= Solution
For the <series-fed parallel LC voltage response>, use <phasors> with time dependence $e^{i\omega t}$. The parallel branch has impedance $iX$, where $X=\omega L/(1-\omega^2LC)$. Therefore
$$
\boxed{Z=R+iX,\quad|Z|=\sqrt{R^2+X^2},\quad\arg Z=\operatorname{atan2}(X,R).}
$$
At the ideal parallel <resonance> $\omega_0=1/\sqrt{LC}$ its input impedance diverges. The input-current phase changes sign on crossing this pole; expressions there are interpreted by limits.
The <capacitor> <voltage> equals the parallel-branch <voltage>. Define $q=R(1-\omega^2LC)/(\omega L)$. Then
$$
\boxed{H(\omega)=\frac{V_C}{V_i}=\frac{iX}{R+iX}=\frac1{1-iq},\quad
|H|=(1+q^2)^{-1/2},\quad\arg H=\arctan q.}
$$
At low <frequency> $H\sim i\omega L/R$; at high <frequency> $H\sim-i/(\omega RC)$; at <resonance> $H=1$. Thus it is a \b[band-pass <voltage> filter or <resonant frequency> selector]. Its phase decreases continuously from $+\pi/2$ through zero to $-\pi/2$. Half-power points satisfy $|q|=1$ and
$$
\omega_\pm=\frac{\sqrt{(RC)^{-2}+4/(LC)}\pm(RC)^{-1}}2,\qquad
\omega_+-\omega_-=\frac1{RC}.
$$
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1-ia2010-lc-response.png]
{title=Voltage response of a parallel resonator}
{description=Original transfer amplitude and phase for the stated series-fed parallel LC circuit. The example has R sqrt(C/L)=1; resonance and both half-power points are marked, with asymptotic phases shown.}
For the supplied component values and $\omega=2\pi(50)\,\mathrm{s^{-1}}$, $X=421.745\,\Omega$ and $V_{C,\rm pk}=99.3046\,\mathrm V$. The element peak currents are
$$
\boxed{I_{R,\rm pk}=\frac{100}{\sqrt{50^2+421.745^2}}=0.23546\,\mathrm A,\quad
I_{L,\rm pk}=\frac{99.3046}{\omega(0.150)}=2.10731\,\mathrm A,\quad
I_{C,\rm pk}=\omega(60\times10^{-6})(99.3046)=1.87185\,\mathrm A.}
$$
<Inductor> and <capacitor> currents are opposite <phasors>; their magnitudes subtract here rather than add. The <resistor> carries the input current, with
$$
\boxed{I_{\rm rms}=I_{R,\rm pk}/\sqrt2=0.16650\,\mathrm A.}
$$
These results assume ideal components and sinusoidal <steady state>.
Back to article page
