and how I write intermediate work since the coefficients and their units are almost always manipulated as a single quantity. But, as already noted, the internationally preferred format for published work is to put the units at the end of the expression.

For demonstration, let's say that the frequency instead was meant to be 500 Hz. In that case the writer could have simply included that information properly

I = 5.0 cos(500 Hz · t + 10°) A

This is perfectly dimensionally sound as the first term is a frequency multiplied by time, which is dimensionless, and the second is an angle, which is dimensionless. But before I can compute the single dimensionless quantity that I need to take the cosine of, I have to do conversions to make the units match. All of the information needed to do so is unambiguously provided.

It is no different than if I said that the length of a room was 12 ft + 7 in. Before I can combine those I have to do whatever conversions are needed to make the units match, but the expression is dimensionally sound because it is a distance added to a distance and all of the information needed is unambiguously provided.