Unformatted text preview: 28. (a) R2 , R3 and R4 are in parallel. By ﬁnding a common denominator and simplifying, the equation
1/R = 1/R2 + 1/R3 + 1/R4 gives an equivalent resistance of
R= (50 Ω)(50 Ω)(75 Ω)
R2 R3 R4
=
= 19 Ω .
R2 R3 + R2 R4 + R3 R4
(50 Ω)(50 Ω) + (50 Ω)(75 Ω) + (50 Ω)(75 Ω) Thus, considering the series contribution of resistor R1 , the equivalent resistance for the network is
Req = R1 + R = 100 Ω + 19 Ω = 1.2 × 102 Ω.
(b) i1 = E /Req = 6.0 V/(1.1875 × 102 Ω) = 5.1 × 10−2 A; i2 = (E − V1 )/R2 = (E − i1 R1 )/R2 =
[6.0 V − (5.05 × 10−2 A)(100 Ω)]/50 Ω = 1.9 × 10−2 A; i3 = (E − V1 )/R3 = i2 R2 /R3 = (1.9 ×
10−2 A)(50 Ω/50 Ω) = 1.9 × 10−2 A; i4 = i1 − i2 − i3 = 5.0 × 10−2 A − 2(1.895 × 10−2 A) = 1.2 × 10−2 A. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Resistance

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