*Notes to a video lecture on http://www.unizor.com*

__Effective Voltage__

and Current (RMS)

and Current (RMS)

We all know the voltage in our outlets at home. In some countries it's

220V, in others 110V or some other. But we also know that electric

current distributed to consumers from power plants is

*alternating (AC)*.

The actual value of

*voltage*and

*amperage*are sinusoidal, changing with time as:

**U(t) = U**_{max}**·sin(ωt)**

**I(t) = U(t)/R = I**_{max}**·sin(ωt)**where

*is some parameter related to the generation of electricity at the power plant,*

**ω***is time,*

**t***is the voltage*

**U**_{max}*amplitude*,

*is the amperage*

**I**_{max}**= U**_{max}**/R***amplitude*,

*is resistance of the circuit where*

**R***alternating current*is running through.

If

*voltage*is variable, what does it mean that in the outlet it is, for example, 220V?

The answer to this question is in comparing energy the electric current is carrying in case of a variable

*voltage*with energy of a

*direct current*.

The AC

*voltage*is sinusoidal, so we can calculate the energy it carries through a circuit during the time of its

*period*

*and calculate the corresponding DC*

**T=2π/ω***voltage*that carries the same amount of energy during the same time through the same circuit.

That corresponding value of the DC

*voltage*is,

__by definition__, the

*effective voltage*in the AC circuit, which is usually called just

*voltage*for alternating current.

First, let's evaluate how much energy the AC with voltage

*carries through some circuit of resistance*

**U(t)=U**_{max}**·sin(ωt)***during the time of its period*

**R***.*

**T=2π/ω**Consider an infinitesimal time interval from

*to*

**t***.*

**t+**d**t**During this interval we can consider the voltage and amperage to be

constant and, therefore, use the expression of the energy flowing

through a circuit of resistance

*for direct current:*

**R***d*

**W(t) = U(t)·I(t)·**d**t = U²(t)·**d**t/R**(see "Electric Heat" topic of this course and "Ohm's Law" for direct current)

To calculate the energy going through a circuit during any period

*, we have to integrate this expression for*

**T=2π/ω***d*on an interval from

**W***to*

**0***.*

**T**

**W**_{[0,T]}=

**∫**

_{[0,T]}

*d*

**W(t) =****= ∫**

_{[0,T]}

*=*

**U²(t)·**d**t/R****= ∫**

_{[0,T]}

*=*

**U²**_{max}**·sin²(ωt)·**d**t/R**= (

*)·*

**U²**_{max}**/R****∫**

_{[0,T]}

**sin²(ωt)·**d**t**To simplify the integration we will use the known trigonometric identity

**cos(2x) = cos²(x) − sin²(x) = 1 − 2sin²(x)**from which follows

*[*

**sin²(x) =***]*

**1 − cos(2x)**

**/2**Next we will do the substitution

**x = ω·t**

**t = x/ω***d*

**t =**d**x/ω***∈ [0,T] ⇒*

**t***∈ [0,ωT]=[0,2π]*

**x**The integral above, expressed in terms of

*, is*

**x****∫**

_{[0,2π]}

**sin²(x)·**d**x/ω =**

==

**∫**

_{[0,2π]}[

*]*

**1−cos(2x)***[*

**·**d**x/(2ω) =**

==

*∫*

**2π−**_{[0,2π]}

*]*

**cos(2x)·**d**x**

**/(2ω) =**

= π/ω= π/ω

Therefore,

**W**_{[0,T]}

**= π·U²**_{max}**/(ω·R)**At the same time we have defined the

*effective*voltage as the one that delivers the same amount of energy as

**W**_{[0,T]}to the same circuit of resistance

*during the same time*

**R***if the current is constant and direct.*

**T=2π/ω**That is

**U²**_{eff}**·2π/(R·ω) = W**_{[0,T]}

**= π·U²**_{max}**/(ω·R)**Cancelling and simplifying the above equality, we conclude

**2·U²**_{eff}**= U²**_{max}or

**√2·U**_{eff}**= U**_{max}For

*effective*voltage 110V the peak voltage (amplitude) is 156V,

for

*effective*voltage 120V the peak voltage (amplitude) is 170V,

for

*effective*voltage 220V the peak voltage (amplitude) is 311V.