Editing RC circuit

{{short description|Electric circuit composed of resistors and capacitors}}
{{more citations needed|date=March 2018}}
{{Linear analog electronic filter|filter1=hide|filter2=hide}}
A '''resistor–capacitor circuit''' ('''RC circuit'''), or '''RC filter''' or '''RC network''', is an [[electric circuit]] composed of [[resistor]]s and [[capacitor]]s. It may be driven by a [[voltage source|voltage]] or [[current source]] and these will produce different responses. A first order RC circuit is composed of one resistor and one capacitor and is the simplest type of RC circuit.

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.

== Natural response ==
[[File:Discharging capacitor.svg|200px|thumb|right| Simplest RC circuit]]
The simplest RC circuit consists of a resistor with [[Electric resistance|resistance]] {{mvar|R}} and a charged capacitor with capacitance {{mvar|C}} connected to one another in a single loop, without an external voltage source. The capacitor will discharge its stored energy through the resistor. If {{Mvar|V(t)}} is taken to be the voltage of the capacitor's top plate relative to its bottom plate in the figure, then the [[Capacitor#Current–voltage relation|capacitor current–voltage relation]] says the current {{Mvar|I(t)}} ''exiting'' the capacitor's top plate will equal {{mvar|C}} multiplied by the ''negative'' time derivative of {{Mvar|V(t)}}. [[Kirchhoff's circuit laws#Kirchhoff's current law (KCL)|Kirchhoff's current law]] says this current is the same current entering the top side of the resistor, which per [[Ohm's law]] equals {{Mvar|V(t)/R}}. This yields a [[linear differential equation]]:

:<math>\overbrace{\Biggl(C\frac{-\mathrm{d}V(t)}{\mathrm{d}t}\Biggr)}^\text{capacitor current} = \overbrace{\Biggl(\frac{V(t)}{R}\Biggr)}^\text{resistor current} ,</math>
which can be rearranged according to the standard form for [[exponential decay]]: 
:<math>\frac{\mathrm{d}V(t)}{\mathrm{d}t} = -\frac{1}{RC}V(t) \, .</math>
This means that the instantaneous rate of voltage decrease at any time is proportional to the voltage at that time. [[Exponential decay#Solution of the differential equation|Solving]] for {{mvar|V(t)}} yields an exponential decay curve that [[asymptotically]] approaches 0: 
:<math>V(t)=V_0 \cdot e^{-\frac{t}{RC}} \,,</math>

where {{math|''V''<sub>0</sub>}} is the capacitor voltage at time {{math|''t'' {{=}} 0}} and '''''{{mvar|e}}''''' is [[E (mathematical constant)|Euler's number]].

The time required for the voltage to fall to {{math|{{sfrac|''V''<sub>0</sub>|''e''}}}} is called the [[RC time constant]] and is given by:<ref>Horowitz & Hill, p. 1.13</ref>

:<math>\tau = RC \,.</math>

When using the [[International System of Units]], {{mvar|R}} is in [[Ohm|ohms]] and {{mvar|C}} is in [[Farad|farads]], so {{mvar|τ}} will be in [[Second|seconds]]. At any time {{mvar|''N''·τ}}, the capacitor's charge or voltage will be {{Sfrac|1|{{Mvar|e}}<sup>N</sup>}} of its starting value. So if the capacitor's charge or voltage is said to start at 100%, then 36.8% remains at {{Math|1·τ}}, 13.5% remains at {{Math|2·τ}}, 5% remains at {{Math|3·τ}}, 1.8% remains at {{Math|4·τ}}, and less than 0.7% remains at {{Math|5·τ}} and later.

