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Unformatted text preview: Version 072/ABACA – midterm 02 – Turner – (59070) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider an RC circuit i C R S E where a battery, a resistor and a capacitor are in series. As the switch is closed, at any instant the current at any part of the circuit is the same. Set the time when we close the switch to t = 0. After the switch is closed the current de creases exponentially with time according to I ( t ) = I e − t/τ where I is the initial current at t = 0, and τ is a constant having dimen sions of time. Consider a fixed observation point within the conductor. How much charge Q ( τ ) passes this point between t = 0 and t = τ ? Notice: The same amount of charge is deposited onto the capacitor plate. 1. Q ( τ ) = 0 . 632 I 2. Q ( τ ) = 1 . 72 I τ 3. Q ( τ ) = . 632 I τ 4. Q ( τ ) = I 2 5. Q ( τ ) = 1 . 72 I 6. Q ( τ ) = 2 . 72 I τ 7. Q ( τ ) = 0 . 632 I 2 τ 8. Q ( τ ) = 2 . 72 I 9. Q ( τ ) = I τ 10. Q ( τ ) = 0 . 632 I τ correct Explanation: Current is △ Q △ t , where Q is the charge and t is the time. In differential form this gives I ( t ) = dQ dt , which can be integrated as follows Q ( t ) = integraldisplay t I dt = I integraldisplay t e − t/τ dt = I τ (1 − e − t/τ ) and Q ( τ ) = I τ (1 − e − 1 ) = . 632 I τ . 002 10.0 points An airfilled cylindrical capacitor has a capac itance of 16 pF and is 5 . 9 cm in length. The coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . If the radius of the outside conductor is 1 cm, what is the required radius of the inner conductor? 1. 1.01899 2. 0.766208 3. 1.43292 4. 0.393585 5. 0.454832 6. 0.893012 7. 1.15044 8. 1.72487 9. 0.814529 10. 1.09696 Correct answer: 0 . 814529 cm. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , ℓ = 5 . 9 cm = 0 . 059 m , b = 1 cm , and C = 16 pF = 1 . 6 × 10 − 11 F . Version 072/ABACA – midterm 02 – Turner – (59070) 2 The capacitance of a cylindrical capacitor is given by C = ℓ 2 k e ln b a where a and b are the inner and outer radii, respectively. Solving for a , we obtain a = b e − ℓ/ (2 k e C ) = (1 cm) e − (0 . 059 m) / [2 k e (1 . 6 × 10 11 F)] = . 814529 cm . 003 10.0 points All the bulbs in the figure below have the same resistance R . The switch S is initially closed. V i A i C i B i D S i If bulb B is removed from the circuit, i.e. , the switch S is opened, what happens to the currents through 1) the battery, 2) bulb A , and 3) bulb D ; Notice that in the diagram the current through the battery, i battery , is labeled as i . Hint: You may find it helpful to work out the currents through bulb A , bulb D , and the battery for both cases by using V = 1 volt and R = 1 Ω....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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