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V 
=
V
oc
– 
ZI
=
V
ocm
-
ZI
m
∠
(
f
+ 
q
)
Then, we may write the following time-domain expressions under sinusoidal steady-state condition.
v
t
V
t
i t
I
t
v t
V
t ZI
t
OC
ocm
ocm
( )
cos
( )
cos(
)
( )
cos
cos(
=
=
+
=
−
+
w
w
f
w
w
f
m
m
++
q
)
The average power delivered to the load circuit is the cycle average of v(t)
×
i(t). 
v t i t
V I
t
t
ZI
t
t
ocm
( ) ( )
cos
cos(
)
cos(
) cos(
)
=
+ −
+
+ +
m
m
w
w
f
w
f
w
f q
2

---

Circuit Theorems in Sinusoidal Steady-State Analysis 
7.33
Hence the average power delivered to the load 
P
V I
ZI
ocm
L
m
m
=
−
0 5
0 5
2
.
cos
.
cos
f
q
P
L
. 
We want to maximize this. But P
L
is a function of I
m
and 
f
since both of them will be influenced by 
the load circuit. Therefore, we set the partial derivatives of P
L
with respect to I
m
and 
f
to zero.
∂
∂
= ⇒
−
=
∂
∂
= ⇒
=
P
I
V
ZI
P
V
ocm
ocm
L
m
m
L
0
0 5
0
0
0 5
0
.
cos
cos
.
sin
f
q
f
f
Second equation leads to 
f
=
0 and with this value of 
f
, the first equation leads to I
V
Z
ocm
m
=
2 cos
.
q
But Z cos
q
=
R
S
. Therefore, I
m
=
V
ocm
/2R
S
.
The condition 
f
=
0 implies that the circuit has to be resistive for maximum power transfer to take 
place. Therefore, the effective reactance of the load at its terminals must be –X
S
such that it will cancel 
the Thevenin’s reactance of the power delivery circuit and make the entire circuit resistive. Moreover, 
the condition that I
m
=
V
ocm
/2R
S
implies that the effective resistance at the load terminals must be R
S
such that the entire circuit becomes a resistor of 2R
S 
in series with 
V
oc
. We now state the Maximum 
power transfer theorem for circuits in sinusoidal steady-state.
Maximum average power is transferred to a load circuit from a power delivery circuit 
under sinusoidal steady-state when the driving-point impedance 
Z
L
=
R
L
+ 
jX
L
of the 
load is the 
conjugate
of Thevenin’s impedance 
Z
S
=
R
S
+ 
jX
S
of the power delivery circuit. 
Therefore 
R
L
=
R
S
and 
X
L
=
-
X
S
are the required conditions. The maximum average power 
transferred under this condition will be 
P
V
R
V
R
ocm
rms
L
S
S
W
max
(
)
(
)
.
=
=
2
2
4
2
example: 7.7-1
(i) Find the Thevenin’s equivalent across 
a
-
b
in the 
circuit in Fig. 7.7-2. (ii) Find the voltage across a 
500
Ω
connected across 
a
-
b. 
(iii)
 
Find the value of 
capacitive reactance to be connected across 
a
-
b
if the 
magnitude of voltage across the 500
Ω
load is to be 
raised to 132kV rms. (iv) If 
a
-
b
gets shorted what is 
the current that flows through the short?
Solution
(i) We use superposition principle to obtain the 
V
oc
phasor.
V
j
j
j
j
oc
=
∠ ° ×
+
+
+
∠ ° ×
+
+
=
∠
°
127 6
1 5
9
3 5
21
127 5
2
12
3 5
21
127 5 43
.
.
.
.
kV rrms
Z
j
j
j
j
j
j
th
= +
+
=
+
+
+
=
+
(
) / /( .
)
(
)( .
)
.
.
.
2
12
1 5
9
2
12 1 5
9
3 5
21
0 857
5 1143 
Ω
.
+
–
+
–
2 + 
j
12 
Ω
1.5 + 
j
9 
Ω
127
∠
6° 
kV rms
127
∠
5° 
kV rms
a
b
Fig. 7.7-2 
Circuit for Example 7.7-1

