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| {{Unreferenced|date=March 2010}}
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| {{Linear analog electronic filter|filter1=hide|filter2=hide}}
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| A '''resistor–capacitor circuit''' ('''RC circuit'''), or '''RC filter''' or '''RC network''', is an [[electric circuit]] composed of [[resistors]] and [[capacitors]] driven by a [[voltage source|voltage]] or [[current source]]. A first order RC circuit is composed of one resistor and one capacitor and is the simplest type of RC circuit.
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| RC circuits can be used to filter a signal by blocking certain frequencies and passing others. The two most common RC filters are the [[high-pass filter]]s and [[low-pass filter]]s; [[band-pass filter]]s and [[band-stop filter]]s usually require [[RLC filter]]s, though crude ones can be made with RC filters.
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| ==Introduction==
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| There are three basic, linear passive [[Lumped element model|lumped]] [[analog circuit]] components: the resistor (R), the capacitor (C), and the [[inductor]] (L). These may be combined in the RC circuit, the [[RL circuit]], the [[LC circuit]], and the [[RLC circuit]], with the abbreviations indicating which components are used. These circuits, among them, exhibit a large number of important types of behaviour that are fundamental to much of [[analog electronics]]. In particular, they are able to act as [[electronic filter#Passive filters|passive filters]]. This article considers the RC circuit, in both [[series and parallel circuits#Series circuits|series]] and [[series and parallel circuits#Parallel circuits|parallel]] forms, as shown in the diagrams below.
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| :''This article relies on knowledge of the complex [[Electrical impedance|impedance]] representation of [[capacitor#AC circuits|capacitors]] and on knowledge of the [[frequency domain]] representation of signals''.
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| == Natural response ==
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| [[File:Discharging_capacitor.svg |200px|thumb|right| RC circuit]]
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| The simplest RC circuit is a capacitor and a resistor in [[series circuit|series]]. When a circuit consists of only a charged capacitor and a resistor, the capacitor will discharge its stored energy through the resistor. The voltage across the capacitor, which is time dependent, can be found by using [[Kirchhoff's circuit laws#Kirchhoff's current law (KCL)|Kirchhoff's current law]], where the current through the capacitor must equal the current through the resistor. This results in the [[linear differential equation]]
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| :<math>
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| C\frac{dV}{dt} + \frac{V}{R}=0
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| </math>.
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| Solving this equation for ''V'' yields the formula for [[exponential decay]]:
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| :<math>
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| V(t)=V_0 e^{-\frac{t}{RC}} \ ,
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| </math>
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| where ''V<sub>0</sub>'' is the capacitor voltage at time ''t = 0.''
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| The time required for the voltage to fall to <math>\frac{V_0}{e}</math> is called the [[RC time constant]] and is given by | |
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| :<math> \tau = RC \ . </math>
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| == Complex impedance ==
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| The [[complex impedance]], ''Z''<sub>''C''</sub> (in [[ohm]]s) of a capacitor with capacitance ''C'' (in [[farads]]) is
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| :<math>Z_C = \frac{1}{sC} </math>
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| The [[complex frequency]] ''s'' is, in general, a [[complex number]],
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| :<math>s \ = \ \sigma + j \omega </math>
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| where
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| *''j'' represents the [[imaginary unit]]:
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| :<math> j^2 = -1</math>
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| *<math>\sigma \ </math> is the [[exponential decay]] constant (in [[radians per second]]), and
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| *<math>\omega \ </math> is the [[sinusoidal]] [[angular frequency]] (also in radians per second).
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| ===Sinusoidal steady state===
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| Sinusoidal steady state is a special case in which the input voltage consists of a pure sinusoid (with no exponential decay). As a result,
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| :<math>
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| \sigma \ = \ 0
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| </math>
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| and the evaluation of ''s'' becomes
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| :<math>
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| s \ = \ j \omega
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| </math>
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| ==Series circuit==
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| [[File:RC Series Filter (with V&I Labels).svg|thumb|frame|right|[[series and parallel circuits#Series circuits|Series]] RC circuit]]
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| By viewing the circuit as a [[voltage divider]], the [[voltage]] across the capacitor is:
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| :<math>
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| V_C(s) = \frac{1/Cs}{R + 1/Cs}V_{in}(s) = \frac{1}{1 + RCs}V_{in}(s)
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| </math>
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| and the voltage across the resistor is:
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| :<math>
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| V_R(s) = \frac{R}{R + 1/ Cs}V_{in}(s) = \frac{ RCs}{1 + RCs}V_{in}(s)
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| </math>.
