Principles Of Electric Circuits
10th Edition
ISBN: 9780134879482
Author: Floyd, Thomas L.
Publisher: Pearson,
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Textbook Question
Chapter 2, Problem 13PQ
A certain current source provides 100 mA to a 1 kΩ load. If the resistance is decreased to 500 Ω, what is the current in the load?
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-E1 + VR1 + VR4 – E2 + VR3 = 0 -------> Loop 1 (a)
R1(I1) + R4(I1 – I2) + R3(I1) = E1 + E2 ------> Loop 1 (b)
R1(I1) + R4(I1) - R4(I2) + R3(I1) = E1 + E2 ------> Loop 1 (c)
(R1 + R3 + R4) (I1) - R4(I2) = E1 + E2 ------> Loop 1 (d)
Now that we have loop 1 equation will procced on finding the equation of I2 current loop. However, a reminder that because we are going in a clockwise direction, it goes against the direction of the current. As such we will get an equation for the matrix that will be:
E2 – VR4 – VR2 + E3 = 0 ------> Loop 2 (a)
-R4(I2 – I1) -R2(I2) = -E2 – E3 ------> Loop 2 (b)
-R4(I2) + R4(I1) - R2(I2) = -E2 – E3 -----> Loop 2 (c)
R4(I1) – (R4 + R2)(I2) = -E2 – E3 -----> Loop 2 (d)
These two equations will be implemented to the matrix formula I = inv(A) * b
R11 R12
(R1 + R3 + R4)
-R4
-R4
R4 + R2
Chapter 2 Solutions
Principles Of Electric Circuits
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