---

7.34
The Sinusoidal Steady-State Response
The Thevenin’s equivalent circuit for sinusoidal steady-state is shown as the circuit in Fig. 7.7-3 (a)
(a)
+
a
b
–
0.857 + 
j
5.143 
Ω
127
∠
5.43° kV rms
(b)
+
a
b
–
0.857 + 
j
5.143 
Ω
500 
Ω
127
∠
5.43° kV rms
Fig. 7.7-3 
(a) Thevenin’s equivalent circuit for the circuit in Fig. 7.7-2. (b) With 500 
w
load
(ii) The voltage phasor across the load of 500 
Ω
=
∠
° ×
+
=
∠
°
127 5 43
500
500 857
5 143
126 77 4 84
.
.
.
.
.
j
kV rms
(iii) We first determine a new Thevenin’s equivalent of the circuit in Fig. 7.7-3 (b) for further load 
connection across 
a
-
b
. This equivalent circuit will have an open-circuit voltage phasor 
=
126.77
∠
4.84
°
kV rms and Thevenin’s equivalent impedance 
=
(0.857
+ 
j5.143)//500 
=
0.908
+ 
j5.125 
Ω
. Let –jX
C
be the capacitive reactance connected now across the output of this new 
equivalent circuit. The magnitude of voltage phasor across the capacitor has to be 132 kV rms.
∴
−
+
−
=
=
+
jX
j
X
i e
X
C
C
C
0 908
5 125
132
126 77
1 0413
0 908
5 1
2
2
.
( .
)
.
.
. .,
.
( . 225
1 0413
1 0843
2
2
−
=
=
X
C
)
.
.
There are two solutions for X
C
. They are 2.7 
Ω
and 129.1 
Ω
. The higher value is accepted. 
(Why?).
Therefore, the required reactance is –129.1
Ω
and the required capacitive reactance 
is 129.1
Ω
. Once we qualify a reactance by using capacitive we do not have to include the 
negative sign.
(iv) The short-circuit current at 
a
-
b
is 
=
126.77
∠
4.84
°
÷
(0.908 
+ 
j 5.125) 
=
126.77
∠
4.84
°÷
5.205
∠-
80
°
=
6.26 
-
j23.54 kA rms 
=
24.36
∠-
75.16
°
kA rms.
example: 7.7-2
Two terminals 
a
and 
b 
are identified as output terminals of a linear circuit containing sinusoidal 
sources at a common frequency. The open-circuit voltage measured across 
a
-
b
is seen to be 100 V 
rms. The voltage phasor across 
a
-
b
goes down to 63.25 V rms when a 20 
Ω
resistor is connected at the 
output and it goes down to 44.71 V rms when a 10 
Ω
resistor is connected at the output. (i) Find the 
resistive load that will draw maximum average power from this circuit if a reactance can be introduced 
in series with the resistive load and can be varied to maximize the power. (ii) Find the resistive 
load that will draw maximum average power for this circuit if no reactance can be added in series
with it.

---

Circuit Theorems in Sinusoidal Steady-State Analysis 
7.35
Solution
We have to find the Thevenin’s equivalent of the circuit with respect to 
a
-
b
first. Let the Thevenin’s 
equivalent impedance be R 
+ 
jX 
Ω
.
Then the magnitude of voltage phasor across 
a
-
b
when a resistor R
L
is connected across the 
terminals 
=
R
R R
jX
R
R R
X
L
L
L
L
+
+
=
+
+
V
V
OC
OC
(
)
2
2
|
V
oc
| is given as 100 V rms. Also known are the rms output voltage values when R
L
=
10 
Ω
and 
R
L
=
20 
Ω
. Substituting the numerical values, we get the following two equations in two unknowns, 
R and X.
20
20
63 25
100
0 6325
10
10
44 71
100
0 44
2
2
2
2
(
)
.
.
(
)
.
.
+
+
=
=
+
+
=
=
R
X
R
X
and
771
Squaring both sides of equations and carrying out the required algebraic manipulation, we get, 
(
)
.
(
)
.
20
999 86
10
500 25
2
2
2
2
+
+
=
+
+
=
R
X
R
X
and 
. Subtracting the second equation from the first 
yields
(
)
(
)
. , . ., (
)
.
.
20
10
499 61
30 2
10 499 61
9 98
2
2
2
+
−
+
+
=
+
×
=
⇒ =
≈
R
R
X
i e
R
R
Ω
110
Ω
. Using the 
value of R 
=
10 
Ω
in 
(
)
.
10
500 25
2
2
+
+
=
R
X
we get X 
=
10.01 
Ω
≈
10 
Ω
.
Thus the Thevenin’s equivalent of the circuit is 100
∠
0
°
 V rms in series with (10 
+ 
j10) 
Ω
.
(i) If a reactance can be put in series with the load resistance and the value of reactance and 
resistance can be independently adjusted, then, maximum power transfer will take place under 
conjugate impedance matching condition. Therefore, R
L
must be 10 
Ω
and X
L
must be –j10 
Ω
for maximum average power transfer to load. The maximum power transferred under this 
condition will be 
=
R(V
oc
/2R)
2
=
V
oc
2
/4R W where V
oc
is the rms open-circuit voltage. Therefore, 
the power transferred to (10
-
j10) 
Ω
is 250 W.
(ii) The condition for maximum power transfer has to be derived for this case. Let R
L
be the load 
resistance. Then the current phasor 
=
V
R R
jX
OC
L
(
)
+
+
and rms value (i.e., magnitude of the 
phasor, assuming V
oc
is the rms open-circuit voltage) of current is 
=
V
R R
X
OC
(
)
.
+
+
L
2
2
Power 
in R
L
=
R
L 
×
I
rms
2
=
R V
R R
X
OC
L
L
2
2
2
(
)
+
+
. Value of R
L
for maximizing this quantity is found by 
equating its derivative with respect to R
L
to zero.
dP
dR
R R
X
R R R
R
R
X
L
L
L
L
L
L
= ⇒
+
+
−
+
= ⇒
=
+
0
2
0
2
2
2
2
(
)
(
)
.
Thus, maximum power transfer takes place in a pure resistive load when load resistance 
is equal to magnitude of Thevenin’s equivalent impedance of the power delivery circuit. The 
required load resistance in this example is 
10
10
14 14
2
2
+
=
.
Ω
and the power transferred is 
=
100
10 14 14
10
14 14 207
2
2
2
(
. )
.
.
+
+
×
=
W

---

7.36
The Sinusoidal Steady-State Response