Page 315 - Schaum's Outline of Theory and Problems of Applied Physics

300 DIRECT-CURRENTCIRCUITS [CHAP. 25

Fig. 25-11

We now solve this equation and the first loop equation for I 1 , set the two expressions for I 1 equal, and then solve
for I:
IR − V e,1 V e,2 − IR − Ir 2
I 1 = and I 1 =
r 1 r 2
IR − V e,1 V e,2 − IR − Ir 2
=
r 1 r 2
I (Rr 2 + Rr 1 + r 2 r 1 ) = V e,2 r 1 + V e,1 r 2
(8V)(0.5 ) + (6V)(0.6 )
V e,2 r 1 + V e,1 r 2
I = = = 0.673 A
Rr 2 + Rr 1 + r 2 r 1 (10 )(0.6 ) + (10 )(0.5 ) + (0.6 )(0.5 )
SOLVED PROBLEM 25.22

Find the currents in the three resistors of the circuit shown in Fig. 25-12(a). The internal resistances of
the emf sources are included in R 1 and R 3 .

We assume the current directions shown in the figure. Applying Kirchhoff’s first rule to junction a yields
I 3 = I 1 + I 2
Next we analyze the two inner loops with the help of Kirchhoff’s second rule. Proceeding counterclockwise in loop
1 yields

V e,1 = I 1 R 1 − I 2 R 2

Fig. 25-12