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oldhw14 - oldhomewk 14 LEE BENJAMIN Due Oct 4 2007 4:00 am...

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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 μ F 7 Ω 9 Ω 1 Ω 15 Ω 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 Ω , R 2 = 9 Ω , R 3 = 1 Ω , R 4 = 15 Ω , and C = 24 μ 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 Ω + 9 Ω = 16 Ω R b = R 3 + R 4 = 1 Ω + 15 Ω = 16 Ω I t = E R t = 16 V 16 Ω = 1 A I b = E R b = 16 V 16 Ω = 1 A Across R 1 E 1 = I t R 1 = (1 A) (7 Ω) = 7 V . Across R 3 E 3 = I b R 3 = (1 A) (1 Ω) = 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 0 = 1 e of its initial voltage? Correct answer: 144 μ 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
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oldhomewk 14 – LEE, BENJAMIN – Due: Oct 4 2007, 4:00 am 2 C R eq I eq where R = R 1 + R 3 = 7 Ω + 1 Ω = 8 Ω , R r = R 2 + R 4 = 9 Ω + 15 Ω = 24 Ω and R eq = parenleftbigg 1 R + 1 R r parenrightbigg - 1 = parenleftbigg 1 8 Ω + 1 24 Ω parenrightbigg - 1 = 6 Ω . Therefore the time constant τ is τ R eq C = (6 Ω) (24 μ F) = 144 μ s . The equation for discharge of the capacitor is Q t Q 0 = e - t/τ , or E t E 0 = e - t/τ = 1 e . Taking the logarithm of both sides, we have - t τ = ln parenleftbigg 1 e parenrightbigg t = - τ ( - ln e ) = - (144 μ s) ( - 1) = 144 μ 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 Ω 23 Ω 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 Ω 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 Ω , R 2 = 23 Ω , E 1 = 6 . 4 V , and E 2 = 14 V . At nodes, we have I 1 - I 2 - I 3 = 0 . (1) Pay attention to the sign of the battery and the direction of the current in the figure. Us- ing the lower circuit in the figure, we get E 2 + I 2 R 2 = 0 so , (2) I 2 = -E 2 R 2 = - (14 V) (23 Ω) = - 0 . 608696 A .
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oldhomewk 14 – LEE, BENJAMIN – Due: Oct 4 2007, 4:00 am 3 Then, for the upper circuit E 1 - I 2 R 2 - I 1 R 1 = 0 (3) E 1 + E 2 - I 1 R 1 = 0 so , I 1 = E 1 + E 2 R 1 (4) = (6 . 4 V) + (14 V) (14 . 7 Ω) = 1 . 38776 A .
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