ID: A5UCSD Physics 2B Unit Exam 3A Current, Resistance, CircuitsAnswer SectionMULTIPLE CHOICE1. ANS: BEach differential area element dAof a circle (cross section) is a ring of thickness drand length (circumference) 2πr. The area of this ring is the product, dA=2πr drI=J•dA∫=J r( )2πr dr0R∫=Kr2ÊËÁÁÁˆ¯˜˜˜2πr dr0R∫=2πKr3dr0R∫=2πKR44=123.14()122Am4ÊËÁÁÁÁÁÁˆ¯˜˜˜˜˜˜166×10−2mÊËÁÁÁˆ¯˜˜˜4=1.45 kATOP: CURRENT DENSITY + INTEGRATION 2. ANS: DThis is a series combination. When all resistors are identicalRSER=NR=3×33Ω =99ΩTOP: SERIES3. ANS: DThis is basically a unit conversion problem if you think about it:Power=energytime⇒energy=power×timeE=P×t=V×I()×t=12.0JCÊËÁÁÁÁÁÁˆ¯˜˜˜˜˜˜5.00CsÊËÁÁÁÁÁÁˆ¯˜˜˜˜˜˜3.00 min()60 sminÊËÁÁÁÁÁÁˆ¯˜˜˜˜˜˜=1.08×104J=10.8kJTOP: ENERGY4. ANS: BImportant to keep track of direction here. Since the protons and electrons are moving in the same direction, you must subtract the electron contribution to the flux since the charge is negative. Calculate the number of particles times the charge/particle and then divide by timeI=Qt=npe++nee−t=npe+−nee+t=np−neÊËÁÁˆ¯˜˜e+t=6.5×1017−3.5×1017ÊËÁÁÁˆ¯˜˜˜1.6×10−19CÊËÁÁÁˆ¯˜˜˜1s=0.048A=48mATOP: CURRENT
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