Exercise:In Figure 4 the two-terminal network of Figure 3 with the parameters of Table 3 isplaced in a circuit. The source is a 2-kHz sinusoid with a 4-V amplitude. Modelthis source voltage as having a phase of zero. Rs = 2 kQ. Find the amplitude andphase of v and i Table 3: Component values for the two-terminal network of Figure 3ValueComponentL100 mH0.022 uFC5 knN-two-terminal networkLSynthesizedCFrequencyv-RGeneratorFigure 4: Testing the two-terminal network-1

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Asked Dec 2, 2019
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Exercise:
In Figure 4 the two-terminal network of Figure 3 with the parameters of Table 3 is
placed in a circuit. The source is a 2-kHz sinusoid with a 4-V amplitude. Model
this source voltage as having a phase of zero. Rs = 2 kQ. Find the amplitude and
phase of v and i
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Exercise: In Figure 4 the two-terminal network of Figure 3 with the parameters of Table 3 is placed in a circuit. The source is a 2-kHz sinusoid with a 4-V amplitude. Model this source voltage as having a phase of zero. Rs = 2 kQ. Find the amplitude and phase of v and i

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Table 3: Component values for the two-terminal network of Figure 3
Value
Component
L
100 mH
0.022 uF
C
5 kn
N-
two-terminal network
L
Synthesized
C
Frequency
v-
R
Generator
Figure 4: Testing the two-terminal network
-1
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Table 3: Component values for the two-terminal network of Figure 3 Value Component L 100 mH 0.022 uF C 5 kn N- two-terminal network L Synthesized C Frequency v- R Generator Figure 4: Testing the two-terminal network -1

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Expert Answer

Step 1

Redrawing the given circuit diagram, using the given points:

two-terminal network
2kQ
+
C
5kQ
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two-terminal network 2kQ + C 5kQ

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Step 2

Evaluating the parameters of the above diagrams using the given frequency of the 2-KHz and taking the sine wave voltage as the reference voltage.

Xj27fL
=j27(2kHz)100mH
j1256.630
j1.25kn
Xc
1
j27(2KH2) (0.022uF
=-j3617.150
j3.61k
V 420
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Xj27fL =j27(2kHz)100mH j1256.630 j1.25kn Xc 1 j27(2KH2) (0.022uF =-j3617.150 j3.61k V 420

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Step 3

Referring the given circuit diagram, the resistance and the capacitance are in parallel to each other and t...

Ze=XZ2¢
eqy
Where
RC
-j3.6x5
5-j3.6
= 2.924-90 +35.75
=2.924-54.25° k2
(1.706-j2.37)k
Hence
ZgrXZzc
Zgrj.25 1.706-j2.37
Zig = (1.706-j1.12) k
eqv
eqv
= 2.04Z-33.28° k2.
Z
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Ze=XZ2¢ eqy Where RC -j3.6x5 5-j3.6 = 2.924-90 +35.75 =2.924-54.25° k2 (1.706-j2.37)k Hence ZgrXZzc Zgrj.25 1.706-j2.37 Zig = (1.706-j1.12) k eqv eqv = 2.04Z-33.28° k2. Z

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