Problem — Ranking RMS current in RLC circuits

Problem

Four series RLC circuits are driven by the same sinusoidal emf of amplitude $V_0$ at the same angular frequency $\omega$, but have different components:

• (A) $R=10\ \Omega$, $X_L-X_C=0$
• (B) $R=10\ \Omega$, $X_L-X_C=10\ \Omega$
• (C) $R=20\ \Omega$, $X_L-X_C=30\ \Omega$
• (D) $R=5\ \Omega$, $X_L-X_C=0$

Rank the magnitude of the RMS current $I_\text{eff}$ from smallest to largest.

Solution

The RMS current is $I_\text{eff}=V_\text{eff}/Z$, with impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$ Since $V_\text{eff}$ and $\omega$ are the same for all, $I_\text{eff}$ is larger the smaller $Z$ is: $I_\text{eff}\propto 1/Z$.

Calculating the impedances: $\text{(A)}\quad Z=\sqrt{10^2+0^2}=10\ \Omega$ $\text{(B)}\quad Z=\sqrt{10^2+10^2}=\sqrt{200}\approx 14{,}1\ \Omega$ $\text{(C)}\quad Z=\sqrt{20^2+30^2}=\sqrt{400+900}=\sqrt{1300}\approx 36{,}1\ \Omega$ $\text{(D)}\quad Z=\sqrt{5^2+0^2}=5\ \Omega$

Ranking the impedances from smallest to largest: $D\ (5)<A\ (10)<B\ (14{,}1)<C\ (36{,}1)$ Since $I_\text{eff}\propto 1/Z$, the order of the currents is reversed. From smallest to largest: $C<B<A<D$

$\ev{C<B<A<D}$

Links

Topics: Electromagnetic induction Concepts: Alternating current · Resonance Skills: Ranking reasoning