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Unformatted text preview: Husain, Zeena – Exam 3 – Due: Apr 13 2004, 10:00 pm – Inst: Sonia Paban 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points At some time after the switch S is closed, there is a current I flowing through the resis tor and inductor which is increasing in time dI dt > 0 (see the figure below). At this time there is an amount of charge of magnitude q on each plate of the capacitor. E S C R L I Which of the following equations is correct? 1. none of these 2. E + I R L dI dt q C = 0 3. E  I R L dI dt q C = 0 correct 4. E  I R + L dI dt q C = 0 5. E + I R L dI dt + q C = 0 6. E  I R L dI dt + q C = 0 7. E + I R + L dI dt q C = 0 8. E + I R + L dI dt + q C = 0 9. E  I R + L dI dt + q C = 0 Explanation: Apply Kirchoff’s loop rule to the above cir cuit. Here we are told that dI dt > 0; therefore the emf induced by the inductor is L dI dt . Summing the potential differences around the loop gives E  I R L dI dt q/C = 0 . 002 (part 2 of 2) 10 points 18 V S 7 . 07 μ F 4 . 18 MΩ 5 . 48 mH In the circuit shown above, what is the charge on the capacitor after the switch has been closed for a long time? Correct answer: 0 . 00012726 C. Explanation: Let : E = 18 V , R = 4 . 18 MΩ , L = 5 . 48 mH , and C = 7 . 07 μ F . After a long time t the capacitor will reach its maximum charge q max . When this occurs the current will drop to zero. Setting I = 0 in the loop equation from Part 1 gives E  q max C = 0 , or q max = C E = (7 . 07 μ F)(1 × 10 6 F /μ F)(18 V) = 0 . 00012726 C , 003 (part 1 of 1) 10 points Assume: The induced emf for the closed loop octagonal CXDY C is E . A solenoid (with magnetic field B ) pro duces a steadily increasing uniform magnetic Husain, Zeena – Exam 3 – Due: Apr 13 2004, 10:00 pm – Inst: Sonia Paban 2 flux through its circular cross section. A octagonal circuit surrounds the solenoid as shown in the figure. The wires connecting in the circuit are ideal, having no resistance. The circuit consists of two identical light bulbs (labeled X and Y ) in series. A wire connects points C and D . The ratio of the solenoid’s area A L left of the wire CD and the solenoid’s area A R right of the wire CD is A L A R = 4 . B B B B Y X D C i 3 i 1 i 2 A L A R The equations for the (right) loop CXDC and the (left) loop CDY C are respectively given by 1. E 5 + i 1 R = 0 and 4 E 5 i 2 R = 0 . 2. 4 E 5 + i 1 R = 0 and E 5 + i 2 R = 0 . 3. E 5 i 1 R = 0 and 4 E 5 i 2 R = 0 . 4. 4 E 5 i 1 R = 0 and E 5 + i 2 R = 0 . 5. 4 E 5 i 1 R = 0 and E 5 i 2 R = 0 . 6....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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