Lesson 27 (1) Root Mean Square The emf from an AC generator has the time dependence given by ℇ = ℇ! !"#$% where ℇ! is the peak emf, ! is the angular frequency. The period is 2!
! = !
The mean square value of the emf is 1 ! !
1 ! !!
1 ! !! 1 + !"#2!
ℇ!!
!
!
!
ℇ =
ℇ !"# !"#" =
ℇ
!"# !"! =
ℇ
!" = ! ! !
2! ! !
2! ! !
2
2
and the root-­‐mean-­‐square emf is ℇ!
ℇ!"# = ℇ! = 2
Similar relations exist between the rms and peak values of AC currents and potential differences (voltages). (2) Current-­‐Voltage Relations Electrical potential can be introduced into an AC circuit. The potential differences across various circuit elements are established by the sources of emf. With the direction of the current indicated by an arrow, the potential difference across an element is defined as !V = V at outlet "V at inlet As shown, for resistor !V = "IR Q
dQ
for capacitor !V = "
I=
C
dt
dI
for inductor !V = "L dt
1 Δ! = ℇ for emf source The voltage drop, or simply voltage, across the resistor, capacitor, and inductor are defined as !"V . These voltages are Q
dI
VR = IR
VC =
VL = L C
dt
For an AC circuit oscillating in a steady state with frequency ! , the relations between the rms values and the phases of the current and voltage for these elements can be found, assuming the current to have the form I = I 0 cos! t V = IR = I 0 R cos! t For a resistor, The current and voltage are in phase. The phasors for them are therefore parallel to each other as shown. The peak and rms values are related by V0 = I 0 R Vrms = I rms R For a capacitor, #
Q 1
1
1
1
"&
V = = ! I dt = ! I 0 cos! t dt =
I 0 sin ! t =
I 0 cos %! t " ( $
C C
C
!C
!C
2'
!
Defining the capacitive reactance by !! = !" the peak and rms values are related by !! = !! !! !!"# = !!"# !! , from which we see that the reactance also has the unit Ω. The phase of the voltage lags the current by 90° . This is illustrated by the phasors as shown. For an inductor, "
dI
"%
V = L = !! LI 0 sin ! t = ! LI 0 cos $! t + ' #
dt
2&
Defning the inductive reactance !! = !" we can write !! = !! !! !!"# = !!"# !! , 2 The phase of the voltage leads the current by 90° , as illustrated by the phasors shown. At high frequency, the capacitive reactance becomes small. So does the voltage. This can be understood because the current reverses itself so rapidly that there is no significant build up of charge on the capacitor, which is proportional to the voltage. The capacitor then behaves as a short circuit. ( a piece of wire with zero resistance). At low frequency, the inductive reactance becomes small. This is because the induced back emf becomes small when the current changes slowly. Again, the inductor behaves as a short circuit. (3) Power The power delivered to the circuit element is given by P = IV , from which the average power over one cycle can be found: 1 T
P = ! IV dt T 0
Using the current-­‐voltage relations, we find IV T
I V 2"
IV
For a resistor, P = 0 0 ! cos2 ! t dt = 0 0 ! cos2 # d# = 0 0 = I rmsVrms T 0
2" 0
2
so that if we use rms values, the formula is the same as that for DC. For a capacitor, "
IV T
"%
I V 2"
I V 2"
P = 0 0 ( cos! t cos $! t ! ' dt = 0 0 ( cos# sin # d# = 0 0 ( sin 2# d# = 0 #
T 0
2&
2" 0
4" 0
A capacitor does not absorb power on the average. If power is absorbed during a half cycle, it is released during the next half. Similarly, it can be shown that an inductor also does not absorb power on the average. For a source of emf, there is no fixed relation between the current and the emf, although an emf ℇ steps up the potential in the circuit by the same amount. If the phase difference between the emf and the current is the angle !
, we can write 3 ℇ = ℇ! cos!" ! = !! cos !" − ! so that the average power delivered by the emf is 1 !
!! ℇ! !!
!=
!ℇ!" =
!"#$!"# ! − ! !" ! !
2! !
The integral is 1 !!
1 !!
!"#$
!"#$!"# ! − ! !" =
!"#$ !"#$!"#% + !"#$!"#% !" =
2! !
2! !
2
Therefore ! = !!"# ℇ!"# !"#$ The quantity !"#! is known as the power factor. It is determined from the combination of elements in the circuit. (4) RL Circuit The loop equation for an RL circuit with an AC source of emf !"
ℇ! !"#$% − !" − ! = 0 !"
