20.4.4 Impedance in Series AC Circuits

When a resistor ($R$), inductor ($L$), and capacitor ($C$) are connected in series to an AC source, the total opposition to current flow is called Impedance ($Z$).

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20.4.4.1 Reactances: A Quick Review

Before finding impedance, recall the individual reactances:

All three have SI unit Ohm ($\Omega$).

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20.4.4.2 Phasor Diagram for a Series RLC Circuit

In a series circuit, the current $I$ is the same through all components and is taken as the reference phasor (along the positive x-axis).

The voltage phasors are:

• $V_R=IR$ — in phase with $I$ (along x-axis)
• $V_L=I{X}_L$ — leads $I$ by $90°$ (along +y axis)
• $V_C=I{X}_C$ — lags $I$ by $90°$ (along −y axis)

Since $V_L$ and $V_C$ are anti-parallel (180° apart), they partially cancel. The resultant total voltage is found by vector (phasor) addition:

$V=\sqrt{V_R^2+(V_L-V_C{)}^2}$

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20.4.4.3 Impedance Formula

Dividing the voltage equation by current $I$:

$Z=\frac{V}{I}=\sqrt{R^2+(X_L-X_C{)}^2}$

Impedance $Z$ is the vector sum of resistance and net reactance. Its SI unit is Ohm ($\Omega$).

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20.4.4.4 Phase Angle

The angle $\varphi$ between the total voltage $V$ and the current $I$ is:

$\tan \varphi =\frac{X_L-X_C}{R}$

• If $X_L>X_C$: circuit is inductive → voltage leads current by $\varphi$
• If $X_L<X_C$: circuit is capacitive → current leads voltage by $\varphi$
• If $X_L=X_C$: resonance → $\varphi =0$, $Z=R$ (minimum impedance)

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20.4.4.5 Worked Example

Problem: A series RLC circuit has $R=6\Omega$, $L=20\text{mH}$, $C=50\mu \text{F}$, connected to a $100\text{V}$, $50\text{Hz}$ AC supply. Find $X_L$, $X_C$, $Z$, and the current $I$.

Solution:

$X_L=2\pi fL=2\pi \times 50\times 0.020=6.28\Omega$

$X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi \times 50\times 50\times 10^{-6}}=63.7\Omega$

$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{6^2+(6.28-63.7{)}^2}=\sqrt{36+3298}\approx 57.7\Omega$

$I=\frac{V}{Z}=\frac{100}{57.7}\approx 1.73\text{A}$

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Summary