## Friday, August 7, 2020

### AC - Effective Voltage: UNIZOR.COM - Physics4Teens - Electromagnetism

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

Effective Voltage
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) = Umax·sin(ωt)

I(t) = U(t)/R = Imax·sin(ωt)

where

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

t is time,

Umax is the voltage amplitude,

Imax = Umax/R is the amperage amplitude,

R is resistance of the circuit where 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 T=2π/ω and calculate the corresponding DC 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 U(t)=Umax·sin(ωt) carries through some circuit of resistance R during the time of its period T=2π/ω.

Consider an infinitesimal time interval from t to t+dt.
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 R for direct current:

dW(t) = U(t)·I(t)·dt = U²(t)·dt/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 T=2π/ω, we have to integrate this expression for dW on an interval from 0 to T.

W[0,T] = [0,T]dW(t) =

= [0,T]U²(t)·dt/R =

= [0,T]max·sin²(ωt)·dt/R =

= (max/R[0,T]sin²(ωt)·dt

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

dt = dx/ω

t ∈ [0,T] x ∈ [0,ωT]=[0,2π]

The integral above, expressed in terms of x, is

[0,2π]sin²(x)·dx/ω =

=
[0,2π][1−cos(2x)]·dx/(2ω) =

=
[2π−[0,2π]cos(2x)·dx]/(2ω) =

= π/ω

Therefore,

W[0,T] = π·U²max/(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 R during the same time T=2π/ω if the current is constant and direct.

That is

eff·2π/(R·ω) = W[0,T] = π·U²max/(ω·R)

Cancelling and simplifying the above equality, we conclude

2·U²eff = U²max

or

2·Ueff = Umax

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.