Physics 241 Lab: RC Circuits – DC Sourcehttp://bohr.physics.arizona.edu/~leone/ua/ua_spring_2010/phys241lab.htmlName:____________________________Section 1:1.1. Today you will investigate two similar RC circuits. The first circuit is the charging up thecapacitor circuit. In this circuit (shown below) the capacitor begins without any charge on it and iswired in series with a resistor and a constant voltage source. The voltage source begins charging thecapacitor until the capacitor is fully charged. The charging up equation that describes the timedependence of the charge on the capacitor is € QCap(t) = Qmax1− e−tRC ⎛ ⎝ ⎜ ⎞ ⎠ ⎟. The final charge on thecapacitor, Qmax is determined by the internal structure of the capacitor (i.e. its capacitance):€ Qmax= C ⋅Vsource.Use a graphing calculator (or mad graphing skills) and make a quick sketch of € QCap(t) vs. t on theaxes. Assume that the source voltage is 9 V, the resistance is 1.0x103 W and the capacitance is 1.0x10-3F. The amount of time that equals the resistance times the capacitance is called the time constant:€ τ =R ⋅C. Create your sketch so that Q(t=t) is sketched above the delineated tic mark. Your sketchbelow:1.2. The second circuit is the discharging the capacitor circuit. In this circuit (shown below) thecapacitor begins with some initial charge and is wired in series with a resistor. The capacitor beginsdischarging through the resistor until no charge remains on the capacitor plates. The dischargingequation that describes the time dependence of the charge on the capacitor is € QCap(t) = Qoe−tRC. (Also think of a switch:Use a graphing calculator (or mad graphing skills) and sketch a graph of € QCap(t) vs. t on the axesbelow. Assume that the resistance is 1.0x103 W and the capacitance is 1.0x10-3 F. Find the initialcharge on the capacitor by assuming the capacitor had been charged to 9 volts by a battery before beingdischarged through the resistor. Create your sketch so that Q(t) is sketched above the delineated ticmark. Be sure to include charge values along the y-axis. Your sketch below:What is the decimal value of e-1 to 3 decimal places? _________ Engineers usually approximate thisnumber as 1/3 (.333) in order to think quickly about exponential decay. For example, if you plug in t=t(one decay time constant), the amount of charge left on the capacitor has decayed to approximately 1/3of its initial value. Approximately how much of the initial charge is left on the capacitor after thecircuit has operated for t=3t seconds? Your work and answer:1.3. Now examine the time dependence of the voltage across the capacitor for the same dischargingcapacitor in part c. As the charge on the capacitor changes, the voltage difference across the capacitorplates also changes. In fact, the definition of capacitance easily relates € VCap(t) and QCap(t) by aconstant: € VCap(t) =QCap(t)C. Therefore, the equation describing the time dependent decay of thevoltage across the capacitor is simply € VCap(t) = Voe−tRC, where € Vo=QoC. You will experimentallytest this equation later in this lab. Sketch a graph of € VCap(t) vs. t on the axes below using youranswer to the previous question (graph of QCAP). Be sure to include voltage values along the y-axis.Your sketch below:As the capacitor discharges, it causes a current to flow through the resistor. Because energy must beconserved, the magnitude of the voltage across the resistor is the same as the voltage across thecapacitor (they are the only circuit components!). Because the resistor is Ohmic, the current throughthe resistor can be related to its voltage and resistance. This gives a time dependent equation for thecurrent through the resistor of € IRes(t) = Ioe−tRC. You should notice that the time dependence of thecharge on the capacitor, the voltage across the capacitor, and the current through the resistor allexhibit the same exponential decay function, and are simply related to each other using properties youalready know. Relate this equation for resistor current to the others by using Ohm’s law to determineIo in terms of R, C and Qo. Your work and answer:Section 2:2.1. A differential equation is an equation that involves derivatives. Most all equations designed tomodel reality in the physical sciences make use of differential equations so a good working knowledgeof this type of mathematics is essential to any working physical scientist or engineer. The followingtable compares an algebraic equations to a differential equations using two examples:Examine the differential equation € d2y(t)dt2= −9y(t). One solution to this differential equation is€ y(t) = 4 sin 3t( ). Check the solution by plugging it into the differential equation to see if it works.Your work and answer:2.2. When analyzing circuits, you often must write a differential equation describing the behavior ofthe circuit. This is most easily done by using conservation of energy to write a voltage equation. Thenuse fundamental concepts to relate voltage to charge on the capacitor to create a differential equationfor Q(t). Examine the discharging circuit for today’s lab and the construction of the differentialequation that describes it:DISCHARGING WITHOUT SOURCE€ 0 = Vres(t) + Vcap(t) (conservation of energy)0 = R ⋅Ires(t) +Qcap(t)C (Ohm's law and definition of capacitance)0 = RdQcap(t)dt+Qcap(t)C (current through resistor is equal to rate of charging on capacitor)dQcap(t)dt= −Qcap(t)R ⋅C (rearrange to get final differential equation) SOLUTION: € QCap(t) = Qoe−tRCCheck the solution function by substituting it into both sides of the differential equation for Qcap(t).Differentiate where appropriate to prove the left-hand side of the equation equals the right-hand sidewith the solution substituted in for Qcap(t). Your work and answer:What trivially happens to the initial charge Qo in this checking process? Your answer to this shouldallow you to see that the initial amount of charge on the capacitor Qo is not determined by thedifferential equation. Basically you choose the initial amount of charge to put on the capacitor platesand the differential equations determines how quickly that charge discharges through the resistor.Your answer:2.3. The
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