= Series RC impedance
{title2=$Z=R-\mathrm i/(\omega C)$}
Consider a stipulated <series circuit> containing an ideal <resistor> of <electrical resistance> $R=100\,\Omega$ and an ideal <capacitor> of <capacitance> $C=1\,\mu\mathrm F$. The same <electric current> passes through both. Their terminal <electric potential differences> add by <Kirchhoff's voltage law>, so
$$
\widehat V=\widehat V_R+\widehat V_C
=R\widehat I+\frac{\widehat I}{\mathrm i\omega C},\qquad
Z=R-\frac{\mathrm i}{\omega C}.
$$
A quick derivation uses $i=C\,dv_C/dt$: taking the <derivative> of the <complex exponential> multiplies the capacitor voltage <phasor> by $\mathrm i\omega$, hence $\widehat V_C=\widehat I/(\mathrm i\omega C)$. Dimensionally, $1/(\omega C)$ has units $\mathrm{s/F}=\mathrm{V/A}=\Omega$, matching $R$.
At <frequency> $f=1000\,\mathrm{Hz}$, the <angular frequency> is $\omega=2\pi f$. The signed capacitive <electrical reactance>, <modulus> and <complex argument> of the total <electrical impedance> are
$$
X_C=-159.154943\,\Omega,\qquad
|Z|=187.963549\,\Omega,\qquad
\phi=-57.858092^\circ.
$$
Apply $V_{\rm rms}=1.00\,\mathrm V$ across the complete <series circuit>. Its <root mean square> <electric current> is
$$
I_{\rm rms}=\frac{1.00\,\mathrm V}{187.963549\,\Omega}
=5.320180\,\mathrm{mA}.
$$
With voltage as the zero-phase reference, the current <phasor> has <complex argument> $+57.858092^\circ$: <electric current> leads. The <phasor> drops have <moduli> $V_{R,\rm rms}=0.532018\,\mathrm V$ and $V_{C,\rm rms}=0.846733\,\mathrm V$; these add as <complex numbers>, not as positive magnitudes. Their squared magnitudes sum to $(1.00\,\mathrm V)^2$.
\Image[/series-rc-impedance-locus-and-frequency.png]
{title=<Series RC impedance> moving in the <complex plane> as <frequency> changes}
{description=Original calculated ideal $R=100\,\Omega$, $C=1\,\mu\mathrm F$ model, not measured data or a filter transfer function. Left: the <electrical impedance> remains on $\operatorname{Re}Z=100\,\Omega$ and approaches the real axis as <frequency> increases. Right: its <modulus> and voltage-minus-current <phase difference>. The marked $1000\,\mathrm{Hz}$ case matches the worked example; $f_*=1/(2\pi RC)=1591.55\,\mathrm{Hz}$ marks $|X_C|=R$, $|Z|=\sqrt2R$, and $\phi=-45^\circ$.}
The mean absorbed <power> uses the <root mean square> values:
$$
P=V_{\rm rms}I_{\rm rms}\cos\phi
=RI_{\rm rms}^2
=2.830432\,\mathrm{mW},\qquad
\cos\phi=\frac R{|Z|}=0.532018.
$$
The ideal <capacitor> has zero cycle-averaged absorbed <power>; all the mean absorbed <power> becomes <Joule heating> in the <resistor>.<University Physics Volume 2 by Samuel J. Ling and others>[§15.4, equations 15.12–15.14 and the capacitor/inductor paragraph: RMS average power and zero ideal reactive average power]{ref} To see why the phase factor appears, average $p(t)=v(t)i(t)$ over a full cycle with $v(t)=\sqrt2V_{\rm rms}\cos\omega t$ and $i(t)=\sqrt2I_{\rm rms}\cos(\omega t-\phi)$. The <cosine addition formula> reduces the average to $V_{\rm rms}I_{\rm rms}\cos\phi$; the term at $2\omega$ averages to zero. Independently, $p_C=d(Cv_C^2/2)/dt$, and the <capacitor energy> returns to its starting value after each cycle.
At fixed applied $V_{\rm rms}$, the <limit> $\omega\to0^+$ gives $|Z|\to\infty$, $\phi\to-\pi/2$ and $I_{\rm rms}\to0$: the ideal <capacitor> blocks steady <electric current>. The <limit> $\omega\to\infty$ gives $Z\to R$, $\phi\to0$ and $I_{\rm rms}\to V_{\rm rms}/R$. These are <limits> of the ideal lumped <electrical circuit>; unwanted lead <inductance> and component losses can invalidate that model at high <frequency>. At $\omega RC=1$, the equal legs of the impedance <right triangle> give $|Z|=\sqrt2R$ and $\phi=-\pi/4$ without numerical evaluation. Measuring both the magnitude and the sign of the <phase difference> distinguishes this <capacitor> response from an equally large positive <electrical reactance>.
Back to article page
