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This problem has been solved satisfactorily. v = 20 volts
i 2 = 2 amps i1 = 1 amp i 4 = 1 amp Find I and Vab in the circuit of Figure 2.1. [2.15] Problem 2.8 i 3 = 4 amps 3Ω 10 V
−+ 5Ω a I +
30 V +
− Vab +
− 8V −
b
Figure 2.1
Applying KVL to the loop,  30 + 3 I − 10 + 5 I + 8 = 0
8 I = 32
I = 4 amps So,  Vab + 5 I + 8 = 0
Vab = 5 I + 8 = (5)(4) + 8  v v Therefore,  Vab = 28 volts eText Main Menu  Textbook Table of Contents  Problem Solving Workbook Contents In Figure 2.1, solve for i1 and i 2 . Problem 2.9 V1 − + + 4V − i2 3A
i1 + 2A
+ +
− 20 V V2 12 V +
− − − Figure 2.1 a b
i2 3A
i1 2A +
− +
− KCL (node a) : KCL (node b) : i1 + 3 + 2 = 0
i1 =  5 amps i2 + 0 = 3
i 2 = 3 amps In Figure 2.1, find V1 and V2 . Problem 2.10 − V1 + + 4V − +
+
20 V +
− Loop 1 12 V Loop 2 −  v v V1 =  8 volts  eText Main Menu  Textbook Table of Contents  V2
+
− − V2 = 8 volts Problem Solving Workbook Contents SERIES AND PARALLEL RESISTORS
Two or more elements are in series if they are cascaded or connected sequentially and
consequently carry the same current. Two or more elements are in parallel if they are connected
to the same two nodes and consequently have the same voltage across them. [2.35]
Problem 2.11
following networks. Find the equivalent resistance at terminals ab for each of the a Req R b R R R R Req a R R a R
b Req R b
(a) (b) (c) R
a
Req a
3R R R Req b (a) 3R R eq = R R + R R = (c) R eq (e) R eq = R 0 = 0 (b) (d) R eq (e) R eq v v 2R b
(d)  R  RR
+ =R
22
= ( R + R ) ( R + R ) = 2R 2R = R
3 (3R ) R 2 9R 2
R
3 = 3R (R + R R ) = 3R R + = 3R R =
=
=R
3 2
2
9R
3R + R
2
2 R (3R )
3 6R2
2
6 (R )(2R ) 3R = R 3R =
= R 2R 3R = =
=R
2 R + 2R 3
11 R 11
R + 3R
3 eText Main Menu  Textbook Table of Contents  Problem Solving Workbook Contents Therefore, R eq = 0 (a) Problem 2.12
resistor. (b), (c), (d) R eq = R 6
R
11 and (e) R eq = Give...
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 Spring '12
 bkav

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