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Unformatted text preview: oldhomewk 14 LEE, BENJAMIN Due: Oct 4 2007, 4:00 am 1 Question 1, chap 28, sect 7. part 1 of 2 10 points The circuit has been connected as shown in the figure for a long time. 16 V S 24 F 7 9 1 1 5 What is the magnitude of the electric po tential across the capacitor? Correct answer: 6 V (tolerance 1 %). Explanation: E S 1 C t b a b I t R 1 I t R 2 I b R 3 I b R 4 Let : R 1 = 7 , R 2 = 9 , R 3 = 1 , R 4 = 15 , and C = 24 F . After a long time implies that the ca pacitor C is fully charged and therefore the capacitor acts as an open circuit with no cur rent flowing to it. The equivalent circuit is I t R 1 I t R 2 R 3 I b I b R 4 a b R t = R 1 + R 2 = 7 + 9 = 16 R b = R 3 + R 4 = 1 + 15 = 16 I t = E R t = 16 V 16 = 1 A I b = E R b = 16 V 16 = 1 A Across R 1 E 1 = I t R 1 = (1 A) (7 ) = 7 V . Across R 3 E 3 = I b R 3 = (1 A) (1 ) = 1 V . Since E 1 and E 3 are measured from the same point a , the potential across C must be E C = E 3 E 1 = 1 V 7 V = 6 V E C  = 6 V . Question 2, chap 28, sect 7. part 2 of 2 10 points If the battery is disconnected, how long does it take for the capacitor to discharge to E t E = 1 e of its initial voltage? Correct answer: 144 s (tolerance 1 %). Explanation: With the battery removed, the circuit is C I R 1 I r R 2 R 3 I I r R 4 r oldhomewk 14 LEE, BENJAMIN Due: Oct 4 2007, 4:00 am 2 C R eq I eq where R = R 1 + R 3 = 7 + 1 = 8 , R r = R 2 + R 4 = 9 + 15 = 24 and R eq = parenleftbigg 1 R + 1 R r parenrightbigg 1 = parenleftbigg 1 8 + 1 24 parenrightbigg 1 = 6 . Therefore the time constant is R eq C = (6 ) (24 F) = 144 s . The equation for discharge of the capacitor is Q t Q = e t/ , or E t E = e t/ = 1 e . Taking the logarithm of both sides, we have t = ln parenleftbigg 1 e parenrightbigg t = ( ln e ) = (144 s)( 1) = 144 s . Question 3, chap 28, sect 4. part 1 of 2 10 points In this problem assume 1) the batteries have zero internal resistance, 2) the currents are flowing in the direction indicated by the arrows. A negative current denotes flow opposite to the direction of the arrow. 14 . 7 23 6 . 4 V 14 V I 1 I 2 I 3 Using the direction of the arrow as posi tive (as shown in the figure), find the current through the 14 . 7 resistor and the 6 . 4 V battery at the top of the circuit. Correct answer: 1 . 38776 A (tolerance 1 %). Explanation: R 1 R 2 E 1 E 2 I 1 I 2 I 3 Let : R 1 = 14 . 7 , R 2 = 23 , E 1 = 6 . 4 V , and E 2 = 14 V ....
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 Fall '07
 Turner

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