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