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PhyIIHW5Solutions - hopkins(tlh982 HW 05 criss(4908 This...

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hopkins (tlh982) – HW 05 – criss – (4908) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 3 . 1 cm length of tungsten filament in a small lightbulb has a resistance of 0 . 031 Ω. Find its diameter. (The resistivity is 5 . 6 × 10 8 Ω · m). Correct answer: 0 . 267023 mm. Explanation: Let : L = 3 . 1 cm = 0 . 031 m , R = 0 . 031 Ω , and ρ = 5 . 6 × 10 8 Ω · m . Resistance is R = ρ L A = ρ L π parenleftbigg d 2 parenrightbigg 2 = 4 ρ L π d 2 , d = radicalbigg 4 ρ L π R = radicalBigg 4 (5 . 6 × 10 8 Ω · m) (0 . 031 m) π (0 . 031 Ω) × parenleftbigg 1000 mm 1 m parenrightbigg = 0 . 267023 mm . 002 10.0 points A length of copper wire has a resistance 32 Ω. The wire is cut into three pieces of equal length. The pieces are then connected as parallel lengths between points A and B. What resistance will this new “wire” of length L 0 3 have between points A and B? Correct answer: 3 . 55556 Ω. Explanation: Let : R 0 = 32 Ω . The new wire has length L = L 0 3 and cross- section A = 3 A 0 , so its resistance is R = ρ L A = ρ parenleftbigg L 0 3 parenrightbigg 3 A 0 = 1 9 parenleftbigg ρ L 0 A 0 parenrightbigg = R 0 9 = 32 Ω 9 = 3 . 55556 Ω . 003 10.0 points A 11 V battery delivers 116 mA of current when connected to a 74 Ω resistor. Determine the internal resistance of the battery. Correct answer: 20 . 8276 Ω. Explanation: Let : V = 11 V , I = 116 mA = 0 . 116 A , and R = 74 Ω . The internal resistance is in series with the given resistor, so V = I ( R + r ) r = V I - R = 11 V 0 . 116 A - 74 Ω = 20 . 8276 Ω . 004 10.0 points A platinum resistance thermometer has a re- sistance of 200 Ω when placed in a ice 0 C bath and 169 Ω when immersed in a crucible containing a melting substance. What is the melting point of the substance? The temperature coefficient of the platinum
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hopkins (tlh982) – HW 05 – criss – (4908) 2 is 0 . 00392( C) 1 . Correct answer: - 36 . 4408 C. Explanation: Let : T 0 = 20 C , R 1 = 200 Ω , T 1 = 0 C , R = 169 Ω , and α = 0 . 00392( C) 1 . The resistance at 20 C is R 0 = R 1 1 + α ( T 1 - T 0 ) = 200 Ω 1 + [0 . 00392( C) 1 ] (0 C - 20 C) = 217 . 014 Ω , and R = R 0 [1 + α ( T - T 0 )] = R 0 + R 0 α ( T - T 0 ) T - T 0 = R - R 0 α R 0 T = T 0 + R - R 0 α R 0 T = 20 C + 169 Ω - 217 . 014 Ω [0 . 00392( C) 1 ] (217 . 014 Ω) = - 36 . 4408 C . 005 (part 1 of 4) 10.0 points An energy plant produces an output potential of 4300 kV and serves a city 176 km away. A high-voltage transmission line carries 690 A to the city. The effective resistance of a trans- mission line [wire(s)] is 4 . 68 Ω / km times the distance from the plant to the city. What is the potential provided to the city, i.e. , at the end of the transmission line? Correct answer: 3731 . 66 kV. Explanation: Let : V plant = 4300 kV , = 176 km , I = 690 A , and ρ = 4 . 68 Ω / km . The potential drop on the wire is V = I R = I ρ ℓ = (690 A) (4 . 68 Ω / km) (176 km) = 568 . 339 kV , so the potential delivered to the city is V city = V plant - V wire = 4300 kV - 568 . 339 kV = 3731 . 66 kV . 006 (part 2 of 4) 10.0 points How much power is dissipated due to resistive losses in the transmission line?
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