Physics · Ch 4 — Electromagnetic Induction and Alternating Current

AC Circuit Containing a Resistor, an Inductor and a Capacitor in Series

4.7.8

AC Circuit Containing a Resistor, an Inductor and a Capacitor in Series

Consider a series circuit containing a resistor R, an inductor L and a capacitor C all connected in series across an alternating source $v=V_m\sin\omega t$ . Because the SAME current i flows through all three elements in series, the voltage drop across each has a fixed, known phase relationship to that shared current: $V_R$ is exactly in phase with i, $V_L$ leads i by $\pi/2$ , and $V_C$ lags i by $\pi/2$ . Drawing the phasor diagram with the current phasor $\vec{OI}$ as the reference direction, the three voltage phasors are $\vec{OA}=V_R$ (along i), $\vec{OB}=V_L$ (rotated $90^{\circ}$ ahead), and $\vec{OC}=V_C$ (rotated $90^{\circ}$ behind) -- with $V_L$ and $V_C$ pointing in exactly OPPOSITE directions on the diagram, since both are $90^{\circ}$ from $V_R$ but on opposite sides of it. Assuming $V_L>V_C$ , their net effect is a single phasor $\vec{OD}=(V_L-V_C)$ pointing in the same sense as $V_L$ ; combining $V_R$ and $(V_L-V_C)$ by the parallelogram law gives the resultant phasor $\vec{OE}$ , whose length equals the applied peak voltage $V_m$ , so

V_m^2 = V_R^2+(V_L-V_C)^2

Expressing each voltage as (current) $\times$ (its own reactance/resistance), $I_m^2Z^2 = I_m^2R^2+I_m^2(X_L-X_C)^2$ , so $I_m=V_m/Z \qquad (4.46)$ , where

Z = \sqrt{R^2+(X_L-X_C)^2} \qquad (4.47)

is the impedance of the circuit -- the effective total opposition (in ohms) that a series RLC circuit offers to alternating current, playing the combined role of R, $X_L$ and $X_C$ together. The corresponding voltage triangle (sides $V_R$ , $(V_L-V_C)$ , hypotenuse $V_m$ ) and its similar impedance triangle (sides R, $(X_L-X_C)$ , hypotenuse Z) share the same angle $\phi$ , giving the phase angle between applied voltage and current as

$\tan\phi = \dfrac{V_L-V_C}{V_R} = \dfrac{X_L-X_C}{R} \qquad (4.48)$ …

Figure 4.46AC circuit containing R, L and C in series

What this figure shows. A resistor R, inductor L and capacitor C are connected one after another in a single series loop across an alternating source labelled $v=V_m\sin\omega t$ , with the voltage drops $V_R$ , $V_L$ and $V_C$ marked across each respective element. Because the same current i flows through all three elements in a series circuit, the figure sets up the phasor-addition problem that follows: $V_R$ must be added to the phasor DIFFERENCE $(V_L-V_C)$ (since $V_L$ and $V_C$ point in exactly opposite directions on a phasor diagram, both being $90^{\circ}$ from $V_R$ but on opposite …

Figure 4.47Phasor diagram for a series RLC circuit when $V_L > V_C$

What this figure shows. With the current phasor $\vec{OI}$ drawn along the reference axis, three voltage phasors are added: $\vec{OA}=V_R$ along the same direction as the current, $\vec{OB}=V_L$ rotated $90^{\circ}$ ahead, and $\vec{OC}=V_C$ rotated $90^{\circ}$ behind; since $V_L>V_C$ here, their difference is represented by phasor $\vec{OD}=(V_L-V_C)$ , pointing in the SAME sense as $V_L$ (upward). The parallelogram law then combines $V_R$ and $(V_L-V_C)$ into the single resultant phasor $\vec{OE}$ , whose length equals the applied peak voltage $V_m$ and whose angle $\phi$ above the current axis is the circuit's overall phase angle -- directly giving $V_m^2=V_R^2+(V_L-V_C)^2$ and confirming that with $V_L>V_C$ the appli …

Figure 4.48Voltage triangle and impedance triangle for $X_L > X_C$

What this figure shows. Two similar right-angled triangles are drawn side by side. The 'voltage triangle' (a) has $V_R$ as its horizontal base, $(V_L-V_C)$ as its vertical side, and the resultant $V_m$ as its hypotenuse, with $\phi$ the angle between $V_m$ and $V_R$ . The 'impedance triangle' (b) is exactly the same shape scaled down by the common factor $I_m$ : R as its horizontal base, $(X_L-X_C)$ as its vertical side, and the impedance Z as its hypotenuse, with the SAME angle $\phi$ between Z and R. Because both triangles share the identical angle $\phi$ , the figure shows directly why $\tan\phi=(X_L-X_C)/R$ can be read off from either triangle equivalently, and why $Z=\sqrt{R^2+(X_L-X_C)^2}$ follows purely fr …

Table 4.1Summary of results of AC circuits

Type of Impedance	Value of Impedance	Phase angle of current with voltage	Power factor
Resistance	$R$	$0^{\circ}$	$1$
Inductance	$X_L=\omega L$	$90^{\circ}$ lag	$0$
Capacitance	$X_C=\dfrac{1}{\omega C}$	$90^{\circ}$ lead	$0$

Misc Example 4.22Impedance and phase angle from given XL, XC and R

Worked out. A series RLC circuit has inductive reactance 184 $\Omega$ , capacitive reactance 144 $\Omega$ and resistance 30 $\Omega$ ; the impedance and the phase angle between voltage and current are required. The impedance is $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{30^2+(184-144)^2}=\sqrt{900+1600}=\sqrt{2500}=50\ \Omega$ . The phase angle is $\tan\phi=(X_L-X_C)/R=(184-144)/30=40/30\approx1.33$ , giving $\phi\approx53.1^{\circ}$ . Since $\phi$ comes out positive (because $X_L>X_C$ ), the voltage leads the current by $53.1^{\circ}$ and the circuit is net inductive, illustrating the direct, two-step application of the impedance and phas …

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