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Unformatted text preview: AP 4.15 [‘a] Redraw the circuit with a helpful voltage and current labeled: Transform the 120 V source in series with the 20 Q resistor into a 6 A source in parallel with the 20 Q resistor. Also transform the #60 V source in series with
the 5 Q resistor into a —12 A source in parallel with the 5 Q resistor. The result is the following circuit: Combine the three current sources into a single current source. using KCL, and
combine the 20 Q, 5 Q. and 6 Q resistors in parallel. The resulting circuit is
shown on the left. To simplify the circuit further. transform the resulting 30 A
source in parallel with the 2.4 Q resistor into a 72 V source in series with the
2.4 Q resistor. Combine the 2.4 Q resistor in series with the 1.6 Q resisor to get
a very simple circuit that still maintains the voltage 11:. The resulting circuit is ' on the right. 40 Use voltage division inthe circuit on the right to calculate c as follows:
8
= —— 72 = 4 V
v 12( ) 8 [b] Calculate 2' in the circuit on the right using Ohm’s law: ., 'v 48
F‘s—E4” Now use i to calculate o, in the circuit on the left:
a, = 6(1.6 + 8) = 57.6 V Returning back to the original circuit, note that the voltage a, is also the
voltage drop across the series combination of the 120 V source and 20 Q resistor. Use this fact to calculate the current in the 120 V source, ta: 12041;, 120—5713 
_ T _ T _. 3.12 A FEW = —(120)’ia = (120)(3.12) = —374.40 W is Thus, the 120 V source delivers 3744 W. AP 4.16 To ﬁnd RTh, replace the 72 V source with a short circuit: 12!} 20;) 6Run" Note that the 5 Q and 20 Q resistors are in parallel, with an equivalent resistance of 520 = 4 Q. The equivalent 4 f2 resistance is in series with the 8 fl resistor for an
equivalent resistance of 4 + 8 = 12 9. Finally, the 12 (2 equivalent resistance is in
parallel with the 12 Q resistor, so RT}, = 1212 = 6 9. Use node voltage analysis to ﬁnd on. Begin by redrawing the circuit and labeling
the node voltages: 12!) 12V 20!) The node voltage equations are 111—72 1:1 illUm] _ a
5 +20+ 8 " 0 ’UTh"’b'1 vTh*72 __
s + 12 ' 0 Place these equations in standard form: UH) + (hi) —6
1 8 “The 12 ' Solving, a, = 60 V and UT}, = 64.8 V. Therefore, the Thévenin equivalent circuit is
a 64.8 V source in series with a 6 Q resistor. : Finally, combine the 20 Q and 12 9 parallel resistOrs to give RN = 20ll12 = 7.5 9. Thus, the Norton equivalent circuit is the parallel combination of a 6 A source and a
7.5 Q resistor. Substitute the constraint equation into the node voltage equations and put the tvvo
equations in standard form: v(1+1) + (5)  4
60 20 W" 20 ' qua) + (—1—+i+£)  o
20 m 40 80 20 '
Solving, o = 172.5 V and on, = 30 V. Now use the test source method to calculate the test current and thus Rn. Replace the current source with a short circuit and apply the test source to get the following
circuit: 600 Write a KCL equation at the rightmost node: .___’U*r ’01" vr+160in
3'“'_‘80+40+ 80 The dependent source constraint equation is ’UT 33,236 Substitute the constraint equation into the KCL equation and simplify the righthand
side: ' ' t = E
T 10
Therefore, RT, = 33 =109
3T Thus, the Thévenin equivalent is a 30 V source in series with a 10 Q resistor. ...
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This note was uploaded on 08/12/2010 for the course ENGR 117 taught by Professor Nickarnold during the Fall '08 term at Santa Barbara City.
 Fall '08
 NickArnold

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