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| ===Transfer functions===
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| The [[transfer function]] from the input voltage to the voltage across the capacitor is
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| :<math>
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| H_C(s) = { V_C(s) \over V_{in}(s) } = { 1 \over 1 + RCs }
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| </math>.
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| Similarly, the transfer function from the input to the voltage across the resistor is
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| :<math>
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| H_R(s) = { V_R(s) \over V_{in}(s) } = { RCs \over 1 + RCs }
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| </math>.
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| ====Poles and zeros====
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| Both transfer functions have a single [[pole (complex analysis)|pole]] located at
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| :<math>
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| s = - {1 \over RC }
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| </math> .
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| In addition, the transfer function for the resistor has a [[zero (complex analysis)|zero]] located at the [[origin (mathematics)|origin]].
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| ===Gain and phase===
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| The magnitude of the gains across the two components are:
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| :<math>
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| G_C = | H_C(j \omega) | = \left|\frac{V_C(j \omega)}{V_{in}(j \omega)}\right| = \frac{1}{\sqrt{1 + \left(\omega RC\right)^2}}
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| </math>
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| and
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| :<math>
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| G_R = | H_R(j \omega) | = \left|\frac{V_R(j \omega)}{V_{in}(j \omega)}\right| = \frac{\omega RC}{\sqrt{1 + \left(\omega RC\right)^2}}
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| </math>,
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| and the phase angles are:
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| :<math>
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| \phi_C = \angle H_C(j \omega) = \tan^{-1}\left(-\omega RC\right)
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| </math>
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| and
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| :<math>
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| \phi_R = \angle H_R(j \omega) = \tan^{-1}\left(\frac{1}{\omega RC}\right)
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| </math>.
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| These expressions together may be substituted into the usual expression for the [[phasor (sine waves)|phasor]] representing the output:
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| :<math>
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| V_C \ = \ G_{C}V_{in} e^{j\phi_C}
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| </math>
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| :<math>
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| V_R \ = \ G_{R}V_{in} e^{j\phi_R}
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| </math>.
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| ===Current===
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| The current in the circuit is the same everywhere since the circuit is in series:
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| :<math>
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| I(s) = \frac{V_{in}(s) }{R + \frac{1}{Cs}} = { Cs \over 1 + RCs } V_{in}(s)
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| </math>
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| ===Impulse response===
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| The [[impulse response]] for each voltage is the inverse [[Laplace transform]] of the corresponding transfer function. It represents the response of the circuit to an input voltage consisting of an impulse or [[Dirac delta function]].
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| The impulse response for the capacitor voltage is
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| :<math>
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| h_C(t) = {1 \over RC} e^{-t / RC} u(t) = { 1 \over \tau} e^{-t / \tau} u(t)
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| </math>
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| where ''u''(''t'') is the [[Heaviside step function]] and
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| :<math>
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| \tau \ = \ RC </math>
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| is the [[time constant]].
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| Similarly, the impulse response for the resistor voltage is
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| :<math>
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| h_R(t) = \delta (t) - {1 \over RC} e^{-t / RC} u(t) = \delta (t) - { 1 \over \tau} e^{-t / \tau} u(t)
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| </math>
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| where ''δ''(''t'') is the [[Dirac delta function]]
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| ===Frequency-domain considerations===
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| These are [[frequency domain]] expressions. Analysis of them will show which frequencies the circuits (or filters) pass and reject. This analysis rests on a consideration of what happens to these gains as the frequency becomes very large and very small.
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| As <math>\omega \to \infty</math>:
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| :<math>G_C \to 0</math>
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| :<math>G_R \to 1</math>.
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| As <math>\omega \to 0</math>:
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| :<math>G_C \to 1</math>
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| :<math>G_R \to 0</math>.
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| This shows that, if the output is taken across the capacitor, high frequencies are attenuated (shorted to ground) and low frequencies are passed. Thus, the circuit behaves as a ''[[low-pass filter]]''. If, though, the output is taken across the resistor, high frequencies are passed and low frequencies are attenuated (since the capacitor blocks the signal as its frequency approaches 0). In this configuration, the circuit behaves as a ''[[high-pass filter]]''.
