Download Problem 7.70 Find Z in the circuit of Fig. P7.70, given that V s = 40 V

Problem 7.70 Find Z in the circuit of Fig. P7.70, given that Vs = 40 V and
◦
Va = 17.22e− j132.2 V.
j4 Ω
Vs
3Ω
+
_
+
_ Va
−j5 Ω
Z
Figure P7.70 Circuit for Problem 7.70.
Solution: In the circuit of Fig. P7.70(a),
Va = Vb
!
− j5
3 − j5
"
.
Hence,
Vb = Va
!
3 − j5
− j5
"
− j132.2◦
= 17.22e
!
3 − j5
− j5
"
= −(3.92 + j19.70) V.
j4 Ω
Vs
+
_
Va
−j5 Ω
Z
j4 Ω
Vs
3Ω
Vb
+
_ Va
Vb
+
_
Z1 = Z || (3 −j5)
Fig. P7.70(a)
By voltage division
Vb =
Vs Z1
j4 + Z1
−(3.92 + j19.70)
Z1
=
,
40
j4 + Z1
Z1 (−3.92 + j19.70 − 40) = − j4(−3.92 + j19.70),
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which leads to
Z1 = (1.36 − j0.97) Ω.
But
Z1 = Z ∥ (3 − j5)
=
Z(3 − j5)
,
Z + 3 − j5
which leads to
(Z + 3 − j5)(1.36 − j0.97) = Z(3 − j5).
Collecting terms in Z, we have
Z(1.36 − j0.97 − 3 + j5) = −(3 − j5)(1.36 − j0.97),
which leads to
Z = (2 − j1) Ω.
All rights reserved. Do not reproduce or distribute. © 2016 National Technology and Science Press