The [[half-life]] ({{math|'''''t''{{sub|{{frac|1|2}}}}'''}}) is the time that it takes for its charge or voltage to be reduced in half:<ref>{{Cite web |last=Hanks |first=Ann |last2=Luttermoser |first2=Donald |title=General Physics II Lab (PHYS-2021) Experiment ELEC-5: RC Circuits |url=https://faculty.etsu.edu/lutter/courses/phys2021/ELEC_5_RC_Circuit.pdf}}</ref>
:<math>\tfrac{1}{2} {=} e^{-\tfrac{t_{1/2}}{\tau}} \longrightarrow t_{1/2} = \ln(2) \, \tau \approx \text{.693} \, \tau \, .</math>
For example, 50% of charge or voltage remains at time {{math|1·''t''{{sub|{{frac|1|2}}}}}}, then 25% remains at time {{math|2·''t''{{sub|{{frac|1|2}}}}}}, then 12.5% remains at time {{math|3·''t''{{sub|{{frac|1|2}}}}}}, and {{Sfrac|1|2<sup>N</sup>}} will remain at time {{math|''N''·''t''{{sub|{{frac|1|2}}}}}}.

=== RC discharge calculator ===
{{calculator
	|id=capacitance
	|formula=capacitance_number*pow(10,capacitance_unitlog10)
	|type=hidden
	|default=.000001
}}

{{calculator
	|id=resistance
	|formula=resistance_number*pow(10,resistance_unitlog10)
	|type=hidden
	|default=1000000
}}

{{calculator
	|id=tau
	|formula=resistance*capacitance
	|type=hidden
	|default=1
}}

{{calculator
	|id=tau_unitlog10div3floor
	|formula=min(max(floor(log10(tau)/3),-5),5)
	|type=hidden
	|default=1
}}

{{calculator
	|id=tau_sigdigits
	|formula=round(tau/pow(10,tau_unitlog10div3floor*3),3)
	|type=hidden
	|default=1
}}

{{calculator
	|id=initialvoltage
	|formula=initialvoltage_number*pow(10,initialvoltage_unitlog10)
	|type=hidden
	|default=1
}}

{{calculator
	|id=decaytime
	|formula=decaytime_in_tau*tau
	|type=hidden
	|default=1
}}

{{calculator
	|id=decaytime_unitlog10div3floor
	|formula=min(max(ifzero(decaytime,0,floor(log10(abs(decaytime))/3)),-5),5)
	|type=hidden
	|default=1
}}

{{calculator
	|id=decaytime_sigdigits
	|formula=round(decaytime/pow(10,decaytime_unitlog10div3floor*3),3)
	|type=hidden
	|default=1
}}

{{calculator
	|id=fraction_of_initialvoltage
	|default=.368
	|type=hidden
	|formula=exp(-decaytime_in_tau)
}}

{{calculator
	|id=percent_of_initialvoltage
	|default=36.8
	|type=hidden
	|formula=round(fraction_of_initialvoltage*100,1)
}}

{{calculator
	|id=initialvoltage
	|type=hidden
	|default=1
	|formula=initialvoltage_number*pow(10,initialvoltage_unitlog10)
}}

{{calculator
	|id=finalvoltage
	|type=hidden
	|size=4
	|NaN-text=?
	|formula=initialvoltage*fraction_of_initialvoltage|default=0.368}}

{{calculator
	|id=finalvoltage_unitlog10div3floor
	|formula=min(max(ifzero(decaytime,0,floor(log10(abs(finalvoltage))/3)),-5),5)
	|type=hidden
	|default=1
}}
{{calculator
	|id=finalvoltage_sigdigits
	|formula=round(finalvoltage/pow(10,finalvoltage_unitlog10div3floor*3),3)
	|type=hidden
	|default=1
}}

{{calculator
	|id=xpix
	|type=hidden
	|formula=87.5+562*max(min(decaytime_in_tau,5.1),-0.1)/5
}}

{{calculator
	|id=ypix
	|type=hidden
	|formula=12+434.5*(1-max(min(percent_of_initialvoltage,101),0)/100)
}}

{{calculator
	|id=cutoff_angular
	|type=hidden
	|size=4
	|NaN-text=?
	|formula=1/tau
	|default=1
}}

{{calculator
	|id=cutoff_ordinal
	|type=hidden
	|size=4
	|NaN-text=?
	|formula=cutoff_angular/(2*PI)
	|default=0.159
}}

