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# hw8 - johnson(rj6247 hw 8 Opyrchal(121014 This print-out...

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johnson (rj6247) – hw 8 – Opyrchal – (121014) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Consider the following circuit. 5 . 4 Ω 3 . 1 Ω 3 . 1 Ω 4 . 2 Ω 1 . 5 Ω 3 . 1 Ω 4 . 2 Ω 16 . 0 Ω 16 . 0 Ω 26 V a) Find the equivalent resistance. Correct answer: 16 . 0843 Ω. Explanation: R 1 R 4 R 7 R 3 R 6 R 9 R 8 R 5 R 2 Δ V Let : R 1 = 5 . 4 Ω , R 2 = 16 . 0 Ω , R 3 = 4 . 2 Ω , R 4 = 3 . 1 Ω , R 5 = 16 . 0 Ω , R 6 = 1 . 5 Ω , R 7 = 3 . 1 Ω , R 8 = 4 . 2 Ω and R 9 = 3 . 1 Ω . For resistors in series, R eq,s = R a + R b R 789 = R 7 + R 8 + R 9 = 3 . 1 Ω + 4 . 2 Ω + 3 . 1 Ω = 10 . 4 Ω . For resistors in parallel, 1 R eq,p = 1 R a + 1 R b R 5789 = parenleftbigg 1 R 5 + 1 R 789 parenrightbigg 1 = parenleftbigg 1 16 Ω + 1 10 . 4 Ω parenrightbigg 1 = 6 . 30303 Ω , R 456789 = R 4 + R 5789 + R 6 = 3 . 1 Ω + 6 . 30303 Ω + 1 . 5 Ω = 10 . 903 Ω , R 2456789 = parenleftbigg 1 R 2 + 1 R 456789 parenrightbigg 1 = parenleftbigg 1 16 Ω + 1 10 . 903 Ω parenrightbigg 1 = 6 . 48434 Ω , and R eq = R 1 + R 2456789 + R 3 = 5 . 4 Ω + 6 . 48434 Ω + 4 . 2 Ω = 16 . 0843 Ω . 002 (part 2 of 2) 10.0 points b) Find the current in the 5 . 4 Ω resistor. Correct answer: 1 . 61648 A. Explanation: Let : Δ V = 26 V Δ V = IR = I 1 = I I 1 = Δ V R eq = 26 V 16 . 0843 Ω = 1 . 61648 A . 003 (part 1 of 3) 10.0 points A circuit contains six 190 Ω lamps and a 9 Ω heater connected in parallel. The voltage across the circuit is 126 V. What is the current in the circuit when four lamps are turned on? Correct answer: 2 . 65263 A. Explanation:

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johnson (rj6247) – hw 8 – Opyrchal – (121014) 2 Let : R l = 190 Ω , R h = 9 Ω , and V = 126 V . For the parallel circuit, the voltage is the same across each branch and the currents add up. Thus each branch receives V volts so that the current for each lamp is I l = V R 1 . With 4 lamps in the circuit, I = 4 I l = 4 V R l = 4 (126 V) 190 Ω = 2 . 65263 A . 004 (part 2 of 3) 10.0 points What is the current when all lamps are on? Correct answer: 3 . 97895 A. Explanation: With six lamps in the circuit, I 2 = 6 I l = 6 V R l = 6 (126 V) 190 Ω = 3 . 97895 A . 005 (part 3 of 3) 10.0 points What is the current if six lamps and the heater are operating? Think about this situation with a 12 A fuse in the line.
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