Page 316 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 25] DIRECT-CURRENTCIRCUITS 301

and proceeding counterclockwise in loop 2 yields

V e,2 = I 2 R 2 + I 3 R 3
We now have three equations that relate the unknown quantities I 1 , I 2 , and I 3 . One way to proceed (there are others,
equally suitable) is to substitute I 3 = I 1 + I 2 in the second loop equation, which gives

V e,2 = I 2 R 2 + I 1 R 3 + I 2 R 3
V e,2 − I 2 R 2 − I 2 R 3
I 1 =
R 3
From the first loop equation,

V e,1 + I 2 R 2
I 1 =
R 1
Since these two expressions must be equal,

V e,1 + I 2 R 2 V e,2 − I 2 R 2 − I 2 R 3
=
R 1 R 3
At this point we substitute the values of the various emf’s and resistances and solve for I 2 . This substitution can also
be done earlier or later in a calculation of this kind, whatever seems most convenient. The calculation proceeds as
follows:
10 V + (5 )(I 2 ) 5V − (5 )(I 2 ) − (6 )(I 2 )
=
3 6
10 V 5 5V 5 6
+ I 2 = − I 2 − I 2
3 3 6 6 6
5 5 6 5 10

+ + I 2 = − A
3 6 6 6 3
I 2 =−0.714 A
The minus sign means that the current I 2 is in the opposite direction to the one shown in the figure. From the first
loop equation,
10 V − (0.714 A)(5 )
V e,1 + I 2 R 2
I 1 = = = 2.143 A
R 1 3
Finally we find I 3 from the junction equation:
I 3 = I 1 + I 2 = 2.143 A − 0.714 A = 1.429 A

The actual currents are shown in Fig. 25-12(b).
As a check on the calculation we can apply Kirchhoff’s second rule to the outside loop of the circuit, shown in
Fig. 25-12(c). Proceeding counterclockwise, we get

V e,2 + V e,1 = I 1 R 1 + I 3 R 3
5V + 10 V = (2.143 A)(3 ) + (1.429 A)(6 )
15 V = 15 V

SOLVED PROBLEM 25.23

Find the currents in the three resistors of the circuit shown in Fig. 25-13(a). The internal resistances of
the emf sources are included in the resistances shown.
Let us now work directly from the numerical values of the emf’s and resistances in the figure. Applying
Kirchhoff’s first rule to junction a, with the current directions shown, yields
I 1 + I 2 + I 3 = 0