quiz5solutions

quiz5solutions - Quiz 5 Solution Guide Here are the answers...

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Quiz 5 Solution Guide Here are the answers and solution guide for quiz 5. Please contact me if you have any questions. Jason [email protected] Quiz 5 Section A Version A Version B Version C Version D 1 B A D C 2 A B A C 3 C B A A 4 E C A C 5 D B B E 6 C D E D Quiz 5 Section B Version A Version B Version C Version D 1 C C B B 2 B B B D 3 E A E B 4 A E A B 5 B D A A 6 A B E B
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Section A Solutions 1. Solve this expression for drift current, I = nAvQ where n is the density of the charge carriers, A is the cross sectional area of the conductior, v is the drift velocity and Q is the charge of each carrier. 2. Remember, R = ρ *L/A. In this instance, A is proportional to r 2 (d 2 ) so doubling both the length and the diameter of the wire will halve the resistance. 3. Remember the power dissipated by a resistor is P = IV. The battery has internal resistance (4 ohms), so the voltage drop across the resistor that completes the circuit will be 36V-4.0 A*4ohms = 20V. Hence P = 20V*4A=80 watts.
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This note was uploaded on 04/06/2008 for the course PHYS 2b taught by Professor Schuller during the Spring '08 term at UCSD.

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quiz5solutions - Quiz 5 Solution Guide Here are the answers...

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