hw11b - aljabr(faa335 – Hw11 – Ross –(89251 1 This...

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Unformatted text preview: aljabr (faa335) – Hw11 – Ross – (89251) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 5 . 1 Ω 4 . 1 Ω 18 . 4 Ω 23 V 11 . 5 V Find the current through the 18 . 4 Ω lower- right resistor. Correct answer: 0 . 804196 A. Explanation: r 1 r 2 R E 1 E 2 A D E B C F i 1 i 2 I Let : E 1 = 23 V , E 2 = 11 . 5 V , r 1 = 5 . 1 Ω , r 2 = 4 . 1 Ω , and R = 18 . 4 Ω . From the junction rule, I = i 1 + i 2 . Applying Kirchhoff’s loop rule, we obtain two equations: E 1 = i 1 r 1 + I R (1) E 2 = i 2 r 2 + I R = ( I- i 1 ) r 2 + I R =- i 1 r 2 + I ( R + r 2 ) , (2) Multiplying Eq. (1) by r 2 , Eq. (2) by r 1 , E 1 r 2 = i 1 r 1 r 2 + r 2 I R E 2 r 1 =- i 1 r 1 r 2 + I r 1 ( R + r 2 ) Adding, E 1 r 2 + E 2 r 1 = I [ r 2 R + r 1 ( R + r 2 )] I = E 1 r 2 + E 2 r 1 r 2 R + r 1 ( R + r 2 ) = (23 V) (4 . 1 Ω) + (11 . 5 V) (5 . 1 Ω) (4 . 1 Ω) (18 . 4 Ω) + (5 . 1 Ω) (18 . 4 Ω + 4 . 1 Ω) = . 804196 A . 002 (part 1 of 2) 10.0 points For a long period of time the switch S is in position “ b ”. At t = 0 s, the switch S is moved from position “ b ” to position “ a ”. 2 μ F 3 MΩ 5 MΩ 17 V S b a Find the voltage across the 3 MΩ center- left resistor at time t 1 = 2 s....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue.

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hw11b - aljabr(faa335 – Hw11 – Ross –(89251 1 This...

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