7- Use Nodal method for the circuit in fig.4, the equation in node a & b will be 2vl 0.5N a 0. 125N 0.2N .25N P =? fig.4 1V, Va(50) – V,(10) = Va(34) – V½(110) = 11 A) -V«(10) + V½(25) = -107 В) -Va(150) + V½(25) = -107 V«(35) – V,(11) = 10 V«(34) – V,(10) = 16 %3D %3D C) -V«(100) + V,(2.5) = -17 D) (-V«(10) + V,(25) = -17 0.050 aker & na Abed alalan 6- A conductor of 0.5 mm diameter wire has a resistance of 300 Q. Find the resistance of the same length of wire if its diameter were doubled. A) 30Q B)150 2 C) 75 N D) 300 Q Use Nodal method for the circuit in fig.4, the equation in node a & b will be
7- Use Nodal method for the circuit in fig.4, the equation in node a & b will be 2vl 0.5N a 0. 125N 0.2N .25N P =? fig.4 1V, Va(50) – V,(10) = Va(34) – V½(110) = 11 A) -V«(10) + V½(25) = -107 В) -Va(150) + V½(25) = -107 V«(35) – V,(11) = 10 V«(34) – V,(10) = 16 %3D %3D C) -V«(100) + V,(2.5) = -17 D) (-V«(10) + V,(25) = -17 0.050 aker & na Abed alalan 6- A conductor of 0.5 mm diameter wire has a resistance of 300 Q. Find the resistance of the same length of wire if its diameter were doubled. A) 30Q B)150 2 C) 75 N D) 300 Q Use Nodal method for the circuit in fig.4, the equation in node a & b will be
Introductory Circuit Analysis (13th Edition)
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Transcribed Image Text:7- Use Nodal method for the circuit in fig.4, the equation in node a & b will be
2vl
0.5N
a
0. 125N
0.2N
.25N
P =?
fig.4
1V,
Va(50) – V,(10) =
Va(34) – V½(110) = 11
A)
-V«(10) + V½(25) = -107
В)
-Va(150) + V½(25) = -107
V«(35) – V,(11) = 10
V«(34) – V,(10) = 16
%3D
%3D
C)
-V«(100) + V,(2.5) = -17
D)
(-V«(10) + V,(25) = -17
0.050
Transcribed Image Text:aker &
na Abed alalan
6- A conductor of 0.5 mm diameter wire has a resistance of 300 Q. Find the resistance of
the same length of wire if its diameter were doubled.
A) 30Q
B)150 2
C) 75 N
D) 300 Q
Use Nodal method for the circuit in fig.4, the equation in node a & b will be
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