Homework 7-9 Solution
1250
EX:
Use either the node-voltage method or superposition to find the phasor voltage V1.
SOL'N: i) Node-voltage method solution:
We have one node equation. We sum up the three currents, iA1, iA2, and
iA3, flowing out of the top node, written in terms of phasors and
impedances.
V − 6e− j 45° V VA − 12e j 45° V
−3e−90° mA + A
+
= 0A
j5 kΩ
− j10 kΩ
We do the algebra, as always, by grouping the terms that multiply the
node voltage and putting constant terms on the right side.
⎛ 1
⎞
1
6e− j 45° V 12e j 45° V
−90°
VA ⎜
+
= 3e
mA +
+
j5 kΩ
− j10 kΩ
⎝ j5 kΩ − j10 kΩ ⎟⎠
or
⎛ 1
⎞
−2
1
6e− j 45° mA 12e j 45° mA
−90°
VA ⎜
⋅ +
= 3e
mA +
+
j5 Ω
− j10 Ω
⎝ j5 kΩ −2 − j10 kΩ ⎟⎠
or
⎛ −2
⎞
1
6e− j 45° mA − j 12e j 45° mA j
−90°
VA ⎜
+
= 3e
mA +
⋅
+
⋅
j5 Ω
−j
− j10 Ω
j
⎝ − j10 kΩ − j10 kΩ ⎟⎠
or
⎛
⎞
1
− j6e− j 45° mA j12e j 45° mA
−90°
VA ⎜ −
= 3e
mA +
+
5Ω
10 Ω
⎝ − j10 kΩ ⎟⎠
We convert to rectangular form for the addition.
e− j 45° =
2
2
−j
2
2
2
2
+j
and e j 45° =
and e− j90° = − j
2
2
We continue:
⎛ 1 ⎞
− j6( 2 − j 2 ) j12( 2 + j 2 )
VA ⎜
= − j3 +
+
mA
⎟
(2)5 Ω
(2)10 Ω
⎝ j10 kΩ ⎠
or
⎡
− j6( 2 − j 2 ) j12( 2 + j 2 ) ⎤
VA = ⎢ − j3 +
+
⎥ j10 kΩ ⋅ mA
(2)5 Ω
(2)10 Ω
⎣
⎦
or
VA = ⎡⎣ 30 + 6( 2 − j 2 ) − 6( 2 + j 2 ) ⎤⎦ V
or
VA = ⎡⎣ 30 − j12 2 ⎤⎦ V
Now we calculate V1 from VA.
⎡
⎛ 2 + j 2⎞⎤
V1 = VA − 12e j 45° V = ⎢ 30 − j12 2 − 12 ⋅ ⎜
⎟⎠ ⎥ V
2
⎝
⎢⎣
⎥⎦
or
V1 = VA − 12e j 45° V = ⎡⎣ 30 − 6 2 − j18 2 ⎤⎦ V
or
V1 ! 21.5 − j24.5V
or
V1 =
( 30 − 6 2 ) + (18 2 ) e
2
2
⎛ −18 2 ⎞
tan −1⎜
⎝ 30−6 2 ⎟⎠
V ! 33.3e − j 49.8° V
ii) Superposition solution:
case I: Only the current source on the left is turned on.
Voltage V11 is the same as the voltage across the entire
equivalent impedance of j5 kΩΩ in parallel with -j10kΩΩ.
V11 = IzEq = 3e− j90° mA ⋅ j5 kΩ || − j10 kΩ
or
V11 = 3e− j90° mA ⋅ 5 kΩ ⋅ j || − j2 = 15e− j90° ⋅
j(− j2)
V
j − j2
or
V11 = 15e− j90° ⋅
2
V = 30V
−j
Note that e–j90° = –j.
case II: Only the middle voltage source is turned on.
This is a voltage divider.
V12 = 6e− j 45° V
− j10 kΩ
−2
= 6e− j 45° V
j5 kΩ − j10 kΩ
1− 2
or
V12 = 6e− j 45° V(2) = 12e− j 45° V
or
V12 = 12
2(1− j)
V = 6 2(1− j)V
2
case III: Only the right voltage source is turned on.
This is also a voltage divider. We have to add a minus sign in
the voltage-divider formula, however, since the polarity of the
voltage measurement is inverted compared to the standard
voltage divider.
− j10 kΩ
−2
V13 = −12e j 45° V
= −12e j 45° V
j5 kΩ − j10 kΩ
1− 2
or
V13 = −12e j 45° V(2) = −24e j 45° V
or
V13 = −24
2(1+ j)
V = −12 2(1+ j)V
2
Now we sum the results
V1 = 30V + 6 2(1− j)V − 12 2(1+ j)V
or
V1 = ⎡⎣ 30 − 6 2 − j18 2 ⎤⎦ V
This matches the result from the node-voltage method, which it must.