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6- A conductor of 0.5 mm diameter wire has a resistance of 300 2. Find the resistance of the same length of wire if its diameter were doubled. A) 302 B)150 2 C) 75 Q D) 300 N 7- Use Nodal method for the circuit in fig.4, the equation in node a & b will be 2v 0.5N а 0. 1250 3V 0.2N 0.25N P =? fig.4 1V. Va(50) – V½(10) = 8 Va(34) – V½(110) = 11 A) -V«(10) + V½(25) = -107 В) -Va(150) + Vp(25) = -107 Va(35) – V,(11) = 10 Va(34) – V,(10) = = 16 %3D C) -V«(100) + V,(2. 5) D) = -17 (-v«(10) + V,(25) = = -17 0.050

6- A conductor of 0.5 mm diameter wire has a resistance of 300 2. Find the resistance of the same length of wire if its diameter were doubled. A) 302 B)150 2 C) 75 Q D) 300 N 7- Use Nodal method for the circuit in fig.4, the equation in node a & b will be 2v 0.5N а 0. 1250 3V 0.2N 0.25N P =? fig.4 1V. Va(50) – V½(10) = 8 Va(34) – V½(110) = 11 A) -V«(10) + V½(25) = -107 В) -Va(150) + Vp(25) = -107 Va(35) – V,(11) = 10 Va(34) – V,(10) = = 16 %3D C) -V«(100) + V,(2. 5) D) = -17 (-v«(10) + V,(25) = = -17 0.050

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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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) 302 B)150 2 C) 75 Q D) 300 N 7- Use Nodal method for the circuit in fig.4, the equation in node a & b will be 2v 0.5N a 0. 1250 3V 0.2N ).25N P =? fig.4 1V, Va(50) – V„(10) = 8 Va(34) – V½(110) = 11 A) -V«(10) + V½(25) = –107 В) -Va(150) + Vp(25) = –107 Va(35) – V,(11) = 10 Va(34) – V,(10) = = 16 C) -V«(100) + V,(2. 5) = D) = -17 (-v«(10) + V,(25) = = -17 0.050
Transcribed Image Text: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) 302 B)150 2 C) 75 Q D) 300 N 7- Use Nodal method for the circuit in fig.4, the equation in node a & b will be 2v 0.5N a 0. 1250 3V 0.2N ).25N P =? fig.4 1V, Va(50) – V„(10) = 8 Va(34) – V½(110) = 11 A) -V«(10) + V½(25) = –107 В) -Va(150) + Vp(25) = –107 Va(35) – V,(11) = 10 Va(34) – V,(10) = = 16 C) -V«(100) + V,(2. 5) = D) = -17 (-v«(10) + V,(25) = = -17 0.050
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