Unformatted text preview: voltages around a closed path (or
loop) is zero. Equivalently,  v v the sum of the voltage drops around a loop = the sum of the voltage rises around the loop  eText Main Menu  Textbook Table of Contents  Problem Solving Workbook Contents Problem 2.7 Determine all the currents and voltages in Figure 2.1. v
i1
8A 20 Ω i2 i3 10 Ω 5Ω i4
20 Ω Figure 2.1
Carefully DEFINE the problem.
Each component is labeled completely. The problem is clear.
PRESENT everything you know about the problem.
The goal of the problem is to determine v , i1 , i 2 , i 3 , and i 4 .
Establish a set of ALTERNATIVE solutions and determine the one that promises the
greatest likelihood of success.
KCL, involving currents entering and leaving a node, is the obvious circuit analysis technique
to use, especially since the circuit has only one node. KVL could be used but would be more
complicated due to the fact that there are five loops in the circuit. Using KVL creates four
equations with four unknowns (since the fifth loop current is the current source), compared to
one equation and four unknowns using KCL.
With one equation and four unknowns, constraint equations must be found. Realizing that v
is the voltage across each component is the key. Ohm’s law will be used to determine the
relationship between the voltage and each of the four unknown currents. Using substitution,
an equation in terms of v can be found and solved.
ATTEMPT a problem solution.
KCL : 8 = i1 + i 2 + i 3 + i 4 Ohm's law : v = 20 i1 = 10 i 2 = 5 i 3 = 20 i 4
i1 = v 20
i 2 = v 10
i3 = v 5 i 4 = v 20 Substitute the current equations found using Ohm's law into the equation found using KCL.
Then, solve for v .  v v 8 = v 20 + v 10 + v 5 + v 20
160 = v + 2 v + 4 v + v
160 = 8 v
v = 20 volts  eText Main Menu  Textbook Table of Contents  Problem Solving Workbook Contents Hence, i1 = v 20 = 20 20 = 1 amp
i 3 = v 5 = 20 5 = 4 amps i 2 = v 10 = 20 10 = 2 amps
i 4 = v 20 = 20 20 = 1 amp EVALUATE the solution and check for accuracy.
Check the node equation found using KCL, 8 = i1 + i 2 + i 3 + i 4
8 = 1+ 2 + 4 +1
8...
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This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.
 Spring '12
 bkav

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