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 V_\mathrm{in} divided by the sum of the complex impedance of the capacitor and resistor: :\begin{align} I(s) &= \frac{V_\mathrm{in}(s) }{R + \frac{1}{Cs}} \\ &= \frac{ Cs }{ 1 + RCs } V_\mathrm{in}(s)\,. \end{align}
Voltage By viewing the circuit as a
voltage divider, the
voltage across the capacitor is: :\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} and the voltage across the resistor is: :\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}
Transfer functions The
transfer function from the input voltage to the voltage across the capacitor is :H_C(s) = \frac{ V_C(s) }{ V_\mathrm{in}(s) } = \frac{ 1 }{ 1 + RCs } \,. Similarly, the transfer function from the input to the voltage across the resistor is :H_R(s) = \frac{ V_R(s) }{ V_{\rm in}(s) } = \frac{ RCs }{ 1 + RCs } \,.
Poles and zeros Both transfer functions have a single
pole located at :s = -\frac{1}{RC} \,. In addition, the transfer function for the voltage across the resistor has a
zero located at the
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 \sigma can be ignored in the complex frequency equation s {=} \sigma {+} j \omega when only the steady state is of interest. The simple substitution of s \Rightarrow j \omega into the previous transfer functions will thus provide the sinusoidal gain and phase response of the circuit.
Gain The magnitude of the gains across the two components are :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}} and :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}}\,, As the frequency becomes very large (), the capacitor acts like a short circuit, so: :G_C \to 0 \quad \mbox{and} \quad G_R \to 1 \,. As the frequency becomes very small (), the capacitor acts like an open circuit, so: :G_C \to 1 \quad \mbox{and} \quad G_R \to 0 \,.
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. 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 :G_C = G_R = \frac{1}{\sqrt 2}. Solving the above equation yields :\omega_\mathrm{c} = \frac{1}{RC} \quad \mbox{or} \quad f_\mathrm{c} = \frac{1}{2\pi RC} which is the frequency that the filter will attenuate to half its original power.
Phase The phase angles are :\phi_C = \angle H_C(j \omega) = \tan^{-1}\left(-\omega RC\right) and :\phi_R = \angle H_R(j \omega) = \tan^{-1}\left(\frac{1}{\omega RC}\right)\,. As : :\phi_C \to 0 \quad \mbox{and} \quad \phi_R \to 90^{\circ} = \frac{\pi}{2}\mbox{ radians}\,. As : :\phi_C \to -90^{\circ} = -\frac{\pi}{2}\mbox{ radians} \quad \mbox{and} \quad \phi_R \to 0\,. While the output signal's phase shift relative to the input depends on frequency, this is generally less interesting than the gain variations. At
DC (0
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 expressions representing the output: :\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}
Impulse response 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 :h_C(t) = \frac{1}{RC} e^{-\frac{t}{RC}} u(t) = \frac{1}{\tau} e^{-\frac{t}{\tau}} u(t)\,, where is the
Heaviside step function and is the
time constant. Similarly, the impulse response for the resistor voltage is :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)\,, where 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 transforms of the expressions for and given above. Assuming a
step input (i.e. before and then afterwards): :\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}
Partial fractions expansions and the
inverse Laplace transform yield: :\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} These equations are for calculating the voltage across the capacitor and resistor respectively while the capacitor is
charging; for discharging, the equations are vice versa. These equations can be rewritten in terms of charge and current using the relationships and (see
Ohm's law). Thus, the voltage across the capacitor tends towards 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 is both the time for and to reach within of their final value. In other words, is the time it takes for the voltage across the capacitor to rise to or for the voltage across the resistor to fall to . This
RC time constant is labeled using the letter
tau (). The rate of change is a
fractional per . Thus, in going from to , the voltage will have moved about 63.2% of the way from its level at toward its final value. So the capacitor will be charged to about 63.2% after , and is often considered fully charged (>99.3%) after about . When the voltage source is replaced with a short circuit, with the capacitor fully charged, the voltage across the capacitor drops exponentially with from towards 0. The capacitor will be discharged to about 36.8% after , and is often considered fully discharged (\begin{align} \frac{V_\mathrm{in} - V_C}{R} &= C\frac{dV_C}{dt} \\ V_R &= V_\mathrm{in} - V_C \,. \end{align} 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. :\omega \gg \frac{1}{RC}\,. 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 I given above: :I = \frac{V_\mathrm{in}}{R+\frac{1}{j\omega C}}\,, but note that the frequency condition described means that :\omega C \gg \frac{1}{R}\,, so :I \approx \frac{V_\mathrm{in}}{R} which is just
Ohm's law. Now, :V_C = \frac{1}{C}\int_{0}^{t}I\,dt\,, so :V_C \approx \frac{1}{RC}\int_{0}^{t}V_\mathrm{in}\,dt\,. 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., :\omega \ll \frac{1}{RC}\,. 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 again, when :R \ll \frac{1}{\omega C}\,, so :\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} Now, :\begin{align} V_R &= IR = C\frac{dV_C}{dt}R \\ V_R &\approx RC\frac{dV_{in}}{dt}\,. \end{align} Therefore, the voltage
across the resistor acts approximately like a
differentiator of the input voltage for low frequencies.
Integration and
differentiation can also be achieved by placing resistors and capacitors as appropriate on the input and
feedback loop of
operational amplifiers (see
operational amplifier integrator and
operational amplifier differentiator). ==Parallel circuit==