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problem26_22

# problem26_22 - 3 2-=-⇒ = Ω-Ω ⇒ I I I I Along with 3...

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26.22: From the circuit in Fig. 26.42, we use Kirchhoff’s Rules to find the currents, 1 I to the left through the 10 V battery, 2 I to the right through 5 V battery, and 3 I to the right through the 10 resistor: Upper loop: ( 29 ( 29 ( 29 ( 29 . A 00 . 1 0 00 . 5 00 . 5 V 0 . 5 0 V 00 . 5 00 . 4 00 . 1 00 . 3 00 . 2 V 0 . 10 2 1 2 1 2 1 = + = - - = - + - + - I I I I I I Lower loop: ( 29 ( 29 0 0 . 10 00 . 4 00 . 1 V 00 . 5 3 2 = - + + I I ( 29 ( 29 A 00 . 1 2 0 0 . 10 00 . 5 V
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Unformatted text preview: 3 2-=-⇒ = Ω-Ω + ⇒ I I I I Along with , 3 2 1 I I I + = we can solve for the three currents and find: . A 600 . , A 200 . , A 800 . 3 2 1 = = = I I I b) ( 29 ( 29 ( 29 ( 29 . V 20 . 3 00 . 3 A 800 . 00 . 4 A 200 .-= Ω-Ω-= ab V...
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