P28_022.PDF

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Chapter 28 - 28.22
22.
•
S 1 ,S 2 and S 3 all open: i a =0 . 00 A.
•
S 1 closed, S 2 and S 3 open: i a =
E
/ 2 R 1 = 120 V / 40 . 0Ω = 3 . 00 A .
•
S 2 closed, S 1 and S 3 open: i a =
E
/ (2 R 1 + R 2 ) = 120 V / 50 . 0Ω = 2 . 40 A .
•
S 3 closed, S 1 and S 2 open: i a =
E
/ (2 R 1 + R 2 ) = 120 V / 60 . 0Ω = 2 . 00 A .
•
S 1 open, S 2 and S 3 closed: R eq = R 1 + R 2 + R 1 ( R 1 + R 2 ) / (2 R 1 + R 2 )=20 . 0Ω + 10 . 0Ω +
(20 . 0 Ω)(30 . 0Ω) / (50 . 0Ω)= 42 . 0Ω, so i a =
E
/R eq = 120 V / 42 . 0Ω = 2 . 86 A .
•
S 2 open, S 1 and S 3 closed: R eq = R 1 + R 1 ( R 1 +2 R 2 ) / (2 R 1 +2 R 2 )=20 . 0Ω + (20 . 0Ω)
×
(40 . 0Ω) / (60 . 0Ω) = 33 . 3Ω, so i a = E/R eq = 120 V / 33 . 3Ω = 3 . 60 A .
•
S 3 open, S 1 and S 2 closed: R eq = R 1 + R 1 ( R 1 + R 2 ) / (2 R 1 + R 2 )=20 . 0Ω+(20 . 0Ω)
×
(30 . 0Ω) / (50 . 0Ω)
=32 . 0Ω, so i a =
E
/R eq = 120 V / 32 . 0Ω = 3 . 75 A .
•
S 1 , S 2 and S 3 all closed: R eq = R 1 + R 1 R / ( R 1 + R )where R = R 2 + R 1 ( R 1 + R 2 ) / (2 R 1 +
R 2 )=22 . 0 Ω, i.e., R eq =20 . 0Ω + (20 . 0 Ω)(22 . 0Ω) / (20 . 0Ω+ 22 . 0Ω) = 30 . 5Ω , so i a =
E
/R eq =
120 V / 30 . 5Ω = 3 . 94 A .

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