Electric Circuits, Fall 2015
Homework #8
Solution
8.1 [8%] Use the maximum power transfer theorem to determine the load impedance
𝑍𝐿 so that it will draw the maximum average power from the circuit.
Fig. 1 (For problem 8.1)
Solution:
To use the maximum power transfer theorem, we first need to find the Thévenin
impedance. Since the circuit contains a dependent source, the easiest way will be to
zero out the independent current source (thus getting rid of the left branch) and attach
a test voltage to the circuit in place of the load.
𝑉
Notice that𝐼𝑡 = 𝐼𝑥 . The impedance we want is given by 𝑍𝑇ℎ = 𝐼𝑡. One thing we can do is to
𝑡
write an equation using Ohm’s law with the two voltage sources in series:
2000𝐼 +𝑉
𝑡
𝑡
𝐼𝑡 = 6000−𝑗4000
[4]
We thus have the following relationship:
𝑍𝑇ℎ =
𝑉𝑡
𝐼𝑡
= 4000 − j4000Ω
[2]
The load impedance is just the complex conjugate:
∗
𝑍𝐿 = 𝑍𝑇ℎ
= 4000 + j4000Ω [2]
1 / 12
Electric Circuits, Fall 2015
Homework #8
Solution
8.2 [10%] Prove that if only the magnitude of the load impedance can be varied, most
average power is transferred to the load when |𝑍𝐿 | = |𝑍𝑇ℎ |. (Hint: In deriving the
expression for the average load power (page 465 in your textbook), write the load
impedance (Z𝐿 ) in the form 𝑍𝐿 = |𝑍𝐿 |𝑐𝑜𝑠 𝜃 + 𝑗|𝑍𝐿 | 𝑠𝑖𝑛 𝜃, and note that only |𝑍𝐿 | is
variable.)
Solution:
𝑍𝐿 = |𝑍𝐿 |𝑐𝑜𝑠 𝜃 + 𝑗|𝑍𝐿 | 𝑠𝑖𝑛 𝜃
Thus |𝐼| =
|𝑉𝑇ℎ |
√(𝑅𝑇ℎ +|𝑍𝐿 |𝑐𝑜𝑠 𝜃)2 +(𝑋𝑇ℎ +|𝑍𝐿 |𝑠𝑖𝑛 𝜃)2
Therefore 𝑃 = (𝑅
0.5|𝑉𝑇ℎ |2 |𝑍𝐿 |𝑐𝑜𝑠 𝜃
2
2
𝑇ℎ +|𝑍𝐿 |𝑐𝑜𝑠 𝜃) +(𝑋𝑇ℎ +|𝑍𝐿 |𝑠𝑖𝑛 𝜃)
[2]
Let 𝐷 = (𝑅𝑇ℎ + |𝑍𝐿 |𝑐𝑜𝑠 𝜃)2 + (𝑋𝑇ℎ + |𝑍𝐿 |𝑠𝑖𝑛 𝜃)2 , then
𝑑𝑃
𝑑|𝑍𝐿
=
|
(0.5|𝑉𝑇ℎ |2 𝑐𝑜𝑠 𝜃)(𝐷−|𝑍𝐿 |𝑑𝐷/𝑑|𝑍𝐿 |)
𝐷2
[2]
Where
𝑑𝐷
𝑑|𝑍𝐿 |
= 2(𝑅𝑇ℎ + |𝑍𝐿 |𝑐𝑜𝑠 𝜃) cos 𝜃 + (𝑋𝑇ℎ + |𝑍𝐿 |𝑠𝑖𝑛 𝜃) sin 𝜃
[2]
Let
𝑑𝑃
=0
𝑑|𝑍𝐿 |
Then
𝑑𝐷
𝐷 = |𝑍𝐿 | 𝑑|𝑍
𝐿|
[2]
Substituting the expressions for D and (𝑑𝐷/𝑑|𝑍𝐿| ) into this equation gives us the
2
2
relationship 𝑅𝑇ℎ
+ 𝑋𝑇ℎ
= |𝑍𝐿 |2 or |𝑍𝑇ℎ | = |𝑍𝐿 | [2]
2 / 12
Electric Circuits, Fall 2015
Homework #8
Solution
8.3 [10%] Find the average power delivered by the ideal current source in the circuit in
Fig. 2 if ig = 4𝑐𝑜𝑠 5000𝑡 mA
Fig. 2 (For problem 8.3)
Solution:
106
1
jωL = j5,000(0.5 × 10−3 ) = j2.5Ω；jωC = 𝑗5,000×1.25 = −𝑗160𝛺
𝑉𝑜
−4 + j2.5 +
𝑉
𝑉𝑜 −30( 𝑜 )
j2.5
30−j160
=0
[2]
[2]
Thus
𝑉𝑜 = 1.904 + 𝑗10.158 = −10.334∠79.38°
1
So 𝑃 = 2 𝑉𝑚 𝐼𝑚 cos(79.38°) = −3.81𝑊
[4]
[2]
8.4 [10%] The three loads in the circuit shown in Fig. 3 are
𝑆1 = (6 + 𝑗3)kVA
𝑆2 = (7.5 − 𝑗4.5)kVA
𝑆3 = (12 + 𝑗9)kVA.