{{calculator
	|id=cutoff_angular_unitlog10div3floor
	|formula=min(max(floor(log10(cutoff_angular)/3),-4),4)
	|type=hidden
	|default=1
}}

{{calculator
	|id=cutoff_angular_sigdigits
	|formula=round(cutoff_angular/pow(10,cutoff_angular_unitlog10div3floor*3),3)
	|type=hidden
	|default=1
}}

{{calculator
	|id=cutoff_ordinal_unitlog10div3floor
	|formula=min(max(floor(log10(cutoff_ordinal)/3),-4),4)
	|type=hidden
	|default=1
}}

{{calculator
	|id=cutoff_ordinal_sigdigits
	|formula=round(cutoff_ordinal/pow(10,cutoff_ordinal_unitlog10div3floor*3),3)
	|type=hidden
	|default=1
}}

For instance, {{nowrap|{{calculator|id=resistance_number|type=number|min=1|size=3|default=1|max=999|step=.1}}&nbsp;{{calculator|id=resistance_unitlog10|type=select|class=cdx-select|mapping={ "teraohms":12, "gigaohms":9, "megaohms":6, "kiloohms":3, "ohms":0, "milliohms":-3 }|style=min-width:1ch|value=6}}}} of resistance with {{nowrap|{{calculator|id=capacitance_number|type=number|min=1|size=3|default=1|max=999|step=.1}}&nbsp;{{calculator|id=capacitance_unitlog10|type=select|class=cdx-select|mapping={ "farads":0, "millifarads":-3, "microfarads":-6, "nanofarads":-9, "picofarads":-12}|style=min-width:1ch|value=-6}}}} of capacitance produces a time constant of approximately {{nowrap|{{calculator
	|id=tau_sigdigits_label
	|type=plain
	|size=4
	|NaN-text=?
	|formula=tau_sigdigits
	|default=1
}} {{calculator
	|id=tau_unit
	|formula=tau_unitlog10div3floor
	|type=plain
	|default=seconds
	|mapping={
		"[unknown time unit]": "default",
		"petaseconds": 5,
		"teraseconds": 4,
		"gigaseconds": 3,
		"megaseconds": 2,
		"kiloseconds": 1,
		"seconds": 0,
		"milliseconds": -1,
		"microseconds": -2,
		"nanoseconds": -3,
		"picoseconds": -4,
		"femtoseconds": -5
	}
}}.}} This {{Mvar|τ}} corresponds to a [[cutoff frequency]] of approximately {{nowrap|{{calculator
	|id=cutoff_ordinal_sigdigits_label
	|type=plain
	|size=4
	|NaN-text=?
	|formula=cutoff_ordinal_sigdigits
	|default=159
}} {{calculator
	|id=cutoff_ordinal_unit
	|formula=cutoff_ordinal_unitlog10div3floor
	|type=plain
	|default=millihertz
	|mapping={ "[unknown frequency unit]":"default", "terahertz":4, "gigahertz":3, "megahertz":2, "kilohertz":1, "hertz":0, "millihertz":-1, "microhertz":-2, "nanohertz":-3, "picohertz":-4}
}}}} or {{nowrap|{{calculator
	|id=cutoff_angular_sigdigits_label
	|type=plain
	|size=4
	|NaN-text=?
	|formula=cutoff_angular_sigdigits
	|default=1
}} {{calculator
	|id=cutoff_angular_unit
	|formula=cutoff_angular_unitlog10div3floor
	|type=plain
	|default=radians per second
	|mapping={ "[unknown frequency unit]":"default", "teraradians per second":4, "gigaradians per second":3, "megaradians per second":2, "kiloradians per second":1, "radians per second":0, "milliradians per second":-1, "microradians per second":-2, "nanoradians per second":-3, "picoradians per second":-4}
}}.}} If the capacitor has an initial voltage {{Math|''V''<sub>0</sub>}} of {{nowrap|{{calculator
	|id=initialvoltage_number
	|type=number
	|min=-999
	|size=3
	|default=1
	|max=999
	|step=.1
}} {{calculator
	|id=initialvoltage_unitlog10
	|type=select
	|class=cdx-select
	|mapping={ "teravolts":12, "gigavolts":9, "megavolts":6, "kilovolts":3, "volts":0, "millivolts":-3, "microvolts":-6, "nanovolts":-9, "picovolts":-12}
	|style=min-width:1ch
	|value=0
}}}}, then after {{nowrap|{{calculator
	|id=decaytime_in_tau
	|default=1
	|type=number
	|size=5
	|step=.1
	|min=0
}}&nbsp;{{Mvar|τ}}}} (approximately {{nowrap|{{calculator
	|id=decaytime_sigdigits_label
	|type=plain
	|size=4
	|NaN-text=?
	|formula=decaytime_sigdigits
	|default=1
}} {{calculator
	|id=decaytime_unit
	|formula=decaytime_unitlog10div3floor
	|type=plain
	|default=seconds
	|mapping={
		"[unknown time unit]": "default",
		"petaseconds": 5,
		"teraseconds": 4,
		"gigaseconds": 3,
		"megaseconds": 2,
		"kiloseconds": 1,
		"seconds": 0,
		"milliseconds": -1,
		"microseconds": -2,
		"nanoseconds": -3,
		"picoseconds": -4,
		"femtoseconds": -5
	}
}}}} or {{nowrap|{{calculator
	|id=decaytime_in_halflives
	|type=plain
	|size=4
	|NaN-text=?
	|formula=round(decaytime_in_tau/log(2),3)
	|default=1.443
}}&nbsp;[[half-lives]]),}} the capacitor's voltage will discharge to approximately {{nowrap|{{calculator
	|id=finalvoltage_sigdigits_label
	|type=plain
	|size=4
	|NaN-text=?
	|formula=finalvoltage_sigdigits
	|default=368
}} {{calculator
	|id=finalvoltage_unit
	|formula=finalvoltage_unitlog10div3floor
	|type=plain
	|default=millivolts
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		"[unknown voltage unit]": "default",
		"petavolts": 5,
		"teravolts": 4,
		"gigavolts": 3,
		"megavolts": 2,
		"kilovolts": 1,
		"volts": 0,
		"millivolts": -1,
		"microvolts": -2,
		"nanovolts": -3,
		"picovolts": -4,
		"femtovolts": -5
	}
}}:}}