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| The range of frequencies that the filter passes is called its [[Bandwidth (signal processing)|bandwidth]]. The point at which the filter attenuates the signal to half its unfiltered power is termed its [[cutoff frequency]]. This requires that the gain of the circuit be reduced to
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| :<math>G_C = G_R = \frac{1}{\sqrt{2}}</math>.
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| Solving the above equation yields
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| :<math>\omega_{c} = \frac{1}{RC}</math>
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| or
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| :<math>f_c = \frac{1}{2\pi RC}</math>
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| which is the frequency that the filter will attenuate to half its original power.
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| Clearly, the phases also depend on frequency, although this effect is less interesting generally than the gain variations.
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| As <math>\omega \to 0</math>:
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| :<math>\phi_C \to 0</math>
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| :<math>\phi_R \to 90^{\circ} = \pi/2^{c}</math>.
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| As <math>\omega \to \infty</math>:
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| :<math>\phi_C \to -90^{\circ} = -\pi/2^{c}</math>
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| :<math>\phi_R \to 0</math>
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| So at [[Direct current|DC]] (0 [[Hertz|Hz]]), the capacitor voltage is in phase with the signal voltage while the resistor voltage leads it by 90°. As frequency increases, the capacitor voltage comes to have a 90° lag relative to the signal and the resistor voltage comes to be in-phase with the signal.
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| ===Time-domain considerations===
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| :''This section relies on knowledge of ''e'', the [[E (number)|natural logarithmic constant]]''.
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| The most straightforward way to derive the time domain behaviour is to use the [[Laplace transform]]s of the expressions for <math>V_C</math> and <math>V_R</math> given above. This effectively transforms <math>j\omega \to s</math>. Assuming a [[Heaviside step function|step input]] (i.e. <math>V_{in} = 0</math> before <math>t = 0</math> and then <math>V_{in} = V</math> afterwards):
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| :<math>
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| V_{in}(s) = V\frac{1}{s}
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| </math>
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| :<math>
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| V_C(s) = V\frac{1}{1 + sRC}\frac{1}{s}
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| </math>
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| and
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| :<math>
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| V_R(s) = V\frac{sRC}{1 + sRC}\frac{1}{s}
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| </math>.
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| [[Image:Series RC capacitor voltage.svg|thumb|right|230px|Capacitor voltage step-response.]]
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| [[Image:Series RC resistor voltage.svg|thumb|right|230px|Resistor voltage step-response.]]
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| [[Partial fraction]]s expansions and the inverse [[Laplace transform]] yield:
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| :<math>
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| \,\!V_C(t) = V\left(1 - e^{-t/RC}\right)
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| </math>
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| :<math>
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| \,\!V_R(t) = Ve^{-t/RC}
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| </math>.
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| These equations are for calculating the voltage across the capacitor and resistor respectively while the capacitor is [[Electric charge|charging]]; for discharging, the equations are vice-versa. These equations can be rewritten in terms of charge and current using the relationships C=Q/V and V=IR (see [[Ohm's law]]).
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| Thus, the voltage across the capacitor tends towards ''V'' as time passes, while the voltage across the resistor tends towards 0, as shown in the figures. This is in keeping with the intuitive point that the capacitor will be charging from the supply voltage as time passes, and will eventually be fully charged.
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| These equations show that a series RC circuit has a [[RC time constant|time constant]], usually denoted <math>\tau = RC</math> being the time it takes the voltage across the component to either rise (across C) or fall (across R) to within <math>1/e</math> of its final value. That is, <math>\tau</math> is the time it takes <math>V_C</math> to reach <math>V(1 - 1/e)</math> and <math>V_R</math> to reach <math>V(1/e)</math>.
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| The rate of change is a ''fractional'' <math>\left(1 - \frac{1}{e}\right)</math> per <math>\tau</math>. Thus, in going from <math>t=N\tau</math> to <math>t = (N+1)\tau</math>, the voltage will have moved about 63.2% of the way from its level at <math>t=N\tau</math> toward its final value. So C will be charged to about 63.2% after <math>\tau</math>, and essentially fully charged (99.3%) after about <math>5\tau</math>. When the voltage source is replaced with a short-circuit, with C fully charged, the voltage across C drops exponentially with ''t'' from <math>V</math> towards 0. C will be discharged to about 36.8% after <math>\tau</math>, and essentially fully discharged (0.7%) after about <math>5\tau</math>. Note that the current, <math>I</math>, in the circuit behaves as the voltage across R does, via [[Ohm's law|Ohm's Law]].
