Past exam of the physics 2 paper of the University of Oxford / 2022 / 8 / a

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 isFor $Z=A+i\omega B$, the condition $Z_{RC}=Z^{\ast }$ equates the real parts and imaginary parts:These formulas require $B>0$, $C>0$ and a passive realization with $A\ge 0$, 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 ${\omega }^{2}BC=-1$, 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.

Let the unknown series circuit contain electrical resistance $R_s>0$ and inductance $L>0$. Its electrical impedance is $Z=R_s+i\omega L$. Thus $A=R_s$, $B=L$, and conjugate impedance matching is possible withAt this angular frequency, the inductor and load capacitor have opposite electrical reactances of equal magnitude. Literal equality of their separate electrical impedances is impossible; for example, $R_s=L=C=1$ gives $1+i\omega$ versus $R-i/\omega$ for every $\omega >0$.

For electrical resistance $R_p>0$ in a parallel circuit with inductance $L>0$, add electrical admittances and invert:Conjugate impedance matching to the series RC impedance requiresIt is possible at a finite positive angular frequency exactly when $C{R}_p^2>L$, in which caseThe expression for $R$ follows by substituting this ${\omega }^2$ into the real part of $Z$. If $C{R}_p^2\le L$, no such finite matching angular frequency exists. Under literal equality, both passive inductive configurations fail by the sign of their imaginary parts.