<div class="floatnone noresize" style="position: relative; width: 900px; height: 550px;">

[[File:Series_RC_resistor_voltage_relabeled.svg|alt=|frameless|720x500px|class=skin-invert-image]]

<div style="position: absolute; left: calc( var( --calculator-xpix, 200)*1px ); top: calc( var( --calculator-ypix, 286.6)*1px ); padding: 0;">[[Image:Blue pog.svg|10px|link=|alt=|class=|]]</div>

<div style="position: absolute; left: calc( (var( --calculator-xpix, 200) + 10)*1px); top: calc( (var( --calculator-ypix, 286.6) - 10)*1px ); padding: 0; background-color: white; border: 3px solid blue; color: blue">&nbsp;{{Math|''V''<sub>C</sub>}}({{calculator|id=decaytime_in_tau_label|default=1|type=plain|formula=round(decaytime_in_tau,1)}}''τ'')&nbsp;≈&nbsp;{{calculator|default=36.8|type=plain|formula=percent_of_initialvoltage}}%&nbsp;of&nbsp;{{Math|''V''<sub>0</sub>}}&nbsp;</div>

</div>

== Complex impedance ==
The RC circuit's behavior is well-suited to be analyzed in the [[Laplace domain]], which the rest of this article requires a basic understanding of. The Laplace domain is a [[frequency domain]] representation using [[complex frequency]] {{mvar|s}}, which is (in general) a [[complex number]]:
:<math>s = \sigma + j \omega \,, </math>

where

*{{mvar|j}} represents the [[imaginary unit]]: {{math|''j''<sup>2</sup> {{=}} −1}},
*{{mvar|σ}} is the [[exponential decay]] constant, and
*{{mvar|ω}} is the [[sinusoidal]] [[angular frequency]].

