• Tags: LCR, impedance, phase-angle
• Difficulty: Moderate

In a series LCR circuit, the same current I flows through all elements. $V_R$ = IR (in phase with I), $V_L$ = $IX_L$ (leads I by 90), $V_C$ = $IX_C$ (lags I by 90). Since $V_L$ and $V_C$ are anti-phase, the net reactive voltage = |$V_L$ - $V_C$|. By phasor addition: V = sqrt($V_R^2$ + (V_{L-V}_C)^2). Impedance Z = $\frac{V}{I}$ = sqrt($R^2$ + (X_{L-X}_C)^2). Phase angle: tan(phi) = $\frac{X_L-X_C}{R}$. If $X_L$ > $X_C$: circuit is inductive, phi > 0, voltage leads current. If $X_C$ > $X_L$: circuit is capacitive, phi < 0, current leads voltage. Special cases: at resonance ($X_L$ = $X_C$): Z = R, phi = 0. The impedance triangle is a right triangle with R as base, (X_{L-X}_C) as height, and Z as hypotenuse. This triangle is the single most useful diagram for solving AC problems.