P28_022

# P28_022 - = 32 . 0 , so i a = E /R eq = 120 V / 32 . 0 = 3...

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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 )=2 0 . 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 )=2 0 . 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
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Unformatted text preview: = 32 . 0 , so i a = E /R eq = 120 V / 32 . 0 = 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 , i.e., R eq = 20 . 0 + (20 . 0 )(22 . 0 ) / (20 . 0 + 22 . 0 ) = 30 . 5 , so i a = E /R eq = 120 V / 30 . 5 = 3 . 94 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.

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