When evaluating circuit equations in the Laplace domain, time-dependent circuit elements of capacitance and inductance can be treated like resistors with [[Complex impedance|complex-valued impedance]] instead of [[Real number|real]] resistance. While the complex impedance {{mvar|Z<sub>R</sub>}} of a resistor is simply a real value equal to its resistance {{mvar|R}}, the [[Electrical impedance#Generalised s-plane impedance|complex impedance of a capacitor]] {{mvar|C}} is instead:
:<math>Z_C = \frac{1}{C s} . </math>

== Series circuit ==
[[File:RC Series Filter (with V&I Labels).svg|thumb|right|[[series and parallel circuits#Series circuits|Series]] RC circuit]]

=== Current ===
Kirchhoff's current law means that the current in the series circuit is necessarily the same through both elements. Ohm's law says this current is equal to the input voltage <math>V_\mathrm{in} </math> divided by the sum of the complex impedance of the capacitor and resistor:
:<math>\begin{align}
I(s) &= \frac{V_\mathrm{in}(s) }{R + \frac{1}{Cs}} \\
     &=  \frac{ Cs }{ 1 + RCs } V_\mathrm{in}(s)\,.
\end{align}</math>

=== Voltage ===
By viewing the circuit as a [[voltage divider]], the [[voltage]] across the capacitor is:
:<math>\begin{align}
V_C(s) &= \frac{\frac{1}{Cs}}{R + \frac{1}{Cs}}V_\mathrm{in}(s) \\
       &= \frac{1}{1 + RCs}V_\mathrm{in}(s)
\end{align}</math>

and the voltage across the resistor is:
:<math>\begin{align}
V_R(s) &= \frac{R}{R + \frac{1}{Cs}}V_\mathrm{in}(s) \\
       &= \frac{RCs}{1 + RCs}V_\mathrm{in}(s)\,.
\end{align}</math>

===Transfer functions===

The [[transfer function]] from the input voltage to the voltage across the capacitor is
:<math>H_C(s) = \frac{ V_C(s) }{ V_\mathrm{in}(s) } = \frac{ 1 }{ 1 + RCs } \,.</math>

Similarly, the transfer function from the input to the voltage across the resistor is

:<math>H_R(s) = \frac{ V_R(s) }{ V_{\rm in}(s) } = \frac{ RCs }{ 1 + RCs } \,.</math>

====Poles and zeros====

Both transfer functions have a single [[pole (complex analysis)|pole]] located at
:<math>s = -\frac{1}{RC} \,.</math>

In addition, the transfer function for the voltage across the resistor has a [[zero (complex analysis)|zero]] located at the [[origin (mathematics)|origin]].

===Frequency-domain considerations===

The sinusoidal steady state is a special case of complex frequency that considers the input to consist only of pure sinusoids. Hence, the exponential decay component represented by <math>\sigma </math> can be ignored in the complex frequency equation <math>s {=} \sigma {+} j \omega </math> when only the steady state is of interest. The simple substitution of <math>s \Rightarrow j \omega </math> into the previous transfer functions will thus provide the sinusoidal gain and phase response of the circuit.

