Phasors and graphs in the two cases

Problem

In a series RLC circuit driven by $V(t) = V_0 \sin(\omega t)$ at steady state, draw on the same plane (one period $T$) the curves $V_R(t)$, $V_L(t)$, $V_C(t)$ and $V(t)$, in the two cases $X_L > X_C$ and $X_L < X_C$. Comment on which leads and which lags the current. Inspired by (Epstein 2002) — see Riferimenti bibliografici.

Solution

With reference current $i(t) = I_0 \sin(\omega t)$, the phasors of the voltage drops are: $\begin{aligned} V_R = R I_0 \sin(\omega t) &\quad \text{(in phase with } i) \\ V_L = X_L I_0 \sin(\omega t + \tfrac{\pi}{2}) &\quad \text{(leads by } +90^\circ) \\ V_C = X_C I_0 \sin(\omega t - \tfrac{\pi}{2}) &\quad \text{(lags by } -90^\circ) \end{aligned}$ The total voltage is $V = V_0 \sin(\omega t + \varphi)$ with $\tan\varphi = \dfrac{X_L - X_C}{R}$.

Case $X_L > X_C$ (inductive): $\varphi > 0$, so $\ev{V \text{ leads } i}$. Case $X_L < X_C$ (capacitive): $\varphi < 0$, so $\ev{V \text{ lags } i}$.

Inductive case: the total voltage $V$ (blue) leads the current $i$ (red).

Capacitive case: the total voltage $V$ (blue) lags the current $i$ (red).

In both cases $V_R$ stays in phase with $i$, $V_L$ leads by $90^\circ$ and $V_C$ lags by $90^\circ$; it is the dominant term ($X_L$ or $X_C$) that fixes the sign of $\varphi$.

Collegamenti

Argomenti: Electromagnetic induction Concetti: Alternating current Competenze: Phasor method · Reading graphs