oldhw15 - oldhomewk 15 – LEE, BENJAMIN – Due: Oct 7...

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Unformatted text preview: oldhomewk 15 – LEE, BENJAMIN – Due: Oct 7 2007, 4:00 am 1 Question 1, chap 28, sect 4. part 1 of 1 10 points 6 Ω 5 Ω 22 Ω 30 V 15 V Find the current through the 22 Ω lower- right resistor. Correct answer: 0 . 882353 A (tolerance ± 1 %). Explanation: r 1 r 2 R E 1 E 2 A D E B C F i 1 i 2 I Let : E 1 = 30 V , E 2 = 15 V , r 1 = 6 Ω , r 2 = 5 Ω , and R = 22 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 ) = (30 V) (5 Ω) + (15 V) (6 Ω) (5 Ω) (22 Ω) + (6 Ω) (22 Ω + 5 Ω) = . 882353 A . Question 2, chap 28, sect 4. part 1 of 3 10 points Consider the circuit in the figure. 9 . 00 V 3 . 1 Ω 3 . 1 Ω 7 . 1 Ω 7 . 1 Ω 4 . 2 Ω 11 . 0 Ω 1 . 7 Ω Find the current in the 1.7 Ω resistor. Correct answer: 0 . 895625 A (tolerance ± 1 %). Explanation: E R 1 R 2 R 3 R 4 R 5 R 6 R 7 oldhomewk 15 – LEE, BENJAMIN – Due: Oct 7 2007, 4:00 am 2 Let : R 1 = 3 . 1 Ω , R 2 = 3 . 1 Ω , R 3 = 7 . 1 Ω , R 4 = 7 . 1 Ω , R 5 = 4 . 2 Ω , R 6 = 11 . 0 Ω , R 7 = 1 . 7 Ω and Δ V = 9 . 00 V . R 3 and R 4 are in parallel: 1 R 34 = 1 R 3 + 1 R 4 R 34 = parenleftbigg 1 R 3 + 1 R 4 parenrightbigg- 1 = parenleftbigg 1 7 . 1 Ω + 1 7 . 1 Ω parenrightbigg- 1 = 3 . 55 Ω R 2 and R 34 are in series: R 234 = R 2 + R 34 = 3 . 1 Ω + 3 . 55 Ω = 6 . 65 Ω R 5 and R 6 are in parallel: 1 R 56 = 1 R 5 + 1 R 6 R 56 = parenleftbigg 1 R 5 + 1 R 6 parenrightbigg- 1 = parenleftbigg 1 4 . 2 Ω + 1 11 Ω parenrightbigg- 1 = 3 . 03947 Ω R 56 and R 7 are in series: R 567 = R 56 + R 7 = 3 . 03947 Ω + 1 . 7 Ω = 4 . 73947 Ω R 234 and R 567 are in parallel: 1 R 234567 = 1 R 234 + 1 R 567 R 234567 = parenleftbigg 1 R 234 + 1 R 567 parenrightbigg- 1 = parenleftbigg 1 6 . 65 Ω + 1 4 . 73947 Ω parenrightbigg- 1 = 2 . 76725 Ω R 1 and R 234567 are in series: R eq = R 1 + R 234567 = 3 . 1 Ω + 2 . 76725 Ω = 5 . 86725 Ω The current in the circuit is I = Δ V R eq = 9 V 5 . 86725 Ω = 1 . 53394 A Δ V 234567 = I R 234567 = (1 . 53394 A) (2 . 76725 Ω) = 4 . 24479 V Here the current is I 7 = I 567 = Δ V 234567 R 567 = (4 . 24479 V) (4 . 73947 Ω) = . 895625 A . Question 3, chap 28, sect 4. part 2 of 3 10 points Find the potential difference across the 1.7 Ω resistor. Correct answer: 1 . 52256 V (tolerance ± 1 %). Explanation: Solution: Δ V 7 = I 7 R 7 = (0 . 895625 A) (1 . 7 Ω) = 1 . 52256 V ....
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This note was uploaded on 04/07/2008 for the course PHY 303K taught by Professor Turner during the Fall '07 term at A.T. Still University.

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oldhw15 - oldhomewk 15 – LEE, BENJAMIN – Due: Oct 7...

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