==== Gain ====
[[Image:Passive_integrator_circuit_transfer_function.png|right|thumb|350px|Amplitude and phase transfer functions for a series RC circuit]]
The magnitude of the gains across the two components are
:<math>G_C = \big| H_C(j \omega) \big| = \left|\frac{V_C(j \omega)}{V_\mathrm{in}(j \omega)}\right| = \frac{1}{\sqrt{1 + \left(\omega RC\right)^2}}</math>
and
:<math>G_R = \big| H_R(j \omega) \big| = \left|\frac{V_R(j \omega)}{V_\mathrm{in}(j \omega)}\right| = \frac{\omega RC}{\sqrt{1 + \left(\omega RC\right)^2}}\,,</math>
As the frequency becomes very large ({{math|''ω'' → ∞}}), the capacitor acts like a short circuit, so:
:<math>G_C \to 0 \quad \mbox{and} \quad G_R \to 1 \,.</math>

As the frequency becomes very small ({{math|''ω'' → 0}}), the capacitor acts like an open circuit, so:
:<math>G_C \to 1 \quad \mbox{and} \quad G_R \to 0 \,.</math>

===== Operation as either a high-pass or a low-pass filter =====
The behavior at these extreme frequencies show that if the output is taken across the capacitor, high frequencies are attenuated and low frequencies are passed, so such a circuit configuration is a ''[[low-pass filter]]''. However, if the output is taken across the resistor, then high frequencies are passed and low frequencies are attenuated, so such a configuration is a ''[[high-pass filter]]''.

===== Cutoff frequency =====
The range of frequencies that the filter passes is called its [[Bandwidth (signal processing)|bandwidth]]. The frequency 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 
:<math>G_C = G_R = \frac{1}{\sqrt 2}</math>.

Solving the above equation yields
:<math>\omega_\mathrm{c} = \frac{1}{RC} \quad \mbox{or} \quad f_\mathrm{c} = \frac{1}{2\pi RC}</math>
which is the frequency that the filter will attenuate to half its original power.

==== Phase ====
The phase angles are
:<math>\phi_C =  \angle H_C(j \omega) =  \tan^{-1}\left(-\omega RC\right)</math>
and
:<math>\phi_R = \angle H_R(j \omega) =  \tan^{-1}\left(\frac{1}{\omega RC}\right)\,.</math>

As {{math|''ω'' → 0}}:
:<math>\phi_C \to 0 \quad \mbox{and} \quad \phi_R \to 90^{\circ} = \frac{\pi}{2}\mbox{ radians}\,.</math>

As {{math|''ω'' → ∞}}:
:<math>\phi_C \to -90^{\circ} = -\frac{\pi}{2}\mbox{ radians} \quad \mbox{and} \quad \phi_R \to 0\,.</math>

While the output signal's phase shift relative to the input depends on frequency, this is generally less interesting than the gain variations. At [[Direct current|DC]] (0&nbsp;[[Hertz|Hz]]), the capacitor voltage is in phase with the input 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 input signal and the resistor voltage comes to be in-phase with the input signal.

==== Phasor representation ====
The gain and phase expressions together may be combined into these [[phasor (sine waves)|phasor]] expressions representing the output:
:<math>\begin{align} V_C &= G_C V_\mathrm{in}  e^{j\phi_C} \\ V_R &= G_R V_\mathrm{in} e^{j\phi_R}\,. \end{align}</math>

===Impulse response===
[[Image:Passive_integrator_circuit_impulse_response_1.png|right|thumb|350px|The impulse response of a series RC circuit]]
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]].

The impulse response for the capacitor voltage is
:<math>h_C(t) = \frac{1}{RC} e^{-\frac{t}{RC}} u(t)  =  \frac{1}{\tau} e^{-\frac{t}{\tau}} u(t)\,,</math>

where {{math|''u''(''t'')}} is the [[Heaviside step function]] and {{math|''τ'' {{=}} ''RC''}} is the [[time constant]].

Similarly, the impulse response for the resistor voltage is
:<math>h_R(t) = \delta (t) - \frac{1}{RC} e^{-\frac{t}{RC}} u(t)  =  \delta (t) - \frac{1}{\tau} e^{-\frac{t}{\tau}} u(t)\,,</math>

where {{math|''δ''(''t'')}} is the [[Dirac delta function]]

===Time-domain considerations===

:''This section relies on knowledge of the [[Laplace transform]]''.

