P28_028.PDF

28.(a) R 2 , R 3 and R 4 are in parallel. By ﬁnding a common denominator and simplifying, the equation

1 /R =1 /R 2 +1 /R 3 +1 /R 4 gives an equivalent resistance of

R =

R 2 R 3 R 4

R 2 R 3 + R 2 R 4 + R 3 R 4 =

(50 Ω)(50 Ω)(75 Ω)

(50 Ω)(50 Ω) + (50 Ω)(75 Ω) + (50 Ω)(75 Ω) =19Ω .

Thus, considering the series contribution of resistor R 1 , the equivalent resistance for the network is

R eq = R 1 + R = 100 Ω + 19 Ω = 1 . 2

10 2 Ω .

(b) i 1 =

/R eq =6 . 0V / (1 . 1875

10 2 Ω) = 5 . 1

10 − 2 A; i 2 =(

E−

V 1 ) /R 2 =(

E−

i 1 R 1 ) /R 2 =

10 − 2 A)(50 Ω / 50Ω)=1 . 9 × 10 − 2 A; i 4 = i 1 −i 2 −i 3 =5 . 0 × 10 − 2 A − 2(1 . 895 × 10 − 2 A) = 1 . 2 × 10 − 2 A .

−

(5 . 05

10 − 2 A)(100 Ω)] / 50Ω = 1 . 9

10 − 2 A; i 3 =(

E−

V 1 ) /R 3 = i 2 R 2 /R 3 =(1 . 9

[6 . 0V