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CU-Boulder PHYS 1120 - RC Circuits

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RC-1RC CircuitsAn RC circuit is a circuit with a resistor (R) and a capacitor (C). RC circuits are used to construct timers and filters. Example 1. Very simple RC circuit: a capacitor C, charged to an initial voltage V0 = Q0/C, attached to a resistor R with a switch. 00QQC VV C= � =Close the switch at time t = 0 , so current I starts to flow. The charged capacitor is acting like a battery: it produces a voltage difference across the resistor which drives the current through the resistor:At t = 0+, 00VIR= . d QId t=- (– sign because Q is decreasing)Vacross C = Vacross R , VC = VR , QI RC=, Q d QRC d t= - � , d Q 1Qd t RC= - RC = "time constant" =  , has units of time is a differential equation of the form d xa xd t= �, where a is a constant. This equation says: (rate of charge of x)  x  exponential solution : ( )0x x exp a t=Check: ( )a t0a t0d x ed xa x e a xd t d t�= = � � = � . It works!a > 0  exponential growth , a < 0  exponential decayThe solution to d Q 1Qd t RC= - is 0 0t tQ(t) Q exp CV expRC� � � �= - = -� � � �t� � � � Notice that at t = 0, the formula gives Q = Q0 .In time  , Q falls by a factor of exp(–1) = 1/e  0.37..Last update: 1/13/2019 Dubson Phys1120 Notes, University of Colorado R C +Q0–Q0switchR C +Q–QIRC-2In time 2, Q falls by a factor of exp(–2) = (1/e)(1/e)  0.14.. Q approaches zero asymptotically, and so does V and I 0 0t tQ(t) Q exp Q expRC� � � �= - = -� � � �t� � � �Example 2: More complex RC circuit: Charging a capacitor with a battery.Let's use symbol E for battery voltage (E short for emf) because there are so many other V's in this example.Before switch is closed, I = 0, Q = 0.Close switch at t = 0.Always true that E = VC + VR , by Kirchhoff's Voltage Law (Loop Law)Last update: 1/13/2019 Dubson Phys1120 Notes, University of Colorado V = Q / Ct0VdQI exp( t / )dt R=- =+ - ttQtQ0Q0/eR C VR = I REswitchVC = Q / CRC-3The charge Q on the capacitor and the voltage VC = Q / C across the capacitor cannot change instantly, since it takes time for Q to build up, so ..At t = 0+ , Q = 0 , VC = 0, E = VC + VR = VR = I R  I0 = E / RAlthough Q on the capacitor cannot change instantly, the current I = dQ/dt can change instantly."Current through a capacitor" means dQ/dt . Even though there is no charge ever passing between the plates of the capacitor, there is a currentgoing into one plate and the same current is comingout of the other plate, so it is as if there is a currentpassing through the capacitor. R C= V VQ d Q= I R , IC d td Q Q= Rd t C++ =++EEE Qualitatively, as t  , Q  , VC = Q/C  , VR  , I = VR / R . After a long time , t >> = RC ,the current decreases to zero: I = 0, VC = E , Q = C E Analytic solution:CV (t) [1 exp( t / RC) ]= - -EQ(t) C [1 exp( t / RC) ]= � - -EThings to remember: Uncharged capacitor acts like a "short" ( a wire ) since VC = Q / C = 0. After a long time, when the capacitor is fully charged, it acts like an "open-circuit" ( a break the wire). We must have IC = 0 eventually, otherwise Q   , VC   .Last update: 1/13/2019 Dubson Phys1120 Notes, University of Colorado II++++----R C EIVc = Q / CtERC-4Last update: 1/13/2019 Dubson Phys1120 Notes, University of


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CU-Boulder PHYS 1120 - RC Circuits

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