Question

Physics 1125 Monday Challenge Homework 7: RC circuits. Due on Monday March 9, 2020 at 8PM Submit a PDF scan of your solution to the PHYS 1125 Canvas site. In this homework you will solve the same kind of first order differential equation you worked with last week. You can refer to that solution, you can even use Mathematica to do the work for you. Note that Mathematica can take care of the initial conditions as well, further simplifying what you have to do. This let's you concentrate on the physics! 1. Every bit stored in memory in your computer or cell phone is, at its core, a capacitor that stores charge. The value of the bit can be determined by measuring the voltage across the capacitor C. A one (1) bit has a voltage of approximately 1 volt, and a zero (0) bit has a voltage of approximately 0 volts. The real circuit is quite complicated, with the charge on the capacitor controlled by multiple transistors that control the flow of current in the circuit and measure the bit value. S1 Figure 1: RC circuit. (a) The chip circuitry can change the bit from 1 to 0 by closing switch S at time t = 0 (with S2 kept open) and discharging the capacitor through the resistor R. Use Kirchhoff's laws to determine an equa- tion for the voltage across the capacitor as a function of time after the switch is closed. Use Kirchhoff's voltage-drop rule around the right loop of the circuit to show that the differential equation for the charge on the capacitor can be written (1) dt RC where the direction of current flow is chosen so that a positive cur- rent I = is defined so as to increase the charge on the capacitor. Determine the units of the quantity RC in two ways: first by exam- ining the units of the physical quantities in the differential equation (1), and separately by expressing the units of resistance and capac- itance in SI units. What is the numerical value of RC for the case C = 100 aF = 100 x 10-18 F and R = 100 k2? (b) The differential equation (1) is homogenous and has the same form as the equation we solved for the drift velocity. Solve the equation

Answer 1

WAT 1(a) Given - Direction of current I=do. CHE at is so as to increase the charge on capacitor. tence of current flow B C DEB that clock wise direction then charge on capacitor will decrease. Hence - Sign is used along I. Now Using Kirchoff's Voltage drop sale in BCDEB. Ve tv = 0 [ Potential potential drop difference in Capatitor atrass R

dt -IR, 7 (-0-e so v=iR IQ=017 IR, to a age Ri+Q- Fi Izde 7 auche + c = = 0 Now do represent current hence its unit is Ampere. Hence by principle of Homogeneity, a should have unit of ampere. Ampere unit of a = Ampere Unit of t = Ampere confort unult of the Angere fordelen unit of R, C = 8 1 Separately unit of RC = AF. RC = look f-x 100 x 108 F = louxo x 100x10 sf do co = 10-12F Ric second -

integrating both sides do = Stadt - 1 do - 1 Set loga = - Act tot at t-o a=0o=107be put in above ean log 10th = - 1 183 loker lot log 1076 = 6 log o - - 1 t t dog 10 6-16 - - t too Looo HC logo - log 10 by = polymotion co= 10-16 RC

(t) - Alt) ve (t) = 1016 10t 100701 When Ve=0.50 0.5= -lo't vo logo.s= dog 107 n logos = -lo't loge logos = -10 t t = logos -10!! += 0.301 Xo's t ) Applying Kirchoff's Voltage I Rule in ABEFA VT To V t = 1, 1 R ++Q) = V 2

at C 1dQ + Q © at R₂C Integrating factor I.t. = e IF = multiply Q by T.F t F2C it to the fare my felicitate at sides (V tot - cui pect to at t=0 a=0 put in above eah -O ell - Cult

o = cu + c -chi-o put in ② elect = co che con a = cuices - 1) 100X163 % look roll Put all known values A = 100008 I (( - Texto x Tooxits) To = 166 (10) V, (t) = 3) Vft) = 10% (1 700e) Velt) = 1- etolt toox 1018 T When Velt) = 0.50 05=1-6 tot Vict) loft lo't = 0.5 loge lot - 1705 - lot = logor 1 t= logos loge (t = 0.301x10ls