what is it
what is it
Realolman , see my answers below . Unfortunately in the real world the correct answer is always "depends" as my old buddy Sam Huertze says. But fortunately in a steady state situation it is kind of like a joke my first calculus prof told me. A good ole boy engineer and a mathmatican were told there was a sweet young thing at the end of the hall willing to do anything they wanted when they got there. The only catch was they could only move half the distance toward her every 6 seconds. The mathmatician said that was impossible and would take to infinity to get to touch her. Technically the mathmatician's answer is the correct and 100% right answer. The engineer pondered for a few seconds, pulled out his slide rule, pencil, and burger wrapper, and said no problem, in about 12 minutes I will be close enough.
1. Wouldn't the frequency, wave form, capacitance, inductance, and whatever else, already have to be known in order to get the rms values?
No. You are measuring the RMS value of a voltage which is kind of like a DC volts average or equivalent. You dont have to know the load characteristics. The RMS value of a sinusoidal 60 hz or 1 gigahertz frequency with a peak voltage of about 170 vac would be about 120 vrms. Other wave forms would be different. You dont need to know if it was capacitive, inductive, resistive, etc. The voltage is measured independently of the current and vice versa. In a way the meter doesnt need to know the frequency but it would have to be designed for all frequencys and wave forms and that just aint possible. As long as the meter is designed to respond to the frequency and wave form the resultant rms value will be ok. Thats why they give frequency ranges and accuracy specifications. At high frequencys the length of the leads and distance apart can cause false readings due to capacitive and inductive effects. The leads may "drop" a lot of the voltage such that only a small fraction of the true voltage gets to the measuring circuit. Thats why some computer cables can be no longer than about 30 feet. You put X watts in at the front and zero gets to the end. The means or way the meter uses to calculate the "true rms" will affect the results. RMS values involve integrating the function over one cycle of the wave form. Which involves a different time for each frequency. A half cycle works for symetrical wave forms.
2. How would the rms values be known without knowing all the relevent information?
You are measuring a voltage or current independently of each other with your meter or at least that is implied by stating 200 vrms and 5 arms. The wave form and frequency do matter to what the meter can respond to and the means used to derive the rms voltage or current by the meter. As long as the meter can respond properly, it will give a true rms value for the voltage or current. The voltage and amps must be measured relative to each other in time to know the power. To know this usualy involves measuring when they reach peak values. Which brings up a phase angle meter or o-scope.
3. If the frequency, wave form, and whatever else are already known, why does there need to be more information?
True power or watts depends on knowing if the load is capacitive, inductive, resistive, or a combination. In the real world there is always a little capacitance, inductance, and resistance, even in a light bulb or a straight piece of wire. Most folks think no load equals no current. And that would be moslty true in low voltage circuits. But that all changes when its 525000 volts and 100 miles of line. A few amps of no load curent at 525000 volts and 60 hertz will create a ten foot arc that sounds like the air is being ripped apart. The wire may have only a fraction of a microhenry or picofarad per foot, but over a hundred miles at a high voltage, it is awesome. At 220 volts and a few hundred feet of house wiring it is barely noticible. The higher the frequency the more the inductive and capacitive components will affect the volts and amps. True power or watts is volts times amps times a factor of the angle between them. In a sinusoidal system at one frequency----- watts = Vrms x Arms x cosine of angle between V and A. If you connect a capacitor to an ac voltage you will measure an rms voltage and an rms current but the watts will be zero. Same with an inductor, the watts will be zero. With a wave form the power depends on the direction the voltage and current are moving in time. Kind of like mules pulling a plow or a game of tug of war. The voltage and current have to be in the same direction at the same time to make power. in a capacitor or inductor they are 90 deg apart. When ever one is giving the other is taking and the net result is zero. In a resistor the current and voltage are always in the same direction and that way they can do work or power. Capacitors and inductors pull current if a voltage is applied but they do not consume power in the good ole boys world. In the mathmaticians world there will be a little power used in the resistance of the leads and internals of the capacitor or inductor.
Now back to the given question of 200 vrms and 5 arms. Given no angle between the the answer is 200 x 5 = 1000 VA or 1 KVA. This is called apparent power or Q. Given only volts and amps rms one can only know the apparent power or Q. To find watts, real power ,or P, at least the angle between the volts and current is needed. Knowing frequencys and wave forms and meter characteristics are needed to give a more acutate answer.
Now in defense of the good ole boy we are also given that the load is an incandesant lamp. Assuming most normal conditions of wanting a lamp on this earth, the frequency involved is 25, 50, or 60 cycles mostly sinusoidal wave. As such it will be almost purely resistive. Which implies that the phase angle between voltage and current is close enough to the sweet young thing above. So in that case 200 vrms x 5 arms is 1000 VA, 1 KVA or 999.999999999 W or 1.00000000 KW.
That also brings up another practical problem. We are given that it is "an" incandesent lamp. Which implies one lamp. While they make them, they are typically expensive and very short lived. Will cost you about a dollar an hour bulb cost plus electricity usage and labor to replace. Due to tungsten characteristics and filament construction, incandesent bulbs more than about 200 or 300 watts arent very cost effective. A 1000 watt halogen lamp is a better choice and is basically an incandesent lamp. But they typically arent called incandesent. Its still expensive but lasts longer and costs a little less to operate. But still nowhere near as good as HID types.
But at the same time in defense of the mathmatician don't come whining to me when your volt meter shows 200 volts rms, your amp meter shows 5 amps rms. and your watt meter shows 900 watts rms for a 1000 watt incandesant bulb. Dont be surprised if I tell you all are probably correct after asking you where its used, what frequency, what power supply, other equipment in the area, etc. Also dont be surprised if your volts and amps for the the TV speaker shows that they are the same rms for a TV commercial as for the program. As my ole buddy Sam Heurtze says the correct answer is always "depends". To be 100% correct you need to know the wave form and frequency or mathmatical equation for the volts, amps, and impedence to calculate the true power.
In the end all 3 answers are close enough to correct as long as you know all the limits. I voted for not enough info.
But then again the only IQ test I ever took said I was a smart moron or a dumb average person. Probably explains why it took me six years and repeating some courses 3 times to get a four year degree. It still amazes me that we started out with sticks, bark, rocks, and dirt and made machines that can craft to less than one thousandth of an inch. I still havent figured out accuracy of reading versus accuracy of full scale. Accuracy to me implies I have some standard to compare an unknown to. Accuracy of full scale fits that. Accuracy of reading to me implies i am using an infinite number of standards. I measured an unknown value, comparted it to itself, and stated it is correct within 0.5% plus or minus 5 digits. How do you compare something to itself and come up with any error at all doesnt make sense, much less measuring something with no standard to compare to and assigning a value.