To find the current, we first rewrite it in terms of the voltages in the form !! + !! = ℇ This can be considered as the x-­‐component of the equation for the phasors: !! + !! = ℇ If we find the solution to this vector equation, we can find the current. We start by drawing a vector representing the current phasor ! , which is common to all elements. We next add the phasors !! and !! , noting that !! is in phase with ! , and !! leads ! by 90º. The emf phasor ℇ is now added as the 4 sum of the vectors !! and !! . We also indicate the magnitudes of the voltage phasors in terms of the peak current. From the diagram, we see that !! ! ! + !! !! ! = ℇ!! ℇ!
!! = !
where we have defined ! = !! + !!! called the impedance. The relation between the impedance, the resistance, and the reactance is shown as a right angled triangle. Also, the angle ! between ℇ and ! is in the first quadrant, and satisfies !!
!"#$ = !
The current lags the emf by this angle, which can also be seen in the triangle for the impedance. The solution for the current is therefore ! = !! !"# !" − ! The power factor is !"#$. Example: The emf of an AC power supply in volts is given by ℇ = 10!"# 1200!" where time ! is in seconds. The power supply is connected in series with a 5.0! resistor and 1.0mH inductor. Find the peak current and the average power delivered by power supply. Write an expression for the current as a function of time. Find the current at ! = 3.003!. !"#
Solution: ! = 1200! = 3770 ! !! = !" = 3770×1.0×10!! = 3.77Ω !!
10
! = !! + !!! = 6.26Ω !! = =
= 1.60! ! 6.26
!!
! = !"#!!
= 37° = 0.20! !"#$ = 0.80 !
5 !=
!! ℇ!
2 2
!"#$ = 6.4! ! = 1.60!"# 1200!" − 0.2! At ! = 3.003!, ! = 1.60!"# 1200!×3.003 − 0.2! = 1.60!"# 1200!×0.003 − 0.2!
= 1.60!"# 3.6! − 0.2! = 1.60!"# 1.6! − 0.2! = 1.60!"# 1.4!
= 1.60!"# 252° = −0.49! Example: Find the voltage across the resistor for the circuit of the previous example when the frequency is (a) 60Hz and (b) 6000Hz. Solution: (a) !"#
! = 2!×60 = 377 ! !! = 377×10!! = 0.377Ω ! = 5! + 0.377! = 5.014Ω 10
!! =
= 1.99! !!! = !! ! = 9.95! 5.014
!"#
(b) ! = 2!×6000 = 37700 ! !! == 37.7Ω ! = 5! + 37.7! = 38.0Ω 10
!! =
= 0.26! !!! = !! ! = 1.3! 38
We see in this example that as the frequency increases, the voltage across the resistor goes down. If the source of emf is a signal with a range of frequency, and the voltage across the resistor is conveyed to another part of the circuit for detection, then the high frequency part of the signal has little contributions. We now have a low-­‐pass filter. The fact that this is a low-­‐
pass filter is obvious from the general relation ℇ! !
!!! = !! ! =
!! + !" !
(5) RC Circuit For an RC circuit with the same emf as for the RL circuit before, the equation for the phasors is !! + !! = ℇ. Noting as before that the current phaser ! is common to all elements and that the capacitor voltage phasor !! = ℇ lags the 6 current phasor by 90º, we construct the phasor diagram with peak voltage magnitude indicated as shown. From this we find ℇ!
!! = ! = !! + !!! !
We also that the current leads the emf by the phase angle ! given by !!
!"#$ = !
so that the current is ! = !! !"# !" + ! . Example: In an RC circuit with an AC power supply, !"""!"#
! = 5.0Ω, ! = 20!", ! =
It is also known that the peak voltage !
across the capacitor is 4.0V. Find the peak voltage across the resistor and the peak emf of the AC power supply. Find also the phase difference between the emf and the voltage across the capacitor. !
!
!
Solution: !! = !" = !"""×!"×!"!! = 8.33Ω !! = !!! = 0.48! !
!!! = !! R = 0.48×5 = 2.4V ℇ! =
!
!
!!!
+ !!!
= 4! + 2.4! = 4.66V The angle between the phasors ℇ and !! is !!!
!
5
! = !"#!!
= !"#!!
= !"#!!
= 31° !!!
!!
8.33
The voltage across the capacitor leads the emf by this phase angle. In an RC circuit with an AC power source, the peak voltage across the resistor is !ℇ!
!!! =
1
!
! +
!" !
which increases with frequency to reach the maximum value of ℇ! at infinite frequency, when the capacitor behaves like a wire of zero resistance. If this voltage is delivered to another circuit, the high frequency component will dominate, and we then have a high-­‐pass filter. A loud speaker system with a tweeter for high frequency sound and woofer for low frequency sound can be based on the circuit as shown. 7