V2 (3) + Vs (2j) = V2 + 2V3 = - 1). Step-2: By applsing Analy sis at Nodal V3 : V3 - V2 V3 V3 - 60C30° -4 jio - ju V, Jio -4 - 60L30" 8 32+ 60C3 %3D 10 Vs (-4i+ 103 + 5) + V2 (u) 32 + S1.96 +. V3 ( 5+63) + Vz (ui) = s[ 83.96 + j30). V3 (s+6] + vz (ui) = 419.8+ jiso + V2 (4L90")= 4us.79 (19.66 ° B solviva e9.. (2) 4 (3): get we - 87.895 -j 279.862 293 34 L25?.56°V - 8.01 + j I09.931 Vマ: 110. 22 /94r167° v volkage values are v, = 60L30° u; 293. 34 (252.56°v V3 = 110.22 94.167 v Hence solved

Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Question
Write a routine in Python to solve the circuits of the images, of up to 5 nodes. (The exercise is already solved)
V2 (j) + Vs (z) =
120 L-60° a).
V2 +
2 V3
12o (-15O
(1).
Step-2:-
By applsing
Analy sis at
Nodal
V3:
V3- Vz
V3
V3- 60C30"
-4
jio
- ju
J.
jio
-4 -
60C30°
jio
8
32+ 60C30"
10
V3 (-4j+ 10j + 5) + Ve (ui)
32+ 51.96+ j30
V3 ( 5+63) + Vz (uj) = s[ 83.96+ j30).
V3 (s+6] + Uz (uj) =
419.8 + ji5o
Vs | 7.81 (50-196) + v2 (4L90")= 4u5.79 (19.66° -(2) ·
By solviug
e9.. (2) 4 (3);
get
-87.895- j 279.862
293· 34 L 252.56°V
- 8.01 + ji09.931
V3 :
110. 22 94•167° v
voltage
values
are
v, = 60L30° u;
V2 = 293. 34 (252.56°v
Vg =
110.22/94167°v
Hence solved
Transcribed Image Text:V2 (j) + Vs (z) = 120 L-60° a). V2 + 2 V3 12o (-15O (1). Step-2:- By applsing Analy sis at Nodal V3: V3- Vz V3 V3- 60C30" -4 jio - ju J. jio -4 - 60C30° jio 8 32+ 60C30" 10 V3 (-4j+ 10j + 5) + Ve (ui) 32+ 51.96+ j30 V3 ( 5+63) + Vz (uj) = s[ 83.96+ j30). V3 (s+6] + Uz (uj) = 419.8 + ji5o Vs | 7.81 (50-196) + v2 (4L90")= 4u5.79 (19.66° -(2) · By solviug e9.. (2) 4 (3); get -87.895- j 279.862 293· 34 L 252.56°V - 8.01 + ji09.931 V3 : 110. 22 94•167° v voltage values are v, = 60L30° u; V2 = 293. 34 (252.56°v Vg = 110.22/94167°v Hence solved
step-1:
Given
Ne kwork-
jior
jion.
60 (30°J
- ju r
-jue
(T) UL0°A
fsom khe
circuit,
60 (30°
By applying
Analysis
Node-ve
->
Nodal
at
V2 - 60 (30°
V2 - V3
Jio
- ju
jio
V2
60 L30°
+
jio
10
jV3
2
- j6 L30°
%3D
jio
10
Vz
jv3
十
6[30-90°.
10
4
Vz (-4j + 5i)
+ 2j v3
6 60°
%3D
20
Transcribed Image Text:step-1: Given Ne kwork- jior jion. 60 (30°J - ju r -jue (T) UL0°A fsom khe circuit, 60 (30° By applying Analysis Node-ve -> Nodal at V2 - 60 (30° V2 - V3 Jio - ju jio V2 60 L30° + jio 10 jV3 2 - j6 L30° %3D jio 10 Vz jv3 十 6[30-90°. 10 4 Vz (-4j + 5i) + 2j v3 6 60° %3D 20
Expert Solution
Step 1 Function to generate the list of all edges:
def generate_edges(graph):
    edges = []
    for node in graph:
        for neighbour in graph[node]:
            edges.append({node, neighbour})

    return edges

print(generate_edges(graph))
 
[{'c', 'a'}, {'c', 'b'}, {'b', 'e'}, {'c', 'd'}, {'c', 'b'}, {'c', 'e'}, {'c', 'a'}, {'c', 'd'}, {'c', 'e'}, {'b', 'e'}]
 

As we can see, there is no edge containing the node "f". "f" is an isolated node of our graph. The following Python function calculates the isolated nodes of a given graph:

 
def find_isolated_nodes(graph):
    """ returns a set of isolated nodes. """
    isolated = set()
    for node in graph:
        if not graph[node]:
            isolated.add(node)
    return isolated
 

If we call this function with our graph, a list containing "f" will be returned: ["f"]

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