= Solution
Use the $e^{i\omega t}$ <phasor> convention and interpret <impedance matching> as conjugate matching for the <maximum power transfer theorem>. The load <electrical impedance> of the <series circuit> is
$$
Z_{RC}=R-\frac{i}{\omega C}.
$$
For $Z=A+i\omega B$, the condition $Z_{RC}=Z^*$ equates the <real parts> and <imaginary parts>:
$$
\boxed{R=A,\qquad \omega^2BC=1,\qquad \omega=\frac1{\sqrt{BC}}.}
$$
These formulas require $B>0$, $C>0$ and a passive realization with $A\geq0$, evaluated at the matching <angular frequency>; if $A,B$ depend on $\omega$, the equation for $\omega$ is implicit.
There is a convention ambiguity in the word “matched”. Literal equality $Z_{RC}=Z$, as required for equal bridge arms in part (b), instead gives $R=A$ and $\omega^2BC=-1$, requiring $B<0$. The inductive cases below use conjugate matching; literal equality is impossible for either passive inductive configuration at a finite positive <angular frequency> because its <electrical reactance> is positive while that of the <series RC impedance> is negative.
Back to article page
