VASSESSMENT PROBLEM Objective 3-Know how to use circuit analysis techniques to solve a circuit in the frequency domain Using the values of resistance and inductance in the circuit of Fig. 9.15, let V, = 125/-60° V and w = 5000 rad/s. Find 9.6 b) the magnitude of the steady-state outp at current i. %3! a) the value of capacitance that yields a steady-state output current i with a phase angle of -105". Answer: (a) 2.86 µF: (b) 0.982 A. NOTE: Also try Chapter Problem 9.24.

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Solve the Assesment problem
Zc = i
= -i
(5000)(5)
j40 2.
Figure 9.16 A The frequency-domain equivalent circuit of the
circuit shown in Fig. 9.15.
VASSESSMENT PROBLEM
Objective 3-Know how to use circuit analysis techniques to solve a circuit in the frequency domain
b) the magnitude of the steady-state outp ut
Using the values of resistance and inductance in
the circuit of Fig. 9.15, let V, = 125/-60° V
and w = 5000 rad/s. Find
9.6
current i.
a) the value of capacitance that yields a
steady-state output current i with a phase
angle of -105".
Answer: (a) 2.86 uF:
(b) 0.982 A.
NOTE: Also try Chapter Problem 9.24.
Transcribed Image Text:Zc = i = -i (5000)(5) j40 2. Figure 9.16 A The frequency-domain equivalent circuit of the circuit shown in Fig. 9.15. VASSESSMENT PROBLEM Objective 3-Know how to use circuit analysis techniques to solve a circuit in the frequency domain b) the magnitude of the steady-state outp ut Using the values of resistance and inductance in the circuit of Fig. 9.15, let V, = 125/-60° V and w = 5000 rad/s. Find 9.6 current i. a) the value of capacitance that yields a steady-state output current i with a phase angle of -105". Answer: (a) 2.86 uF: (b) 0.982 A. NOTE: Also try Chapter Problem 9.24.
V.b Series, Parallel, and Delta-to-Wye Simplirk atiors
323
Example 9.6
Combining Impedances in Series
A 90 1 resistor, a 32 mH inductor, and a SaF
capacitor are connected in series across the termi-
The phasor transforin of , is
Fig. 9.15. The steady-state expTession for the seurce
I ANOS se asnos adeyor epiosnuns e jo speu
Figure 9.16 illustrates the frequency-domain
cquivalent circut of the circuit shown in Fig 915.
b) We cumpute the plusor current simply by divid-
ing the voltage of the voltage source by the equiv-
alent impedance between the terminals a,b. From
Eq 9.45,
A GOE + 0OUS) sea psL S 'a aieyon
circuit.
uapAnbo ureurop-kouanbaij ay 1anasuo) (e
b) Calculate the steady-state current i by the plhasor
method.
0-f - U1 + cx = "7
Thus
- $/-23, 13 A.
Figure 9.15 A The circui: for Exam ple 9.5.
150 /53.13
We may now write the steady-state expression
for directly:
Solution
=Scos (50 - 23.13) A.
w = S0XX0 rad/s. Therefore the impedance of the
32 mH inductor is
and the impedance of the capacitor is
Figure 9.16 A TIhe frequenzy demain equivalent circuit ef the
CIcui: shewn in fig 9.15.
" = z
VASSESSMENT PROBLEM
Objective 3-Know how to use circuit analysis techniques to solve a circuit in the frequency domain
Using the values of resistance and inductance in
the circuit of Fig. 9.15, let V, 125/-60" V
and ow 5000 rad/s. Find
9.6
b) the magnitude of the steady-state outpat
carrent i.
a) the value of capacitance that yields a
steady-state oulpat curreat i with a phase
angle of -105.
98z (e) uaMsuy
V Z560 (4)
NOTE Also try Chopter Problem 9.24.
Transcribed Image Text:V.b Series, Parallel, and Delta-to-Wye Simplirk atiors 323 Example 9.6 Combining Impedances in Series A 90 1 resistor, a 32 mH inductor, and a SaF capacitor are connected in series across the termi- The phasor transforin of , is Fig. 9.15. The steady-state expTession for the seurce I ANOS se asnos adeyor epiosnuns e jo speu Figure 9.16 illustrates the frequency-domain cquivalent circut of the circuit shown in Fig 915. b) We cumpute the plusor current simply by divid- ing the voltage of the voltage source by the equiv- alent impedance between the terminals a,b. From Eq 9.45, A GOE + 0OUS) sea psL S 'a aieyon circuit. uapAnbo ureurop-kouanbaij ay 1anasuo) (e b) Calculate the steady-state current i by the plhasor method. 0-f - U1 + cx = "7 Thus - $/-23, 13 A. Figure 9.15 A The circui: for Exam ple 9.5. 150 /53.13 We may now write the steady-state expression for directly: Solution =Scos (50 - 23.13) A. w = S0XX0 rad/s. Therefore the impedance of the 32 mH inductor is and the impedance of the capacitor is Figure 9.16 A TIhe frequenzy demain equivalent circuit ef the CIcui: shewn in fig 9.15. " = z VASSESSMENT PROBLEM Objective 3-Know how to use circuit analysis techniques to solve a circuit in the frequency domain Using the values of resistance and inductance in the circuit of Fig. 9.15, let V, 125/-60" V and ow 5000 rad/s. Find 9.6 b) the magnitude of the steady-state outpat carrent i. a) the value of capacitance that yields a steady-state oulpat curreat i with a phase angle of -105. 98z (e) uaMsuy V Z560 (4) NOTE Also try Chopter Problem 9.24.
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