The most straightforward way to derive the time domain behaviour is to use the [[Laplace transform]]s of the expressions for {{mvar|V<sub>C</sub>}} and {{mvar|V<sub>R</sub>}} given above. Assuming a [[Heaviside step function|step input]] (i.e. {{math|''V''<sub>in</sub> {{=}} 0}} before {{math|''t'' {{=}} 0}} and then {{math|''V''<sub>in</sub> {{=}} ''V''<sub>1</sub>}} afterwards):
:<math>\begin{align}
V_\mathrm{in}(s) &= V_1\cdot\frac{1}{s} \\
V_C(s) &= V_1\cdot\frac{1}{1 + sRC}\cdot\frac{1}{s} \\
V_R(s) &= V_1\cdot\frac{sRC}{1 + sRC}\cdot\frac{1}{s} \,.
\end{align}</math>

[[Image:Series RC capacitor voltage.svg|thumb|right|230px|Capacitor voltage step-response.]]
[[Image:Series RC resistor voltage.svg|thumb|right|230px|Resistor voltage step-response.]]

[[Partial fraction]]s expansions and the [[inverse Laplace transform]] yield:
:<math>\begin{align} V_C(t) &= V_1 \cdot \left(1 - e^{-\frac{t}{RC}}\right) \\ V_R(t) &= V_1 \cdot \left( e^{-\frac{t}{RC}} \right)\,. \end{align}</math>

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 {{math|''C'' {{=}} ''{{sfrac|Q|V}}''}} and {{math|''V'' {{=}} ''IR''}} (see [[Ohm's law]]).

Thus, the voltage across the capacitor tends towards {{mvar|V<sub>1</sub>}} 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.

The product {{math|''RC''}} is both the time for {{mvar|V<sub>C</sub>}} and {{mvar|V<sub>R</sub>}} to reach within {{math|{{sfrac|1|''e''}}}} of their final value. In other words, {{math|''RC''}} is the time it takes for the voltage across the capacitor to rise to {{math|''V''<sub>1</sub>·(1 − {{sfrac|1|''e''}})}} or for the voltage across the resistor to fall to {{math|''V''<sub>1</sub>·({{sfrac|1|''e''}})}}. This [[RC time constant]] is labeled using the letter [[tau]] ({{math|''τ''}}).

The rate of change is a ''fractional'' {{math|1 − {{sfrac|1|''e''}}}} per {{mvar|τ}}. Thus, in going from {{math|''t'' {{=}} ''Nτ''}} to {{math|''t'' {{=}} (''N'' + 1)''τ''}}, the voltage will have moved about 63.2% of the way from its level at {{math|''t'' {{=}} ''Nτ''}} toward its final value. So the capacitor will be charged to about 63.2% after {{mvar|τ}}, and is often considered fully charged (>99.3%) after about {{math|5''τ''}}. When the voltage source is replaced with a short circuit, with the capacitor fully charged, the voltage across the capacitor drops exponentially with {{mvar|t}} from {{mvar|V}} towards 0. The capacitor will be discharged to about 36.8% after {{mvar|τ}}, and is often considered fully discharged (<0.7%) after about {{math|5''τ''}}. Note that the current, {{mvar|I}}, in the circuit behaves as the voltage across the resistor does, via [[Ohm's law|Ohm's Law]].

These results may also be derived by solving the [[differential equation]]s describing the circuit:
:<math>\begin{align} \frac{V_\mathrm{in} - V_C}{R} &= C\frac{dV_C}{dt} \\ V_R &= V_\mathrm{in} - V_C \,. \end{align}</math>
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.

====Integrator====

Consider the output across the capacitor at ''high'' frequency, i.e.
:<math>\omega \gg \frac{1}{RC}\,.</math>

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:
:<math>I = \frac{V_\mathrm{in}}{R+\frac{1}{j\omega C}}\,,</math>
but note that the frequency condition described means that
:<math>\omega C \gg \frac{1}{R}\,,</math>
so
:<math>I \approx \frac{V_\mathrm{in}}{R}</math>
which is just [[Ohm's law|Ohm's Law]].

