0195136047.pdf

1.3 KIRCHHOFF’S LAWS 43

veb=− 8 − 7 =−15 V

Alternatively,

veb=vea+vab=− 20 + 5 =−15 V

or

veb=vec+vcf+vfb=−vx+vcf+ 0 =− 12 − 3 =−15 V

The student should observe thatvbe=−veb=15 V and nodebis at a higher potential than

nodee.

EXAMPLE 1.3.3

Referring to Figure 1.3.2, let boxes 2, 3, and 5 consist of a 0.2-H inductor, a 5-resistor, and a
0.1-F capacitor, respectively. GivenA=5, andv 1 =10 sin 10t,i 2 =5 sin 10t, andi 3 =2 sin
10 t−4 cos 10t, findi 5.

Solution

From Equation (1.2.19),

v 2 =L

di 2

dt

= 0. 2

d

dt

(5 sin 10t)=10 cos 10tV

From Equation (1.2.1),

v 3 =Ri 3 = 5 (2 sin 10t−4 cos 10t)=10 sin 10t−20 cos 10tV

From Equation (1.3.2),

v 5 =v 1 +v 3 +v 4 −v 2 =v 1 +v 3 + 5 v 2 −v 2

=10 sin 10t+(10 sin 10t−20 cos 10t)+ 4 (10 cos 10t)

=20 sin 10t+20 cos 10tV

From Equation (1.2.9),

i 5 =C

dv 5

dt

= 0. 1

d

dt

(20 sin 10t+20 cos 10t)

=20 cos 10t−20 sin 10tA

Note the consistency of voltage polarities and current directions in Figure 1.3.2.

EXAMPLE 1.3.4

Consider the network shown in Figure E1.3.4(a).

(a) Find the voltage drops across the resistors and mark them with their polarities on the

circuit diagram.

(b) Check whether the KVL is satisfied, and determineVbfandVec.