a) Calculate the complex power associated with each voltage source, 𝑉𝑔1 and 𝑉𝑔2 .
b) Verify that the total real and reactive power delivered by the sources equals the total
real and reactive power absorbed by the network.
3 / 12
Electric Circuits, Fall 2015
Homework #8
Solution
Fig. 3 (For problem 8.4)
Solution:
a)
6000 − 𝑗3000
= (40 − 𝑗20)A (𝑟𝑚𝑠)
150
7500 + 𝑗4500
𝐼2 =
= (50 + 𝑗30)A (𝑟𝑚𝑠)
150
𝐼1 =
𝐼1 =
12,000−𝑗9000
300
= (40 − 𝑗30)A (𝑟𝑚𝑠) [2]
𝐼𝑔1 = 𝐼1 + 𝐼3 = (80 − 𝑗50)A (𝑟𝑚𝑠)
𝐼𝑔2 = 𝐼2 + 𝐼3 = (90 + 𝑗0)A (𝑟𝑚𝑠)
𝐼𝑛 = 𝐼1 − 𝐼2 = (−10 − 𝑗50)A (𝑟𝑚𝑠)
𝑉𝑔1 = 0.1𝐼𝑔1 + 150 + 0.2𝐼𝑛 = (156 − 𝑗15)V (𝑟𝑚𝑠)
𝑉𝑔2 = 0.1𝐼𝑔2 + 150 − 0.2𝐼𝑛 = (161 + 𝑗10)V (𝑟𝑚𝑠) [2]
Thus,
𝑆𝑔1 = −𝑉𝑔1 𝐼1∗ = (−13230 − 𝑗6600)VA
𝑆𝑔2 = −𝑉𝑔2 𝐼2∗ = (−14490 − 𝑗900)VA [2]
b)
2
𝑃0.1 = |𝐼𝑔1 | (0.1) = 890W
𝑃0.2 = |𝐼𝑛 |2 (0.2) = 520W
2
𝑃0.1 = |𝐼𝑔2 | (0.1) = 810W
[2]
∑ 𝑃𝑎𝑏𝑠 = 890 + 520 + 810 + 6000 + 7500 + 12,000 = 27,720 W
4 / 12
Electric Circuits, Fall 2015
Homework #8
Solution
∑ 𝑃𝑑𝑒𝑙 = 13,230 + 14,490 = 27,720 W = ∑ 𝑃𝑎𝑏𝑠
∑ 𝑄𝑎𝑏𝑠 = 3000 + 9000 = 12,000VAR
∑ 𝑄𝑑𝑒𝑙 = 4500 + 6600 + 900 = 12,000VAR = ∑ 𝑄𝑎𝑏𝑠
[2]
8.5 [8%] For the circuit in Fig. 4, find 𝑉𝑆 .
Fig. 4 (For problem 8.5)
Solution:
15
𝑆2 = 15 − 𝑗 0.8 sin(cos−1(0.8)) = 15 − 𝑗11.25[1]
But
𝑆2 = 𝑉2 𝐼2∗
𝑆2 15 − 𝑗11.25
𝐼2∗ = =
𝑉2
120
𝐼2 = 0.125 + 𝑗0.009375 [1]
𝑉1 = 𝑉2 + 𝐼2 (0.3 + 𝑗0.15) = 120.02 + 𝑗0.0469
And
[2]
10
𝑆1 = 10 + 𝑗 0.9 sin(cos −1(0.9)) = 10 + 𝑗4.843[1]
𝑆1 = 𝑉1 𝐼1∗
But
𝑆1
= 0.0837 + 𝑗0.0405.