Now,
:<math>V_C = \frac{1}{C}\int_{0}^{t}I\,dt\,,</math>
so
:<math>V_C \approx \frac{1}{RC}\int_{0}^{t}V_\mathrm{in}\,dt\,.</math>
Therefore, the voltage ''across the capacitor'' acts approximately like an [[integrator]] of the input voltage for high frequencies.

====Differentiator====

Consider the output across the resistor at ''low'' frequency i.e.,
:<math>\omega \ll \frac{1}{RC}\,.</math>

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 {{mvar|I}} again, when
:<math>R \ll \frac{1}{\omega C}\,,</math>
so
:<math>\begin{align} I &\approx \frac{V_\mathrm{in}}\frac{1}{j\omega C} \\ V_\mathrm{in} &\approx \frac{I}{j\omega C} = V_C \,.\end{align}</math>

Now,
:<math>\begin{align} V_R &= IR = C\frac{dV_C}{dt}R \\ V_R &\approx RC\frac{dV_{in}}{dt}\,. \end{align}</math>
Therefore, the voltage ''across the resistor'' acts approximately like a [[differentiator]] of the input voltage for low frequencies.

[[integral|Integration]] and [[derivative|differentiation]] can also 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]]'').



==Parallel circuit==

[[File:RC Parallel Filter (with I Labels).svg|thumb|right|250px|[[series and parallel circuits#Parallel circuits|Parallel]] RC circuit]]

The parallel RC circuit is generally of less interest than the series circuit. This is largely because the output voltage {{math|''V''<sub>out</sub>}} is equal to the input voltage {{math|''V''<sub>in</sub>}} — as a result, this circuit acts as a filter on a current input instead of a voltage input.

With complex impedances:
:<math>\begin{align} I_R &= \frac{V_\mathrm{in}}{R} \\ I_C &= j\omega C V_\mathrm{in}\,. \end{align}</math>

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:
:<math>\begin{align} I_R &= \frac{V_\mathrm{in}}{R} \\ I_C &= C\frac{dV_\mathrm{in}}{dt}\,. \end{align}</math>

When fed by a current source, the transfer function of a parallel RC circuit is:
:<math>\frac{V_\mathrm{out}}{I_\mathrm{in}} = \frac{R}{1+sRC}\,.</math>

== Synthesis ==
It is sometimes required to [[network synthesis|synthesise]] an RC circuit from a given [[rational function]] in ''s''. For synthesis to be possible in passive elements, the function must be a [[positive-real function]]. To synthesise as an RC circuit, all the critical frequencies ([[poles and zeroes]]) must be on the negative real axis and alternate between poles and zeroes with an equal number of each. Further, the critical frequency nearest the origin must be a pole, assuming the rational function represents an impedance rather than an admittance.

The synthesis can be achieved with a modification of the [[Foster synthesis]] or [[Cauer synthesis]] used to synthesise [[LC circuit]]s. In the case of Cauer synthesis, a [[ladder network]] of resistors and capacitors will result.<ref>Bakshi & Bakshi, pp. 3-30–3-37</ref>

== See also ==
*[[RC time constant]]
*[[RL circuit]]
*[[LC circuit]]
*[[RLC circuit]]
*[[Electrical network]]
*[[List of electronics topics]]
*{{Slink|Johnson–Nyquist noise|Thermal noise on capacitors}} – gives the derivation of ''kTC'' noise caused by a resistor with a capacitor to be:
:: <math>
V_\text{rms} = \sqrt{ k_\text{B} T \over C }.
</math>
*[[Step response]]

==References==
{{reflist}}

==Bibliography==
* Bakshi, U.A.; Bakshi, A.V., ''Circuit Analysis - II'', Technical Publications, 2009 {{ISBN|9788184315974}}.
* Horowitz, Paul; Hill, Winfield, ''The Art of Electronics'' (3rd edition), Cambridge University Press, 2015 {{ISBN|0521809266}}.

{{Authority control}}

[[Category:Analog circuits]]
[[Category:Electronic filter topology]]