# 0195136047.pdf

(Joyce) #1

80 CIRCUIT ANALYSIS TECHNIQUES

``V 1 2 Ω 0.5 V^15 Ω^ V^ =?``

``I =?``

``10 Ω``

``10 Ω``

``4 Ω``

``(a)``

``5 V``

``+``

``−``

``+``

``−``

``+``

``−``

``Figure E2.2.5``

``V 1 2 Ω 5 Ω``

``````0.5 V 1 × 10 = 5 V 1
10 Ω 4 Ω 10 Ω``````

``(b)``

``5 V``

``+``

``−``

``+−``

``+``

``−``

``I 1 I 2``

``0.5 V 1``

``(c)``

``O Reference``

``A B C``

(^12) = 0.5 S (^15) = 0.2 S
1
4
105 = 0.5 A = 0.1 S
= 0.25 S
1
10
101 = 0.1 S
V 1

• Combining the constraint equation with the loop equations, one gets
16 I 1 − 2 I 2 = 5 ;− 2 I 1 + 17 I 2 =− 10 (I 1 −I 2 ), or 8 I 1 + 7 I 2 = 0
from which
I 1 = 35 / 128 A; I 2 =− 5 / 16 A
Thus, the current through the 5-V source isI=I 1 = 35 / 128 = 0 .273 A, and the
voltage across the 5-resistor isV= 5 I 2 = 5 (− 5 / 16 )=− 1 .563 V.
(b) Node-Voltage Method: The 5-V voltage source with its 10-series resistor is replaced
by its Norton equivalent. Resistances are converted into conductances and the circuit is
redrawn in Figure E2.2.5(c) with the nodes shown.
The nodal equations are
A: ( 0. 1 + 0. 25 )VA− 0. 25 VB= 0. 5
B: − 0. 25 VA+( 0. 25 + 0. 5 + 0. 1 )VB− 0. 1 VC= 0. 5 V 1
C: − 0. 1 VB+( 0. 1 + 0. 2 )VC=− 0. 5 V 1