spring_06_exam2_sol_phy2049

spring_06_exam2_sol_phy2049 - PHY2049 Spring 2006 Exam 2...

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PHY2049 Spring 2006 Exam 2 Solutions Prof. Darin Acosta Prof. Greg Stewart March 27, 2006 Exam 2 Solutions a ______ _____________ b ____________ __________ | 5 Ω ->i 1 | <- i 2 5 Ω | 3V 10 Ω 6 V - | - | | i 3 | d ___________________________ c ___________ _____________ 10 Ω 1. In the multiloop circuit shown, what is the current in the left hand loop (abcd), in units of Amperes? a. 3/55 b. 3/11 c. 12/55 d. 0 e. 9/55 Solution: Use Kirchhoff's rules (sum of currents at a point = 0 and sum of EMF's around a loop = 0) to obtain: (direction of currents chosen - remember these can be 180 o wrong, which will give a minus sign in the solution - are shown in red ) eq. #1: i 1 + i 2 = i 3 (sum of currents at point b) eq. #2: -5i 1 - 10i 3 +3V=0 (sum of EMF's around left hand loop starting at point a and going clockwise) eq. #3: +5i 2 - 6V +10i 2 + 10i 3 = 0 (sum of EMF's around right hand loop starting at point b and going clockwise) Solve these 3 eqns for i 1 : #3 15 i 2 + 10i 3 =6 #2 10 i 3 = -5i 1 + 3 put this in eq. above * 15i 2 + (-5i 1 + 3) = 6 also, put eq. above as i 3 = (1/10)(-5i 1 + 3) in eq. #1 to get ** i 1 + i 2 = (1/10)(-5i 1 + 3) use * to express i 2 = (6 -3 +5i 1 )/15 and put in ** to get i 1 + 1/5 + (1/3) i 1 = -(1/2)i 1 + 3/10 solve for i 1 , get i 1 = 3/55 Amp However, the wording of the problem was found to be ambiguous, as the current through the middle branch (10 ohm resistor) is not the same as the left current (through the 3V battery). Both were marked correct, so if one chose i 3 =3/11 that is also correc t. 1

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PHY2049 Spring 2006 Exam 2 Solutions ⎯⎯⎯⎯ / ⎯⎯⎯ | | | q 0 | |____________________| 2. A capacitor starts fully charged with charge q 0 . It is then, through the closing of a switch, discharged through a series resistor, see circuit in Figure. If C=10 μ F and R=100 Ω , how long in units of milliseconds (1 millisecond = 10 -3 s) does it take the charge on the capacitor to reach 0.5 q 0 ? (1 μ F = 10 -6 F) a. 0.69 b. 0.63 c. 0.37 d. 0.31 e. 0.5 Solution: q = q 0 e -t/RC if want q/q 0 = 0.5, then ln(0.5) = -t/RC, or -0.6931 = -t/(100* 10 -5 ) or t=0.69 10 -3 s, so answer (in units of milliseconds) is 0.69. 3. An electron is accelerated from rest by a potential difference of 20,000V.
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This test prep was uploaded on 04/17/2008 for the course PHY 2054 taught by Professor Avery during the Spring '08 term at University of Florida.

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spring_06_exam2_sol_phy2049 - PHY2049 Spring 2006 Exam 2...

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