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Unformatted text preview: Solution for Practice Test 5 for Test 2 Solution to Practice Test Problem 5.1(Potential Difference Along x Axis) Problem: A region of space contains an electric field vector E = x 2 x where = +150 N Cm 2 . Compute the potential difference, V ab (including the appropriate sign) between the point a = 1m and the point b = 2m along the x-axis. Select One of the Following: (a) +350V (b) +3150V (c)- 3150V (d-Answer)- 350V Solution To move from point a to point b , you move in the direction of the field line, so the potential is negative. By the definition of electric potential, | V ab | = vextendsingle vextendsingle vextendsingle vextendsingle- integraldisplay a b vector E d vector vextendsingle vextendsingle vextendsingle vextendsingle In this case, d vector = xdx . Substitute, | V ab | = vextendsingle vextendsingle vextendsingle vextendsingle- integraldisplay b a ( x 2 x ) ( xdx ) vextendsingle vextendsingle vextendsingle vextendsingle Since x x = 1 , this simplifies to | V ab | = vextendsingle vextendsingle vextendsingle vextendsingle integraldisplay b a ( x 2 dx vextendsingle vextendsingle vextendsingle vextendsingle = parenleftbigg x 3 3 vextendsingle vextendsingle vextendsingle vextendsingle b a parenrightbigg | V ab | = 3 ( b 3- a 3 ) Since we have already figured out the sign, V ab =- 3 ( b 3- a 3 ) =- 150 N Cm 2 3 ((2m) 3- (1m) 3 ) V ab =- 350V which was choice d . Total Points for Problem: 3 Points Solution to Practice Test Problem 5.2(Series/Parallel Light Bulb Problem) 1 Problem: The figure to the right shows a circuit containing identical light bulbs connected to a perfect battery. At time t = 0 , bulb C is connected by closing the switch. Let I batt,closed be the current through the battery with the switch closed and I batt,open the current through the battery when the switch is open. Also let B A,closed be the brightness of bulb A when the switch is closed and B A,open be the brightness of bulb A when the switch is open. Which of the following set of inequalities correctly describe the current through the battery and the brightness of bulb A . Select One of the Following: (a) I batt,before > I batt,after and B A,before < B A,after (b-Answer) I batt,before < I batt,after and B A,before < B A,after (c) I batt,before > I batt,after and B A,before > B A,after (d) I batt,before < I batt,after and B A,before > B A,after (e) I batt,before = I batt,after and B A,before = B A,after A B C Solution When the switch is closed the resistance of the parallel combination is decreased, so the total resistance of the system of light bulbs is decreased. Since the total resistance of the system is decreased, the battery delivers more current. All the battery current flows through bulb A , therefore bulb A is brighter with the switch closed....
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This note was uploaded on 05/04/2008 for the course PHYS 2074 taught by Professor Stewart during the Spring '08 term at Arkansas.
- Spring '08