For understanding what RMS (root square value) of AC current is you will have to first understand what AC current is. AC current is basically the type of current in which the charges flowing through a conductor periodically reverse their directio
n.
So instead of flowing in one direction the electrons vibrate about their mean position. The vibration of the electron is similar to that of the movement of a pendulum. The current is maximum when the electron is at the center and is zero when the electron is at the extreme position.
RMS Value Of AC
In
AC the magnitude of electricity changes continuously and the
direction changes periodically.
Alternating current has a maximum value and a minimum value (only their signs differ). The maximum current is called the peak value of current and is denoted by Io. (read as I knot).
Let’s consider a bulb. We connect this bulb to an AC supply. Let the maximum current (Io) flowing through this circuit be 10A. Does this mean that a steady 10A current is flowing though the bulb? No, the magnitude of current keeps changing. At one instant it can be 1A and at another it can be 8A. So if someone asks you how much current is flowing though this bulb, then what will your answer be? If your thinking 10A then your wrong because it is not a steady flow and the current keeps changing. Lets find the answer to this question.
AC shows all the effects that DC shows. These effects include heating effect (I2RT), magnetic effect etc. For all these effect we have a formula to calculate the magnitude of the effect in which one of the factor is current. So for calculating the magnitude of heat produced by AC, which value of AC will you put? The maximum value or the minimum value? The answer is none of these. We use the average current flowing through the appliance and this average current is known as RMS value AC.
Consider the following graph. The x-axis represents time and the y-axis represents the EMF. Since we are using AC current the EMF changes at every given moment.
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Work done by this current = Power x time = I2RT
At time T1, the current is less and so the work done is less. At time T2, the current is more and hence the work done is more.
Now, we calculate the average work done. When we calculate the average current, we notice that in I2RT, the RT part is common. Only the current changes. Hence we actually calculate the average of I2
∴ Iavg2 = I12 + I22 + I32 . . . . In2 / n
∴ Iavg = √I12 + I22 + I32 . . . . In2 /n (under root of Iavg2 )
Iavg is know as root mean square, that is, mean of root of squares.
We took the squares of current in the beginning as in the work formula, it is I2 and not I.
→ Iavg = IRNS
Alternating current has a maximum value and a minimum value (only their signs differ). The maximum current is called the peak value of current and is denoted by Io. (read as I knot).
Let’s consider a bulb. We connect this bulb to an AC supply. Let the maximum current (Io) flowing through this circuit be 10A. Does this mean that a steady 10A current is flowing though the bulb? No, the magnitude of current keeps changing. At one instant it can be 1A and at another it can be 8A. So if someone asks you how much current is flowing though this bulb, then what will your answer be? If your thinking 10A then your wrong because it is not a steady flow and the current keeps changing. Lets find the answer to this question.
AC shows all the effects that DC shows. These effects include heating effect (I2RT), magnetic effect etc. For all these effect we have a formula to calculate the magnitude of the effect in which one of the factor is current. So for calculating the magnitude of heat produced by AC, which value of AC will you put? The maximum value or the minimum value? The answer is none of these. We use the average current flowing through the appliance and this average current is known as RMS value AC.
Consider the following graph. The x-axis represents time and the y-axis represents the EMF. Since we are using AC current the EMF changes at every given moment.
---------
Work done by this current = Power x time = I2RT
At time T1, the current is less and so the work done is less. At time T2, the current is more and hence the work done is more.
Now, we calculate the average work done. When we calculate the average current, we notice that in I2RT, the RT part is common. Only the current changes. Hence we actually calculate the average of I2
∴ Iavg2 = I12 + I22 + I32 . . . . In2 / n
∴ Iavg = √I12 + I22 + I32 . . . . In2 /n (under root of Iavg2 )
Iavg is know as root mean square, that is, mean of root of squares.
We took the squares of current in the beginning as in the work formula, it is I2 and not I.
→ Iavg = IRNS
Consider the following example.
We want to heat a bucket of water upto a certain temperature in 1 hour. For this purpose we use DC and AC separately.
→ CASE 1 : 10A AC
The AC used is given by the following graph.
---------
Since the current is not constant in an AC supply, the heating will also be different. That is, sometimes it will heat the bucket using 10A current and sometimes 0A current. We observe that the bucket is heated up to the specific temperature in 1 hour.
→ CASE 2 : 10A DC
We want to heat a bucket of water upto a certain temperature in 1 hour. For this purpose we use DC and AC separately.
→ CASE 1 : 10A AC
The AC used is given by the following graph.
---------
Since the current is not constant in an AC supply, the heating will also be different. That is, sometimes it will heat the bucket using 10A current and sometimes 0A current. We observe that the bucket is heated up to the specific temperature in 1 hour.
→ CASE 2 : 10A DC
Since in DC, there
is constant current, the bucket is heated by the same amount of power. We
observe that it took the bucket 45 minutes to heat up to the specific
temperature.
∴ We conclude that 10A DC is doing more work as compared
to 10A AC.
→ CASE 3 : 8A DC
The current supply
has decreased from 10A to 8A DC. We observe that it took 50 minutes to heat the
bucket to the specific temperature.
∴ We conclude that 8A DC is doing more work as compared to
10A AC.
→ CASE 4 : 6A DC
The current is now
reduced to 6A. We observe that it took 1 hour 10 minutes to heat the bucket.
∴ We conclude that 10A AC is doing more work as compared
to 6A DC.
By doing this
several times, i.e. by trial and error, we find that 7A DC heats the bucket in
1 hour. Hence we can say that 7A DC is doing the same amount of work as
compared to 10A DC.
So if someone asks
us the magnitude of AC we used to heat the bucket we won’t tell them 0-10A or
anything like that. We tell them that we used 7A AC to heat the bucket. This is
because the 0-10A AC did the same amount of work the 7A DC did
∴ 0-10A AC = 7A DC
This comparison is
done on the basis of the work done.
∴ 7A is called the
RMS (root mean square) value of AC because we used work to derive it and
average work is the average of electric current.
∴For calculating any effect of AC current like magnetic effect and heating
effect, we put the RMS value (7A in this case) in the formulas.
NOTE: If you go to
the electrical store and buy an electric heater and someone asks you how much
electric current this heater uses, then you will tell the RMS value of the
devise (which is usually specified).
Now,
V=IR (Ohm's Law)
∴VRNS = IRNS X R
In India the
domestic voltage is 220V. This is the RMS voltage.
The derivation of
the formula to calculate the RMS AC includes integration. For now I'm only
giving you the direct formulas and not the integration part, however
comment down below if you want me to show the derivation.
Where I= peak value of current of AC current.
Where V= peak value of voltage in AC.
Very well explained, my concepts regarding this topic are very clear now. Thanks.
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