Electrical Circuits II
AC Power Analysis
AC Power Analysis:
• Every industries and household electronic device has a power
rating that indicates how much power the equipment requires;
exceeding the power rating can do permanent damage to an
appliance.
• Power is the most important quantity in electric utilities,
electronic and communication systems, because such
systems involve transmission of power from one point to
another.
• The choice of AC over DC allowed high-voltage power
transmission from the power generating plant to the
consumer.
a. INSTANTEOUS AND AVERAGE POWER
Instantaneous power (in watts) is
the power at any instant and time.
𝑷 𝒕 =𝒗 𝒕 𝒊 𝒕
Where: 𝒗 𝒕 = Instantaneous voltage
across the element
Consider the general case of
instantaneous power absorbed by
an arbitrary combination of circuit
element under sinusoidal
excitation.
𝒗 𝒕 = 𝑽𝒎 𝒄𝒐𝒔 𝒘𝒕 + 𝑶𝒗
𝒊 𝒕 = 𝑰𝒎 𝒄𝒐𝒔(𝒘𝒕 + 𝑶𝒊 )
𝒊 𝒕 = Instantaneous current through it
•
We can also think of the instantaneous
power as the power absorbed by the
element at a specific instant of time.
Instantaneous quantities are demoted
by lower case letter.
It is the rate at where can element
absorbed energy.
Where 𝑽𝒎 𝒂𝒏𝒅 𝑰𝒎 = amplitudes (peak values)
𝑶𝒗 𝒂𝒏𝒅 𝑶𝒊 = phase angles of the voltage and
current
𝒑 𝒕 = 𝒗 𝒕 𝒊 𝒕 = 𝑽𝒎 𝑰𝒎 𝒄𝒐𝒔 𝒘𝒕 + 𝑶𝒗 𝒄𝒐𝒔(𝒘𝒕 + 𝑶𝒊 )
Trigonometric identity
cos(Wt+tv)
p(t) VmIm
=
𝒄𝒐𝒔 𝑨 𝒄𝒐𝒔 𝑩 =
COS
(at
+fi)
=Wt Ov
+
A
B
Wt+8;
=
𝟏
𝒄𝒐𝒔 𝑨 − 𝑩 + 𝒄𝒐𝒔 𝑨 + 𝑩
𝟐
-
𝟏
𝟏
𝒑 𝒕 = 𝑽𝒎 𝑰𝒎 𝒄𝒐𝒔 𝑶𝒗 − 𝑶𝒊 + 𝑽𝒎 𝑰𝒎 𝒄𝒐𝒔(𝟐𝒘𝒕 + 𝑶𝒗 + 𝑶𝒊 )
𝟐
𝟐
Part 1
The instantaneous power 𝒑(𝒕)
entering a circuit.
Part 2
Two parts:
1. Is constant or time independent;
values depends on the phase
difference between the voltage and
current.
2. Is a sinusoidal function whose
frequency is 𝟐𝒘 which is twice the
angular frequency of the voltage or
current.
Observation: 𝒑 𝒕 = 𝒑 𝒕 + 𝑻𝟎
•
𝒑 𝒕 is periodic and has a period of
𝑻𝟎 = 𝑻/𝟐, since its frequency is twice
that of voltage or current.
•
𝒑(𝒕) is positive for some part of each
cycle and negative for the rest of the
cycle.
Trigonometric identity
•
When 𝐩 𝐭 is positive, power is absorbed by the circuit.
•
When 𝐩 𝐭 is negative, power is absorbed by the source, that is power transferred from the
circuit to the source. This is possible because of the storage element in the circuit (capacitors
and inductors).
The instantaneous power changes with time and is therefore difficult to measure. The average
power is more convenient to measure. The wattmeter, the instrument for measuring power,
responds to average power.
The average power in watts is the average of the instantaneous power over one period.
𝟏 𝑻
𝑷 = න 𝒑 𝒕 𝒅𝒕
𝑻 𝟎
𝟏 𝑻𝟏
𝟏 𝑻𝟏
𝑷 = න 𝑽𝒎 𝑰𝒎 𝒄𝒐𝒔 𝑶𝒗 − 𝑶𝒊 𝒅𝒕 + න 𝑽𝒎 𝑰𝒎 𝒄𝒐𝒔 𝟐𝒘𝒕 + 𝑶𝒗 + 𝑶𝒊 𝒅𝒕
𝑻 𝟎 𝟐
𝑻 𝟎 𝟐
𝟏
𝟏 𝑻
𝟏
𝟏 𝑻
𝑷 = 𝑽𝒎 𝑰𝒎 𝒄𝒐𝒔 𝑶𝒗 − 𝑶𝒊 න 𝒅𝒕 + 𝑽𝒎 𝑰𝒎 න 𝒄𝒐𝒔(𝟐𝒘𝒕 + 𝑶𝒗 + 𝑶𝒊 ) 𝒅𝒕
𝟐
𝑻 𝟎
𝟐
𝑻 𝒐
Constant
Sinusoidal
Trigonometric identity
The average of sinusoid over its period is zero because the area
under the sinusoid during a positive half-cycle is canceled by the
area under it during the following negative half-cycle. Thus, the term
vanishes and the average power becomes.
