Page 317 - Schaum's Outline of Theory and Problems of Applied Physics

302 DIRECT-CURRENTCIRCUITS [CHAP. 25

Fig. 25-13

Clearly one or two current directions are incorrect, but this makes no difference since the result is a negative current
in those cases. Next we apply Kirchhoff’s second rule to loop 1 and proceed clockwise:

−2V − 6V = (10 )(I 2 ) − (4 )(I 1 )
The emf’s are considered negative because we encountered their + terminals first. Solving for I 2 gives
−8V 4
I 2 = + (I 1 ) =−0.8A + 0.4I 1
10 10
Now we proceed clockwise in loop 2 and obtain

6V + 5V = (8 )(I 3 ) − (10 )(I 2 )
Substituting I 3 =−I 1 − I 2 and solving for I 2 yields

11 V =−(8 )(I 1 ) − (8 )(I 2 ) − (10 )(I 2 )
−11 V 8
I 2 = − (I 1 ) =−0.611 A − 0.444I 1
18 18
Setting equal the two expressions for I 2 and solving for I 1 , we find

−0.8A + 0.4I 1 =−0.611 A − 0.444I 1
0.844I 1 = 0.189 A

I 1 = 0.224 A
From the first loop equation

I 2 =−0.8A + 0.4I 1 =−0.710 A
and so

I 3 =−I 1 − I 2 =−0.224 A + 0.710 A = 0.486 A
The actual currents are shown in Fig. 25-13(b).
Again we check the results by using the outside loop of the circuit as shown in Fig. 25-13(c). Proceeding
clockwise yields
−2V + 5V = (8 )(0.486 A) − (4 )(0.224 A)

3V = 3V