{VERSION 2 3 "APPLE_68K_MAC" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }1 0 0 0 6 6 0 0 0 0 0 0 -1 0 }{PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "Autho r" 0 19 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 8 8 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 26 "Computer Intensive Physic s" }}{PARA 18 "" 0 "" {TEXT -1 23 "Physics 212E: Fall 1997" }}}{EXCHG {PARA 19 "" 0 "" {TEXT -1 16 "Steven R. Dunbar" }}{PARA 19 "" 0 "" {TEXT -1 40 "Department of Mathematics and Statistics" }}{PARA 19 "" 0 "" {TEXT -1 30 "University of Nebraska-Lincoln" }}{PARA 19 "" 0 "" {TEXT -1 23 "Lincoln, NE, USA 68588" }}{PARA 19 "" 0 "" {TEXT -1 20 " sdunbar@math.unl.edu" }}{PARA 0 "" 0 "" {TEXT -1 32 "http://www.math.u nl.edu/~sdunbar" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 468 "\n6-1: RC Cir cuits, \n\nKEYWORDS: RC Circuits, Voltage, Current, Decay Curve, Time \+ Constant\n\nOBJECTIVE: Write the differential equation that results fr om applying the loop equation to a single loop containing both capacit ance and resistance and derive the solution to the differential equati on. For a single RC loop, write the equation for current, charge, or voltage of a capacitor as a function of time, and use the equation to derive the parameters of the solution.." }}{PARA 0 "" 0 "" {TEXT -1 204 "\nPREREQUISITE: Know the defintion of capacitance in terms of cha rge, know Ohm's law, know the laws for applying the loop equation, ele mentary knowledge of differnetal equations and initial value problems. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "LENGT H: One class period, about 40-50 minutes." }}{PARA 0 "" 0 "" {TEXT -1 59 "AUDIENCE: Instructor, student use in computer lab setting." }} {PARA 0 "" 0 "" {TEXT -1 21 "SOFTWARE: Maple V R 4" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 152 "FUTURE: Add the figures of the circuits, and also add some further explanations and tips and \+ pointers on setting and solving the differnetial equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "HISTORY: Created October 8, 1997" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 143 "Suggested InfoMall Reading: Dudley Williams and \+ John Spangler, Physics for Science and Engineering, Suggested Keywords : RC circuit, capacitor" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 46 "Solving for Capacitor Charge in the RC-Ci rcuit" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 221 "Consider the RC circuit that you studied in the lab. A \+ capacitor is charged with a battery and then immediately switched to a resistor in series. We want to mathematically analyze the charge and current in this circuit." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 107 "Write th e equation for the voltage difference across the resistor of resistanc e R in terms of the current i" }}{PARA 0 "" 0 "" {TEXT -1 119 " flowin g through it. (Remember Ohm's law!) (Current flows through the resist or due to the discharge of the capacitor.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 125 "Note: I in Maple is reserved \+ for the imaginary constaant, the square root of -1. Therefore we use \+ lower-case i for current." }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "Vr := ;" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "Write the voltage in terms of the charge in the capacitor and its capacitance C." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "Vc := ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 175 "Write the circuit equation that decribes the total vol tage around the loop. (Remember that an equation must have an equal s ign in it. We will call this equation voltage_eqn)" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "voltage_e qn := ; " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 260 "The charge on the capacitor is changing with time. Expl icitly recognize this fact by substituting in charge as a function of \+ time into the voltage drop equation. (Fill in the blanks in the follo wing substitution command, and reassign to the name voltage_eqn)" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "voltage_eqn := subs( = q(t), );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 169 "Now use the definition o f current as rate of change of charge to express the current I in term s of q(t). (Fill in te blaanks and reassign again to the name voltage_ eqn.)" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "voltage_eqn := subs( = diff( q(t), t), );" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 124 "N ow assign (mathematically, arbitrarily) the initial amount of charge i n the capacitor at the moment of throwing the switch." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "init_con d := q(0)= q0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 156 "Finally, solve the resulting symbolic differential \+ equation by using Maple. (Fill in the blanks to get a properly format ted differential equation problem.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 " soln := dsolve( \{ , \+ \}, );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "charge := rhs (soln);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 153 "Discussion Question: What is the mathematical form of t he solution, that is what kind of function is it? Answer in a concise , complete sentence or two." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 173 "Of course, we can 't measure the charge very easily, but we can measure the current. Di fferentiate the charge with respect to time and get the current as a f unction of time." }}{PARA 0 "" 0 "" {TEXT -1 1 "," }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 20 "current := diff(, );" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 101 "How long does it take for the current to decrease to 1/2 of its original value? Solve symbolically." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "half_time :=" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 145 "Substitu te in the values of C = 0.47 micro-Farad and R = 2000 Ohms, to derive \+ the time it take for the circuit you studied to be half-discharged." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "26 0 0" 12 }{VIEWOPTS 1 1 0 1 1 1803 }