𝑉1
𝐼1 = 0.0837 − 𝑗0.0405 [1]
𝐼 = 𝐼1 + 𝐼2 = 0.2087 + 𝑗0.053
𝑉𝑠 = 𝑉1 + 𝐼(0.2 + 𝑗0.04) = 120.06 + 𝑗0.0658 = 120.06∠0.03°V [2]
𝐼1∗ =
5 / 12
Electric Circuits, Fall 2015
Homework #8
Solution
8.6 [10%]
a) Find 𝐼𝑜 in the circuit in Fig. 5.
b) Find 𝑉𝐴𝑁 .
c) Find 𝑉𝐴𝐵 .
d) Is the circuit a balanced or unbalanced three-phase system? Why?
Fig. 5 (For problem 8.6)
Solution:
a)
277∠0°
= 2.77∠(−36.87°)A (𝑟𝑚𝑠)
80 + 𝑗60
277∠ − 120°
𝐼𝑏𝐵 =
= 2.77∠(−156.87°)A (𝑟𝑚𝑠)
80 + 𝑗60
277∠120°
𝐼𝑐𝐶 =
= 2.77∠(83.13°)A (𝑟𝑚𝑠)
80 + 𝑗60
𝐼𝑜 = 𝐼𝑎𝐴 + 𝐼𝑏𝐵 + 𝐼𝑐𝐶 = 0
[4]
𝐼𝑎𝐴 =
b)
𝑉𝐴𝑁 = (78 + 𝑗54)𝐼𝑎𝐴 = 262.79∠(−2.17°)V (𝑟𝑚𝑠)
[2]
c)
𝑉𝐵𝑁 = (77 + 𝑗56)𝐼𝑏𝐵 = 263.73∠(−120.84°)V (𝑟𝑚𝑠)
𝑉𝐴𝐵 = 𝑉𝐴𝑁 − 𝑉𝐵𝑁 = 452.89∠(28.55°)V (𝑟𝑚𝑠) [2]
d) Unbalanced. Because sources and loads are not respectively balanced. [2]
6 / 12
Electric Circuits, Fall 2015
Homework #8
Solution
8.7 [8%] Determine the line currents for the three-phase circuit of Fig. 6. Let 𝑉𝑎 =
110∠0°， 𝑉𝑏 = 110∠ − 120°，𝑉𝑐 = 110∠120°
Fig. 6 (For problem 8.7)
Solution:
We apply mesh analysis to the circuit shown below:
(100 + 𝑗80)𝐼1 − (20 + 𝑗30)𝐼2 = 𝑉𝑎 − 𝑉𝑏 = 165 + 𝑗95.263
−(20 + 𝑗30)𝐼1 + (80 − 𝑗10)𝐼2 = 𝑉𝑏 − 𝑉𝑐 = −𝑗190.53
Solving (1) and (2) gives
𝐼1 = 1.8616 − 𝑗0.6084 A
𝐼2 = 0.9088 − 𝑗1.722 A [2]
Thus
𝐼𝑎 = 𝐼1 = 1.9585∠(−18.1)°A
𝐼𝑏 = 𝐼2 − 𝐼1 = 1.4656∠(−130.55)°A
𝐼𝑐 = −𝐼2 = 1.947∠(117.8)°A [2]
7 / 12
(1) [2]
(2) [2]
Electric Circuits, Fall 2015
Homework #8
Solution
8.8 [10%]
a) Show that the impedance seen looking into the terminals 𝑎 − 𝑏 in the circuit in Fig.
7 is given by the expression:
𝑍𝐿
𝑍𝑎𝑏 =
𝑁 2
(1 + 𝑁1 )
2
b) Show that if the polarity terminal of either one of the coils is reversed that:
𝑍𝐿
𝑍𝑎𝑏 =
𝑁 2
(1 − 𝑁1 )
2
Fig. 7 (For problem 8.8)
Solution:
a)
𝑁1 𝐼1 = 𝑁2 𝐼2 ,
𝑁1
𝐼2 =
𝐼
𝑁2 1
𝑉𝑎𝑏
𝑍𝑎𝑏 = 𝐼
1 +𝐼2
𝑉1
𝑉2
=𝐼
𝑉2
1 +𝐼2
𝑁
=
𝑉2
[2]
𝑁
(1+ 1 )𝐼1
𝑁2
𝑁
= 𝑁1, 𝑉1 = 𝑁1 𝑉2
2
2
𝑁
𝑉1 + 𝑉2 = 𝑍𝐿 𝐼1 =(1 + 𝑁1 )𝑉2[2]
2
𝑍𝑎𝑏 =
𝑍𝐿 𝐼1
𝑁
𝑁
(1+ 1 )(1+ 1 )𝐼1
𝑁2
𝑁2
8 / 12
=
𝑍𝐿
𝑁
(1+ 1 )2
𝑁2
[2]
Electric Circuits, Fall 2015
Homework #8
Solution
Q. E. D.
b) Assume dot on the 𝑁2 coil is moved to the lower terminal. Then:
𝑁1
𝑉1 = − 𝑉2
𝑁2
𝑁
𝑉1 + 𝑉2 = 𝑍𝐿 𝐼1 =(1 − 𝑁1 )𝑉2
2
𝑍𝑎𝑏 =
𝑉2
𝑁
(1− 1 )𝐼1
=
𝑁2
𝑍𝐿
𝑁
(1− 1 )2
[4]