𝟏
𝑷 = 𝑽𝒎 𝑰𝒎 𝒄𝒐𝒔 𝑶𝒗 − 𝑶𝒊
𝟐
Phasor form of 𝒗 𝒕 𝒂𝒏𝒅 𝒊 𝒕 𝒂𝒓𝒆 𝑽 =
𝑽𝒎 ∠𝑶𝒗 𝒂𝒏𝒅 𝑰 = 𝑰𝒎 ∠𝑶𝒊
𝟏 ∗ 𝟏
𝑽𝑰 = 𝑽𝒎 𝑰𝒎 ∠𝑶𝒗 − 𝑶𝒊
𝟐
𝟐
=
𝟏
𝑽 𝑰 [𝒄𝒐𝒔 𝑶𝒗 − 𝑶𝒊 + 𝒋 𝒔𝒊𝒏 𝑶𝒗 − 𝑶𝒊 ]
𝟐 𝒎 𝒎
Note: as 𝑶𝒗 − 𝑶𝒊 = 𝒄𝒐𝒔 𝑶𝒊 − 𝑶𝒗 , what is
important is the difference in the
phases of the voltage and current.
Trigo identity
Real part of average power:
𝟏
𝟏
𝑷 = 𝑹𝒆 𝑽𝑰∗ = 𝑽𝒎 𝑰𝒎 𝒄𝒐𝒔(𝑶𝒗 − 𝑶𝒊 )
𝟐
𝟐
Consider two special cases:
a. For a resistive circuit purely resistive circuit absorb power at all times. 𝑶𝒗 = 𝑶𝒊
𝟏
𝟏 𝟐
𝟏 𝟐
𝑷 = 𝑽𝒎 𝑰𝒎 = 𝑰𝒎 𝑹 = ฬ𝑰ቚ 𝑹
𝟐
𝟐
𝟐
Where |𝑰|𝟐 = 𝑰 𝒙 𝑰∗
b. When 𝑶𝒗 − 𝑶𝒊 = ±𝟗𝟎𝒐 ; a purely reactive
circuit
𝑷=
𝟏
𝑽𝒎 𝑰𝒎 𝒄𝒐𝒔𝟗𝟎𝒐 = 𝟎
𝟐
Shows that a purely reactive
circuit absorbs no average
power.
PROBLEMS
1. Given that,
i)
AkaAit
𝒗 𝒕 = 𝟏𝟐𝟎 𝒄𝒐𝒔 𝟑𝟕𝟕𝒕 + 𝟒𝟓𝒐 𝑽 𝒂𝒏𝒅 𝒊 𝒕 = 𝟏𝟎 𝒄𝒐𝒔(𝟑𝟕𝟕𝒕 − 𝟏𝟎𝒐 ) 𝑨
find the instantaneous power and the average power
absorbed by the passive linear network.
mm(cos(tr-8i) cos(2Nt+Ov+8i)]
p(t):
p
+
VmIm cos
=
I
p(t)
((10)(cos(45
=
p(t)
P
-110)) cos(547 45 110)))
244.2 400 cos(7547
+
+
=
244.2W
=
(tv-oi)
+
+
750)
W
+
PROBLEMS
1.1. Given that,
𝒗 𝒕 = 𝟏𝟔𝟓 𝒄𝒐𝒔 𝟏𝟎𝒕 + 𝟐𝟎𝒐 𝑽 𝒂𝒏𝒅 𝒊 𝒕 = 𝟐𝟎 𝒔𝒊𝒏(𝟏𝟎𝒕 + 𝟔𝟎𝒐 ) 𝑨
find the instantaneous power and the average power
absorbed by the passive linear network.
(20) (cos(20-40)
p(t)-
=
4
1247.97
1244.97W
=
+
cos(20t
+
1650 as (20t + 80°
W
20
+
+60))
2. Calculate the average power absorbed by an impedance
Z=30-j70Ω when voltage V=120∠0° V is applied across it.
I
->
-
1
6
-04
I 280
I
P
=
=
1.58)
=
P 77.45W
=
z
20=j40
=
1.58(64.800A
=
cs
(0-44.0)
2.2. Calculate the average power absorbed by an impedance
Z= 40∠-22° Ω when current I=20∠30° A flows through it.
0 1z
=
P
P
(20(30%)(4(( -22) 8002800
=
=
x20)cos(8 -90)
=
4414,47
=
N
3
3. For the circuit shown, find the average power supplied by the
source and the average power absorbed by the resistor.
91 1
=
Pource
b
(12)
=
Po 1,
=
v
IR
=
PR
=
=
=
1.12256.59"A
cos(90-55.57)
2.5W
=
5W
(1.12(5.57)(4) 4.47(56.5%V
=
=
1447)
c0sO
2.5W
=
3.3 For the circuit shown, calculate the average power absorbed
by the resistor and inductor. Find the average power supplied
by the voltage source.
Vi
PL
I
I(j)
=
50.4(116.87
=
(0.4) cosi-24.
=
41 0
=
I
VR
OR
b
=
160245
-
=
50.6226.54A
·
=
37
E.(R) (50.6126.5) [7) 151.8(24.500
=
=
=
1.8)
=
cos
(24.57-24.57) 7840W
=
Pus-1.4)
dVs
=
20s
2840.38 W
(45-26.5)
4. Determine the average power generated by each source and
the average power absorbed by each passive element in the
circuit.
Iz 4UF (49.11.A
=
24)
45
-
=
V, +V2
V, v2
=
Vjio
(S
(30-99.11) 287.83N
=
0
=
+
Vji0
+
(4)(20 (I, I2) (510)
-
+
=
V,
185.0126.21°
=
·9
4,
KVL
2)(4)
=
=
-
367.85N
mesh 1:
11 4L0°A
=
RVL
+60(40
meiha:(j10-j5) I.2-f10I,
0
=
15-j4)r50
=
1
=
as
ic.21)
4. Determine the average power generated by each source and
the average power absorbed by each passive element in the
circuit.
PG P4 0
=
I,
=
L0·A
A
=
Va 12k (4)(207 80V
=
=
=
42
47
=
10W
=