How to solve problems 3 and 4 (the second option in the table)? Table 2Var# V,v R,e R2,24,MHL2, nH Ci, juF|C2 ,AF|640 52052.20102.2.2015260160037.330 20Joo30043.イo|301808001216411705301210595 15800236727イイ子35371453002727イ2757201o0095133240イ5020010401435150650イイ40156.1567イOo400イ2401616036265イ301340イ710486480105146018イO11560320go156019イイ251001303601660201イイ0554210400イ7602112806865450イ20|イイO23イイのイ1018221260205801910134012566265202413854525070142514イ032260|イ80イイ0| 26220 27224254382252315175 96230752422028151258040165252202916190 65100200|の Problem 3For the given variant of alterna tingCuzrent circuit (Table 2):1) To deter mine the current in the citcuitand voltage across each of the elements.2) To built current and vobtage voetordiagram.3)Ta determine the ralue of the regctanceat which voltage resonance, arise and the valueof resonance currens through the circuit.C2Problem 4For, the given variaut of altezna tingcurrent cirEuit (Table 2):1) To determine the currents in all brenchesby gymbol method.2) To built cur rent and voltage veetordiágram.3) To determine the value of ore ofreactance at which current resonancewillarise anod the 'value of resohance currentin un branched part of the circuit34

Answered: For the given variant of alternating…

Introductory Circuit Analysis (13th Edition)

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ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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Question

How to solve problems 3 and 4 (the second option in the table)?

Table 2
Var# V,v R,e R2,24,MHL2, nH Ci, juF|C2 ,AF|
640 520
5
2.
20
10
2.
2.
20
15
260
1600
3
7.
3
30 20
Joo
300
4
3.
イo|30
180
800
12
16
41
170
530
12
10
5
95 15
800
236
7
27
イイ
子
35
37
145
300
27
27
イ2
7
57
20
1o00
95
13
32
40
イ50
200
10
40
14
35
150
650
イイ
40
15
6.
15
67
イOo
400
イ2
40
16
160
36
265
イ30
13
40
イ7
10
48
64
80
105
14
60
18
イO
115
60
320
go
15
60
19
イイ
25
100
130
360
16
60
20
1イ
イ05
54
210
400
イ7
60
21
12
80
68
65
450
イ20
|イイO
23
イイの
イ10
18
22
12
60
205
80
19
10
13
40
125
66
265
20
24
13
85
45
250
70
14
25
14
イ0
32
260|イ80
イイ0| 26
220 27
22
42
54
38
225
23
15
175 96
230
75
24
220
28
15
12580
40
165
25
220
29
16
190 65
100
200
|の

Problem 3
For the given variant of alterna ting
Cuzrent circuit (Table 2):
1) To deter mine the current in the citcuit
and voltage across each of the elements.
2) To built current and vobtage voetor
diagram.
3)Ta determine the ralue of the regctance
at which voltage resonance, arise and the value
of resonance currens through the circuit.
C2
Problem 4
For, the given variaut of altezna ting
current cirEuit (Table 2):
1) To determine the currents in all brenches
by gymbol method.
2) To built cur rent and voltage veetor
diágram.
3) To determine the value of ore of
reactance at which current resonancewill
arise anod the 'value of resohance current
in un branched part of the circuit
34

Expert Solution

Step 1

We are authorized to answer one question at a time, since you have not mentioned which question you are looking for, so we are answering the first one(problem no.3). Please repost your question separately for the remaining question.

Here supply voltage frequency is not given so we will assume supply frequency as 'f' and obtain the answers in terms of s

' f '

Step 2

Problem 3:

From the second option of the table

$R_{1} = 6 Ω R_{2} = 2 Ω L_{1} = 20 m H L_{2} = 15 m H C_{1} = 260 μ F C_{2} = 1600 μ F$

V=6 V

Step 3

1) Given circuit is RLC series circuit and hence current will be remain same in all the elements.

$X_{L} = w L = 2 π f L = 2 π f (L_{1} + L_{2}) X_{L} = 2 π f (20 \times 10^{- 3} + 15 \times 10^{- 3}) X_{L} = 0.22 f X_{c} = \frac{1}{2 π f C} = \frac{1}{2 π f (C_{1} + C_{2})} X_{c} = \frac{1}{2 π f (260 μ F + 1600 μ F)} X_{c} = \frac{85.57}{f}$

Step 4

1)Current in the circuit

$I = \frac{V}{Z} = \frac{6}{R + j (X_{L} - X_{c})} R = R_{1} + R_{2} = 6 + 2 = 8 Ω I = \frac{6}{8 + j (0.22 f - \frac{85.57}{f})}$

voltage across each element

$V_{L 1} = I X_{L 1} = I (2 π f L_{1}) = I (2 π f \times 20 \times 10^{- 3}) = \frac{6}{8 + j (0.22 f - \frac{85.57}{f})} \times (0.1256 f) V_{L 1} = \frac{0.75 f}{8 + j (0.22 f - \frac{85.57}{f})} V_{L 2} = I X_{L 2} = I (2 π f L_{2}) = I (2 π f \times 15 \times 10^{- 3}) = \frac{6}{8 + j (0.22 f - \frac{85.57}{f})} \times (0.0942 f) V_{L 2} = \frac{0.565 f}{8 + j (0.22 f - \frac{85.57}{f})}$

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