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Exam2_2007Spring_Solutions

Exam2_2007Spring_Solutions - MASSACHUSETTS INSTITUTE OF...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2007 Exam Two Spring 2007 Solutions PART I Question 1: (5 points) An ideal battery is hooked to a light bulb with wires (see diagram). A second identical light bulb is connected in series with the first light bulb. After the second light bulb is connected, the current from the battery compared to when only one bulb was connected is: 1. higher by a factor of two 2. the same 3. lower by a factor of two 4. lower by a factor of four Solution: 3. The light bulbs are identical so the resistances are equal 1 2 R R R = . The original current when light bulb is connected is / I R ε = . After the second light is connected, the total resistance 2 R R ′ = doubles hence the current decreases by a factor of two, / 2 / 2 I R I ε ′ = = . 1

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Question 2: (5 points) In the multi-loop circuit shown on the right, which of the following set of equations is correct. Solution: 3 is the correct solution . 1. 1. 2 3 1 2 2 1 1 1 2 2 1 1 3 3 0 0 I I I I R I R I R I R ε ε ε + = + + = = The first equation is incorrect and should be 1 2 3 I I I + = .The second equation is incorrect and should be 2 2 1 1 1 2 0 I R I R ε ε + = . 2. 2 3 1 2 2 1 3 3 2 1 1 3 3 0 0 I I I I R I R I R I R ε ε + = = + = The first equation is incorrect and should be 1 2 3 I I I + = .The third equation is incorrect and should be 2 1 1 3 3 0 I R I R ε = . 3. 1 2 3 2 2 1 1 1 2 2 1 1 3 3 0 0 I I I I R I R I R I R ε ε ε + = + = = CORRECT 2
4. 2 3 1 2 2 1 3 3 2 1 1 3 3 0 0 I I I I R I R I R I R ε ε + = + + = + = The first equation is incorrect and should be 1 2 3 I I I + = .The second equation is incorrect and should be 2 2 1 3 3 0 I R I R ε = . The third equation is also incorrect and should be 2 1 1 3 3 0 I R I R ε = . 5. 1 2 3 2 2 1 1 1 2 2 1 1 3 3 0 0 I I I I R I R I R I R ε ε ε + = + + = + = The second equation is incorrect and should be 2 2 1 1 1 2 0 I R I R ε ε + = . The third equation is incorrect and should be 2 1 1 3 3 0 I R I R ε = . 3

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Question 3: (5 points) In Experiment 6, Faraday’s Law, a coil moves up from underneath a magnet (with its north pole pointing upward) and then back to its starting point. In the figures below, positive current is in the counterclockwise direction when looking down on the coil. (A) (B) (C) (D) Moving from below to above and back, you measured a current of: 1) A then A 5) B then B 2) C then C 6) D then D 3) A then C 7) B then D 4) C then A 8) D then B Solution: 2. As the coil approaches the magnet from below the magnetic flux is pointing up and increasing. Therefore a current flows clockwise (negative current by the choice of sign convention) to produce downward pointing magnetic field to oppose the change.
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