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Unformatted text preview: Version 044 – midterm 02 – chiu – (56565) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Calculate R eq for the circuit below. Given that R = 170 Ω , R 1 = 76 Ω and R 2 = R 3 = R 4 = 8 Ω . 1. 120.0 2. 209.95 3. 133.302 4. 191.115 5. 188.24 6. 166.857 7. 173.909 8. 203.739 9. 224.422 10. 169.134 Correct answer: 188 . 24 Ω. Explanation: We know that R = 170 Ω R 1 = 76 Ω R 2 = 8 Ω R 3 = 8 Ω R 4 = 8 Ω First note that R 1 is in parallel with the series combination of R 2 , R 3 , and R 4 . Then note that this parallel element is in series with R . All together, we may calculate, R eq = R + 1 R 1 + 1 R 2 + R 3 + R 4 1 R eq = R + 3 R 2 + R 1 3 R 1 R 2 1 R eq = R + 3 R 1 R 2 R 1 + 3 R 2 R eq = 170 Ω + 3(76 Ω) (8 Ω) 76 Ω + 3(8 Ω) R eq = 188 . 24 Ω 002 (part 1 of 2) 5.0 points From a microscopic point of view, the im portant quantities in a circuit analysis are the fields and the electron currents. This problem Version 044 – midterm 02 – chiu – (56565) 2 concerns the circuits shown in figures (a) and (b). The bulbs in both circuits are identical and have a filament length L, while the bat teries are also identical with emf, ε . Assume the potential difference along the connecting wires in both circuits is negligible. For the case of Fig(a), the field in the fila ment is E. The circuit satisfies the loop equa tion: ε = E L and the electron current is i . Now consider the circuit of Fig(b). Define E 1 to be the electric field through the top bulb, E 2 L be the electric field in the lowerleft bulb, and E 2 R to be the electric field in the lowerright bulb. Choose the answer that identifies the cor rect statements from the following list. (Ia) E 1 = E 2 L (Ib) E 1 > E 2 L (IIa) E 2 L > E 2 R (IIb) E 2 L = E 2 R (IIIa) E 2 L + E 2 R = E 1 (IIIb) E 2 L + E 2 R > E 1 1. Ib, IIb, IIIb 2. Ia, IIb, IIIb 3. Ib, IIa, IIIa 4. Ia, IIa, IIIb 5. Ib, IIb, IIIa correct 6. Ia, IIa, IIIa 7. Ib, IIa, IIIb 8. Ia, IIb, IIIa Explanation: From Fig(b), the loop equations imply that ε = E 1 L and ε = E 2 L L + E 2 R L . This implies that E 2 L + E 2 R = E 1 and E 1 > E 2 L . So Ib and IIIa are correct. In the steady state, the currents through the two lower bulbs must be the same. In turn the corresponding driving fields through the two bulbs are also the same, i.e. E 2 L = E 2 R . So IIb is correct. 003 (part 2 of 2) 5.0 points Denote the current through the top branch be i 1 and the current through the bottom branch to be i 2 . Choose the answer that identifies the cor rect statements from the following list. (Ia) i 1 = i (Ib) i 1 < i (IIa) i 2 = i (IIb) i 2 = i 2 (IIIa) i battery = 3 2 i (IIIb) i battery = 2 i 1. Ib, IIa, IIIb 2. Ia, IIa, IIIa 3. Ia, IIa, IIIb 4. Ib, IIa, IIIa 5. Ib, IIb, IIIb 6. Ia, IIb, IIIb 7. Ia, IIb, IIIa correct 8. Ib, IIb, IIIa Explanation: The loop equations from Fig(a) and Fig(b) imply that ε = E L =...
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 Spring '08
 Turner
 Energy, Magnetic Field, Correct Answer, Electric charge

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