# RMS or Average Current Homework - Tips & Tricks

• jaus tail
In summary, the author claims that the average value of a dc current is of no use, and that it is always the RMS value that is used in rectifier/converter problems. Additionally, the author recommends that if you are studying speed control of dc motors, you should not bother about rms, and should only focus on average values. Finally, the author notes that transformer also measures RMS value in rating.
jaus tail

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## Homework Equations

I calculated for RMS current.

## The Attempt at a Solution

How to know whether they've asked RMS or Average current? Isn't Output always RMS? Average value is of no use anyways.

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In rectifier/converter problems, always find the average dc output voltage/current unless it is specifically asked that the rms value be found.
jaus tail said:
Average value is of no use anyways.
Why would you think so?
Have you studied speed control of dc motor using rectifier, converter and chopper? You never bother about rms there, it's all about average values.

jaus tail said:
How to know whether they've asked RMS or Average current? Isn't Output always RMS? Average value is of no use anyways.
The specifications tell you that this arrangement is outputting a load current they describe as constant and ripple-free DC current as a result of inductive smoothing.

Purists may insist on inclusion of the word "approximately", but in power engineering that goes without saying. Everything in power electronics is geared towards an approximation of convenience.

cnh1995 said:
In rectifier/converter problems, always find the average dc output voltage/current unless it is specifically asked that the rms value be found.

Why would you think so?
Have you studied speed control of dc motor using rectifier, converter and chopper? You never bother about rms there, it's all about average values.
I've left that out for exam. There is too much to study so I've not studied some parts in each subject.
NascentOxygen said:
The specifications tell you that this arrangement is outputting a load current they describe as constant and ripple-free DC current as a result of inductive smoothing.

Purists may insist on inclusion of the word "approximately", but in power engineering that goes without saying. Everything in power electronics is geared towards an approximation of convenience.
If it's ripple free constant DC current then RMS = Average value. So I should get same answer by rms formula.

jaus tail said:
If it's ripple free constant DC current then RMS = Average value. So I should get same answer by rms formula.
First determine the waveform of the current, then use a formula appropriate to this waveform. If you misapply a formula you will get a wrong answer.

Yes, if you determine the RMS value of a steady current you must arrive at its average (or DC) value.

I found this from link:
http://protorit.blogspot.in/2013/01/power-electronics-thyristor-single-phase-bridge-rectifier.html

I know that since current is said to be constant, there'll be one horizontal line for current. But in the answer in post#1, they've assumed current waveform to be sine wave. That's how they got the equation for average value. The equation of 1 + cos firing angle is for average value.
I still don't understand why average value is taken whereas rms value should be considered. RMS value means DC value means the value that will give same power. Like for heating coils rms value is taken. DC value is never used. Even in network equations rms value is taken.
Transformer also measures RMS Value in rating. Power = V rms times I rms times power factor.

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jaus tail said:
But in the answer in post#1, they've assumed current waveform to be sine wave. That's how they got the equation for average value. The equation of 1 + cos firing angle is for average value.
No, they assumed the voltage to be sinusoidal with a portion missing and used a formula to find the average value of that voltage rectified. The large series inductor performs an averaging function, and the steady current through the load has its value determined by the average of the voltage across L + R. (The average of a periodic voltage across an inductor is zero.)

It would be instructive for you to sketch the expected current waveform in one thyristor, and also the current in the diode. When the thyristor is conducting, it is supplying all the load current. When the thyristor is not conducting, the backswing diode is supplying all the load current (a fixed value).

(I think they have glossed over something in determining that average voltage, those negative-going portions, but I won't mention it either.)

jaus tail

## What is RMS or Average Current?

RMS or Root Mean Square current is a measure of the average current in an alternating current (AC) circuit. It takes into account both the magnitude and direction of the current, unlike average current which only looks at the magnitude.

## Why is it important to calculate RMS or Average Current?

Calculating RMS or Average Current is important because it gives us a more accurate representation of the current in an AC circuit. This is important for properly sizing components and ensuring the safe and efficient operation of the circuit.

## How do you calculate RMS or Average Current?

The formula for calculating RMS or Average Current is: Irms = Imax / √2, where Imax is the maximum or peak current in the circuit. Alternatively, you can also use the formula Irms = Iavg x 1.11, where Iavg is the average current in the circuit.

## What are some tips for solving RMS or Average Current homework problems?

1. Make sure you understand the difference between RMS and average current.2. Pay attention to the units in the problem and make sure they are consistent.3. Use the appropriate formula for the given problem.4. Double-check your calculations to ensure accuracy.5. If you are stuck, try breaking down the problem into smaller parts and tackling them one at a time.

## Are there any tricks to solving RMS or Average Current problems?

One trick is to use the fact that the RMS value of a sinusoidal wave is equal to the peak value divided by the square root of 2. This can save you time in calculations if the problem involves a sinusoidal wave.

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