problem26_21

Problem26_21 - 26.21 a) The sum of the currents that enter the junction below the 3 Ω resistor equals 3.00 A 5.00 A = 8.00 A b) Using the

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Unformatted text preview: 26.21: a) The sum of the currents that enter the junction below the 3 - Ω resistor equals 3.00 A + 5.00 A = 8.00 A. b) Using the lower left loop: ε1 − ( 4.00 Ω ) ( 3.00 A ) − ( 3.00 Ω ) ( 8.00 A ) = 0 ⇒ ε1 = 36.0 V. Using the lower right loop: ε2 − ( 6.00 Ω ) ( 5.00 A ) − ( 3.00 Ω ) ( 8.00 A ) = 0 ⇒ ε2 = 54.0 V. c) Using the top loop: 54.0 V − R ( 2.00 A ) − 36.0 V = 0 ⇒ R = 18.0 V = 9.00 Ω. 2.00 A ...
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This homework help was uploaded on 04/19/2008 for the course PHYS 4 taught by Professor Stuart during the Spring '08 term at UCSB.

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