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problem26_24

# problem26_24 - -Ω ⇒ I I I I Along with 3 2 1 I I I = we...

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26.24: After switching the 5-V battery for a 20-V battery in the circuit of Fig. 26.42, there is a change in the equations from Problem 26.22 in both the upper and lower loops: Upper loop: ( 29 ( 29 0 V 00 . 20 00 . 4 00 . 1 00 . 3 00 . 2 V 0 . 10 2 1 = - + - + - I I ( 29 ( 29 . A 00 . 2 0 00 . 5 00 . 5 V 0 . 10 2 1 2 1 - = + = - - - I I I I Lower loop: ( 29 ( 29 0 0 . 10 00 . 4 00 . 1 V 00 . 20 3 2 = - + + I I ( 29 ( 29 . A 00 . 4 2 0 0 . 10 00 . 5 V 00 . 20 3 2 3 2 - = - =
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Unformatted text preview: -Ω + ⇒ I I I I Along with , 3 2 1 I I I + = we can solve for the three currents and find: . A 2 . 1 , A 6 . 1 , A 4 . 3 2 1 + =-=-= I I I b) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 V 6 . 7 3 A 4 . 4 A 6 . 1 3 4 1 2 = Ω + Ω = Ω-Ω I I...
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