𝑁2
Q. E. D.
8.9 [8%] Find the Norton equivalent for the circuit in Fig. 8 at terminals 𝑎 − 𝑏.
Fig. 8 (For problem 8.9)
Solution:
The first step is to replace the mutually coupled circuits with the equivalent circuits
using dependent sources. To obtain 𝐼𝑁 , short-circuit 𝑎– 𝑏 as shown in figure below
and solve for 𝐼𝑠𝑐 .
Loop 1:
−100∠60° + 20𝐼1 + 𝑗10(𝐼1 − 𝐼2 ) + 𝑗5𝐼2 = 0 [2]
Loop 2:
−𝑗5𝐼2 + 𝑗5(𝐼1 − 𝐼2 ) + 𝑗20𝐼2 − 𝑗10(𝐼1 − 𝐼2 ) = 0
Substituting back into the first equation, we can get
𝐼2 = 𝐼𝑠𝑐 = 𝐼𝑁 = 1.1452∠6.37°A
9 / 12
[2]
Electric Circuits, Fall 2015
Homework #8
Solution
𝐼2 = 0
−100∠30° + (20 + 𝑗10)𝐼1 = 0
𝑉𝑜𝑐 = 𝑗10𝐼1 − 𝑗5𝐼1 = 22.36∠93.43°V [2]
Thus
𝑍𝑁 =
𝑉𝑜𝑐
𝐼𝑠𝑐
= 19.525∠87.06°Ω
[2]
8.10 [10%] Calculate the average power dissipated by the 20-Ω resistor in Fig.9
Fig. 9 (For problem 8.10)
Solution:
At node 1,
200−𝑉1
10
=
𝑉1 −𝑉4
40
At node 2,
10 / 12
+ 𝐼1 [1]
Electric Circuits, Fall 2015
Homework #8
𝑉1 −𝑉4
Solution
𝑉
= 204 + 𝐼3 [1]
40
At the terminals of the first transformer,
𝑉2
= −2
𝑉1
𝐼2
1
= − 2[2]
𝐼1
For the middle loop,
−𝑉2 + 50𝐼2 + 𝑉3 = 0[1]
At the terminals of the second transformer,
𝑉4
=3
𝑉3
𝐼2
𝐼1
1
= − 3[2]
We can solve for 𝑉4 from the equations above
𝑉4 = 14.87V (rms) [1]
𝑃=
𝑉4 2
20
= 11.05W[2]
8.11 [8%] If 𝑀 = 0.2H and 𝑣𝑠 = 12 cos 10𝑡 V in the circuit of Fig.10, find 𝑖1 and 𝑖2 .
Calculate the energy stored in the coupled coils at 𝑡 = 15ms.
Fig. 10 (For problem 8.11)
Solution:
The frequency-domain equivalent circuit is shown below.
11 / 12
Electric Circuits, Fall 2015
Homework #8
For mesh 1,
−12 + 𝑗5𝐼1 + 𝑗2𝐼2 − 𝑗4(𝐼1 − 𝐼2 ) = 0[1]
For mesh 2,
𝑗4(𝐼1 − 𝐼2 ) + 𝑗10𝐼2 + 𝑗2𝐼1 + 5𝐼2 = 0[1]
We can solve for 𝐼1 and 𝐼2 from the equations above
𝐼1 = 3.082∠40.73°A
𝐼2 = 2.367∠ − 99.46°A[2]
Thus,
𝑖1 (𝑡) = 3.082 cos(10𝑡 + 40.73°) A
𝑖2 (𝑡) = 2.367 cos(10𝑡 − 99.46°) A[2]
At 𝑡 = 15𝑚𝑠
𝑖1 = 2.00789A
𝑖2 = −0.03594A
1
1
w = 2 (0.5)𝑖1 2 + 2 (1)𝑖2 2 − (0.2)𝑖1 𝑖2 = 1.0230J [2]
12 / 12
Solution