RC-1 RC Circuits An 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. 00QQCVVC=⇒ = 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= . dQIdt=− (– sign because Q is decreasing) Vacross C = Vacross R , VC = VR , QIRC=, QdQRCdt=−⋅ , dQ 1Qdt RC=− RC = "time constant" = τ , has units of time is a differential equation of the form dxaxdt=⋅, where a is a constant. This equation says: (rate of charge of x) ∝ x ⇔ exponential solution : ()0xxexpat=Check: ()at0at0dx edxax e axdt dt⋅==⋅⋅=⋅ . It works! a > 0 ⇒ exponential growth , a < 0 ⇒ exponential decay The solution to dQ 1Qdt RC=− is 00ttQ(t) Q exp CV expRC⎛⎞ ⎛=−= −⎜⎟ ⎜τ⎝⎠ ⎝⎞⎟⎠ Notice that at t = 0, the formula gives Q = Q0 . In time τ , Q falls by a factor of exp(–1) = 1/e ≅ 0.37.. switch +Q0 C R –Q0R C +QI –Q Last update: 10/8/2009 Dubson Phys1120 Notes, ©University of ColoradoRC-2 In time 2τ, Q falls by a factor of exp(–2) = (1/e)(1/e) ≅ 0.14.. ⇒ Q approaches zero asymptotically, and so does V and I Q 00ttQ(t) Qexp QexpRC⎛⎞ ⎛=−= −⎜⎟ ⎜τ⎝⎠ ⎝⎞⎟⎠ Q0 Q0/e t τ 0VdQI exp( t / )dt R=−=+ −τ V = Q / C t 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. ©University of Colorado 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) The 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 / R Although 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 current going into one switch R C VC = Q / C E VR = I R - + I I - + - + - +Last update: 10/8/2009 Dubson Phys1120 Notes,RC-3 plate and the same current is coming out of the other plate, so it is as if there is a current passing through the capacitor. RC=V VQd=IR , ICddQ Q=Rdt C++=+EQt+EE C Qualitatively, as t ↑ , Q ↑ , VC = Q/C ↑ , VR ↓ , I = VR / R ↓. After a long time , t >> τ = RC , the current decreases to zero: I = 0, VC = E , Q = C E Analytic solution: CV (t) [1 exp( t / RC)]=−−E Q(t) C [1 exp( t / RC)]=⋅−−E Things 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 → ∞ . R E I Vc = Q / C E t Last update: 10/8/2009 Dubson Phys1120 Notes, ©University of
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