14exam3 - Husain Zeena Exam 3 Due 10:00 pm Inst Sonia Paban...

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Husain, Zeena – Exam 3 – Due: Apr 13 2004, 10:00 pm – Inst: Sonia Paban 1 This print-out should have 16 questions. Multiple-choice 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 d I 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 d I dt - q C = 0 3. E - I R - L d I dt - q C = 0 correct 4. E - I R + L d I dt - q C = 0 5. E + I R - L d I dt + q C = 0 6. E - I R - L d I dt + q C = 0 7. E + I R + L d I dt - q C = 0 8. E + I R + L d I dt + q C = 0 9. E - I R + L d I dt + q C = 0 Explanation: Apply Kircho ff ’s loop rule to the above cir- cuit. Here we are told that d I dt > 0; therefore the emf induced by the inductor is - L d I dt . Summing the potential di ff erences around the loop gives E - I R - L d I dt - q/C = 0 . 002 (part 2 of 2) 10 points 18 V S 7 . 07 μ F 4 . 18 M Ω 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 M Ω , 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 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

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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. E 5 - i 1 R = 0 and 4 E 5 + i 2 R = 0 . 7. E 5 + i 1 R = 0 and 4 E 5 + i 2 R = 0 . cor- rect 8. 4 E 5 + i 1 R = 0 and E 5 - i 2 R = 0 . Explanation: By definition, the areas of the left and right loops are related by A = A L + A R . Since A L A R = 4, we can solve for A L and A R in terms of A . A L = 4 A 5 A R = A 5 . Then we can compute the magnitude of the induced emf around the right and left loops.
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