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Version 001 – Circuits I – tubman – (2016winter1) This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Circuits 03 001 10.0 points When electrons move back and forth, reversing their direction regularly, the current is called 1. direct current. 2. a short circuit. Let : I = 13.0 × 10−3 A ∆Q = 15.0 C . 1 and ∆Q ∆t 15 C ∆Q = = 1153.85 s . ∆t = I 0.013 A I= 3. alternating current. correct 4. induced current. Explanation: Meters 02 002 10.0 points An instrument used to detect the current in a circuit is called 1. an ammeter. correct 2. a motor. 3. a voltmeter. Hewitt CP9 23 03 004 10.0 points What is drift velocity? 1. The lowest speed of an electron in a metal 2. The speed of an electric field 3. The average speed of atoms in a liquid 4. The average speed of electrons in a conductor in an electric field correct 5. The highest speed of an electron in a metal 4. a transformer. 5. an ohmmeter. 6. an electroscope. 7. a generator. Explanation: Holt SF 19A 01 003 10.0 points If the current in a wire of a CD player is 13.0 mA, how long would it take for 15.0 C of charge to pass a point in this wire? Correct answer: 1153.85 s. Explanation: Explanation: When an electric field is applied to a conductor, the electrons continue their random motions while simultaneously being nudged by this field. The collisions interrupt the motion of the electrons along the field lines. The average speed at which they migrate along a wire is known as drift velocity. Comparisons in Conductors 01 005 (part 1 of 2) 10.0 points Consider two conductors 1 and 2 made of the same ohmic material; i.e., ρ1 = ρ2 . Denote the length by ℓ, the cross sectional area by A. The same voltage V is applied across the ends of both conductors and the field E is inside of the conductor. Version 001 – Circuits I – tubman – (2016winter1) ~E 1 ~E 2 I1 V1 I2 3. V2 4. b b ℓ1 r1 ℓ2 5. r2 If A2 = 2 A1 , ℓ2 = 2 ℓ1 and V2 = V1 , find E2 of the electric fields. the ratio E1 E2 1. =1 E1 E2 2. =8 E1 1 E2 = 3. E1 4 E2 1 4. = E1 8 E2 1 5. = correct E1 2 E2 =4 6. E1 E2 7. =2 E1 1 E2 = 8. E1 12 E2 1 9. = E1 16 E2 1 10. = E1 3 Explanation: V 1 E= ∝ when V1 = V2 , so ℓ ℓ 6. 7. 8. 9. 10. I2 I1 I2 I1 I2 I1 I2 I1 I2 I1 I2 I1 I2 I1 I2 I1 2 1 4 1 = 2 1 = 3 1 = 12 = =8 =4 = 1 correct = 1 8 Explanation: V 1 ℓ ℓ I = ∝ and R = ρ ∝ when R R A A V1 = V2 and ρ1 = ρ2 , so R1 ℓ 1 A2 ℓ1 (2 A1) I2 = = = = 1. I1 R2 ℓ 2 A1 (2 ℓ1 ) A1 Tipler PSE5 25 43 007 10.0 points A carbon rod with a radius of 1 mm is used to make a resistor. What length of the carbon rod should be used to make a 15.8 Ω resistor? The resistivity of this material is 7.7 × 10−5 Ω · m . Correct answer: 644.638 mm. Explanation: E2 ℓ1 ℓ1 1 = = = . E1 ℓ2 2 ℓ1 2 006 (part 2 of 2) 10.0 points I2 of the currents. Find the ratio I1 I2 1. =2 I1 I2 1 2. = I1 16 Let : r = 1 mm = 0.001 m , ρ = 7.7 × 10−5 Ω · m , R = 15.8 Ω . and The cross-sectional area of the rod is A = π r 2 = π(0.001 m)2 = 3.14159 × 10−6 m2 and Version 001 – Circuits I – tubman – (2016winter1) 3 The resistance is ρℓ A AR ℓ= ρ (3.14159 × 10−6 m2 )(15.8 Ω) 103 mm = · 7.7 × 10−5 Ω · m 1m R= = 644.638 mm . Serway CP 17 16 008 10.0 points A length of copper wire has a resistance 22 Ω. The wire is cut into three pieces of equal length, which are then connected as parallel lengths between points A and B. What resistance will this new “wire” of L0 length have between points A and B? 3 R= Current in Tungsten Wire 010 10.0 points A 0.98 V potential difference is maintained across a 1.8 m length of tungsten wire that has a cross-sectional area of 0.49 mm2 . What is the current in the wire? The resistivity of the tungsten is 5.6 × 10−8 Ω · m . Correct answer: 4.76389 A. Explanation: Let : Correct answer: 2.44444 Ω. Explanation: Let : R0 = 22 Ω . L0 and crossThe new wire has length L = 3 sectional area A = 3 A0 , so its resistance is L0 ρ ρL 1 ρ L0 3 R= = = A 3 A0 9 A0 R0 22 Ω = = = 2.44444 Ω . 9 9 Current in a TV 009 10.0 points A typical color television draws about 2.6 A when connected to a 106 V source. What is the effective resistance of the T.V. set? Correct answer: 40.7692 Ω. Explanation: Let : V 106 V = = 40.7692 Ω . I 2.6 A V = 0.98 V , A = 0.49 mm2 = 4.9 × 10−7 m2 , ℓ = 1.8 m , and ρ = 5.6 × 10−8 Ω · m . The resistance is V ρℓ = I A VA (0.98 V)(4.9 × 10−7 m2 ) I= = ρℓ (5.6 × 10−8 Ω · m)(1.8 m) R= = 4.76389 A . Ohms Law 01 011 10.0 points According to Ohm’s Law, if the resistance in a circuit is 50 Ω and the voltage is 7.1 V, what will be the current flow in the circuit? 1. 57.1 A 2. 355 A 3. 0.142 A correct 4. 7.04225 A I = 2.6 A and V = 106 V . 5. 42.9 A Explanation: Version 001 – Circuits I – tubman – (2016winter1) Let : R = 50 Ω and V = 7.1 V . From Ohm’s Law, ∆V = I R, so I= V 7.1 V = = 0.142 A . R 50 Ω Electric Shock 012 (part 1 of 3) 10.0 points The damage caused by electric shock depends on the current flowing through the body; 1 mA can be felt and 5 mA is painful. Above 15 mA, a person loses muscle control, and 70 mA can be fatal. A person with dry skin has a resistance from one arm to the other of about 1.1 × 105 Ω. When skin is wet, the resistance drops to about 5800 Ω. What is the minimum voltage placed across the arms that would produce a current that could be felt by a person with dry skin? Correct answer: 110 V. Explanation: Let : Imin = 1 mA and Rdry = 1.1 × 105 Ω . The minimum voltage depends on the minimum current for a given resistance, so Vmin = Imin Rdry = (1 mA) 1A 1000 mA (1.1 × 105 Ω) = 110 V . 013 (part 2 of 3) 10.0 points For the same electric potential what would be the current if the person had wet skin? Correct answer: 18.9655 mA. Explanation: Let : Rwet = 5800 Ω . 110 V Vmin = I= Rwet 5800 Ω 4 1000 mA 1A = 18.9655 mA . 014 (part 3 of 3) 10.0 points What would be the minimum voltage that would produce a current that could be felt when the skin is wet? Correct answer: 5.8 V. Explanation: V1 = Imin Rwet = (1 mA) = 5.8 V . 1A (5800 Ω) 1000 mA