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| These results may also be derived by solving the [[differential equation]]s describing the circuit:
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| :<math> | |
| \frac{V_{in} - V_C}{R} = C\frac{dV_C}{dt}
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| </math>
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| and
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| :<math>
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| \,\!V_R = V_{in} - V_C
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| </math>.
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| The first equation is solved by using an [[integrating factor]] and the second follows easily; the solutions are exactly the same as those obtained via Laplace transforms.
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| ====Integrator====
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| Consider the output across the capacitor at ''high'' frequency i.e.
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| :<math>\omega \gg \frac{1}{RC}</math>.
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| This means that the capacitor has insufficient time to charge up and so its voltage is very small. Thus the input voltage approximately equals the voltage across the resistor. To see this, consider the expression for <math>I</math> given above:
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| :<math>
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| I = \frac{V_{in}}{R+1/j\omega C}
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| </math>
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| but note that the frequency condition described means that
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| :<math>
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| \omega C \gg \frac{1}{R}
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| </math>
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| so
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| :<math>
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| I \approx \frac{V_{in}}{R}
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| </math> which is just [[Ohm's law|Ohm's Law]].
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| Now,
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| :<math>
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| V_C = \frac{1}{C}\int_{0}^{t}Idt
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| </math>
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| so
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| :<math>
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| V_C \approx \frac{1}{RC}\int_{0}^{t}V_{in}dt
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| </math>,
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| which is an [[integrator]] ''across the capacitor''.
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| ====Differentiator====
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| Consider the output across the resistor at ''low'' frequency i.e.,
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| :<math>
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| \omega \ll \frac{1}{RC}
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| </math>.
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| This means that the capacitor has time to charge up until its voltage is almost equal to the source's voltage. Considering the expression for <math>I</math> again, when
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| :<math>
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| R \ll \frac{1}{\omega C}
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| </math>,
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| so
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| :<math>
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| I \approx \frac{V_{in}}{1/j\omega C}
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| </math>
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| :<math>
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| V_{in} \approx \frac{I}{j\omega C} = V_C
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| </math>
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| Now,
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| :<math> | |
| V_R = IR = C\frac{dV_C}{dt}R
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| </math>
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| :<math> | |
| V_R \approx RC\frac{dV_{in}}{dt}
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| </math>
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| which is a [[derivative|differentiator]] ''across the resistor''.
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| More accurate [[integral|integration]] and [[derivative|differentiation]] can be achieved by placing resistors and capacitors as appropriate on the input and [[feedback]] loop of [[operational amplifier]]s (see ''[[Operational amplifier applications#Inverting_integrator|operational amplifier integrator]]'' and ''[[Operational amplifier applications#Inverting_differentiator|operational amplifier differentiator]]'').
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| ==Parallel circuit==
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| [[File:RC Parallel Filter (with I Labels).svg|thumb|right|250px|[[series and parallel circuits#Parallel circuits|Parallel]] RC circuit]]
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| The parallel RC circuit is generally of less interest than the series circuit. This is largely because the output voltage <math>V_{out}</math> is equal to the input voltage <math>V_{in}</math> — as a result, this circuit does not act as a filter on the input signal unless fed by a [[current source]].
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| With complex impedances:
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| :<math>
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| I_R = \frac{V_{in}}{R}\,
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| </math>
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| and
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| :<math>
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| I_C = j\omega C V_{in}\,
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| </math>.
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| This shows that the capacitor current is 90° out of phase with the resistor (and source) current. Alternatively, the governing differential equations may be used:
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| :<math>
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| I_R = \frac{V_{in}}{R}
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| </math>
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| and
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| :<math>
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| I_C = C\frac{dV_{in}}{dt}
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| </math>.
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| When fed by a current source, the transfer function of a parallel RC circuit is:
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| :<math>
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| \frac{V_{out}}{I_{in}} = \frac{R}{1+sRC}
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| </math>.
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| == See also ==
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| *[[RL circuit]]
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| *[[LC circuit]]
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| *[[RLC circuit]]
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| *[[Electrical network]]
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| *[[List of electronics topics]]
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| *[[Step response]]
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| *[[Continuous-repayment_mortgage#Comparison_with_similar_physical_systems|RC Circuit and continuous-repayment mortgage]]
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| [[Category:Analog circuits]]
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| [[Category:Electronic filter topology]]
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