Applying Complex Numbers to Series AC Circuits

Simple AC circuits may be analyzed by using phasor diagrams. However, when circuits become more complicated, analysis is considerably simplified by using complex numbers. It is essential that the basic operations used with complex numbers, as outlined in this chapter thus far, are thoroughly understood before proceeding with AC circuit analysis.

7.5.1 Series AC Circuits

7.5.1.1 Pure Resistance

In an AC circuit containing resistance R only (see Figure 7.4(a)), the current I_R is in phase with the applied voltage V_R as shown in the phasor diagram of Figure 7.4(b). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 7.4(c). The impedance Z of the circuit is given by:

Figure 7.4 (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram

7.5.1.2 Pure Inductance

In an AC circuit containing pure inductance L only (see Figure 7.5(a)), the current I_L lags the applied voltage V_L by 90° as shown in the phasor diagram of Figure 7.5(b). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 7.5(c). The impedance Z of the circuit is given by:

Figure 7.5 (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram

where X_L is the inductive reactance given by:

where f is the frequency in hertz and L is the inductance in henrys.

7.5.1.3 Pure Capacitance

In an AC circuit containing pure capacitance only (see Figure 7.5(a)), the current I_C leads the applied voltage V_C by 90° as shown in the phasor diagram of Figure 7.5(b). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 7.5(c). The impedance Z of the circuit is given by:

where X_C is the capacitive reactance given by:

where C is the capacitance in farads. FIGURE 7.6

Figure 7.6 (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram

7.5.1.4 R–L Series Circuit

In an AC circuit containing resistance R and inductance L in series (see Figure 7.7(a)), the applied voltage V is the phasor sum of V_R and V_L as shown in the phasor diagram of Figure 7.7(b). The current I lags the applied voltage V by an angle lying between 0° and 90°—the actual value depending on the values of V_R and V_L, which depend on the values of R and L. The circuit phase angle, that is, the angle between the current and the applied voltage, is shown as angle ϕ in the phasor diagram. In any series circuit the current is common to all components and is taken as the reference phasor in Figure 7.7(b). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 7.7(c), where it may be seen that in complex form the supply voltage V is given by:

Figure 7.7 (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram

Figure 7.8(a) shows the voltage triangle that is derived from the phasor diagram of Figure 7.8(b) (triangle Oab). If each side of the voltage triangle is divided by current I, then the impedance triangle of Figure 7.8(b) is derived. The impedance triangle may be superimposed on the Argand diagram, as shown in Figure 7.8(c), where it may be seen that in complex form the impedance Z is given by:

Figure 7.8 (a) Voltage triangle; (b) Impedance triangle; (c) Argand diagram

For example, an impedance expressed as (3+j4) Ω means that the resistance is 3 Ω and the inductive reactance is 4 Ω.

In polar form, Z=|Z| ∠ϕ where, from the impedance triangle, the modulus of impedance  and the circuit phase angle ϕ=tan^-1 (X_L/R) lagging.

7.5.1.5 R-C Series Circuit

In an AC circuit containing resistance R and capacitance C in series (see Figure 7.9(a)), the applied voltage V is the phasor sum of V_R and V_C as shown in the phasor diagram of Figure 7.9(b). The current I leads the applied voltage V by an angle lying between 0°and 90°—the actual value depending on the values of V_R and V_C, which depend on the values of R and C. The circuit phase angle is shown as angle ϕ in the phasor diagram. The phasor diagram may be superimposed on the Argand diagram as shown in Figure 7.9(c), where it may be seen that in complex form the supply voltage V is given by:

Figure 7.9 (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram

Figure 7.10(a) shows the voltage triangle that is derived from the phasor diagram of Figure 7.10(b). If each side of the voltage triangle is divided by current I, the impedance triangle is derived as shown in Figure 7.10(b). The impedance triangle may be superimposed on the Argand diagram as shown in Figure 7.10(c), where it may be seen that in complex form the impedance Z is given by:

Figure 7.10 (a) Voltage triangle; (b) Impedance triangle; (c) Argand diagram

Thus, for example, an impedance expressed as (9 –j14) Ω means that the resistance is 9 Ω and the capacitive reactance X_C is 14 Ω.

In polar form, Z=|Z| ∠ϕ where, from the impedance triangle, angle,  and ϕ=tan^-1 (X_C/R) leading.

7.5.1.6 R-L-C Series Circuit

In an AC circuit containing resistance R, inductance L and capacitance C in series (see Figure 7.10(a)), the applied voltage V is the phasor sum of V_R, V_L and V_C as shown in the phasor diagram of Figure 7.10(b) (where the condition V_L >V_C is shown). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 7.10(c), where it may be seen that in complex form the supply voltage V is given by:

From the voltage triangle the impedance triangle is derived and superimposing this on the Argand diagram gives, in complex form,

Impedance

where,

When V_L=V_C, X_L=X_C and the applied voltage V and the current I are in phase. This effect is called series resonance. FIGURE 7.11

Figure 7.11 (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram

7.5.1.7 General Series Circuit

In an AC circuit containing several impedances connected in series, say, Z₁, Z₂, Z₃, … , Z_n, then the total equivalent impedance Z_T is given by:

Example 7.5

Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances: (a) (12+j5) Ω; (b) –j40 Ω; (c) 30∠60°Ω; (d) 2.20×10⁶∠–30°Ω. Assume for each a frequency of 50 Hz.

Solution

.

(a) From Section 24.2(d), for an R–L series circuit, impedance Z=R+jX_L.

Thus, Z=(12+j5) Ω represents a resistance of 12 Ω and an inductive reactance of 5 Ω in series.

Since inductive reactance X_L=2πfL,

Inductance 

So, the inductance is 15.9 mH.

Thus, an impedance (12+j5) Ω represents a resistance of 12 Ω in series with an inductance of 15.9 mH

(b) For a purely capacitive circuit, impedance Z=–jX_C.

Thus, Z=–j40 Ω represents zero resistance and a capacitive reactance of 40 Ω.

Since capacitive reactance X_C=1/(2πfC),

Thus, an impedance –j40 Ω represents a pure capacitor of capacitance 79.6 μF.

(c) 30∠60°=30(cos 60°+j sin 60°)=15+j25.98

Thus, Z=30∠60°Ω=(15+j25.98) Ω represents a resistance of 15 Ω and an inductive reactance of 25.98 Ω in series.

Since X_L=2πfL,

Thus, an impedance 30∠60°Ω represents a resistance of 15 Ω in series with an inductance of 82.7 mH.

represents a resistance of 1.905×10⁶ Ω (i.e., 1.905 MΩ) and a capacitive reactance of 1.10×10⁶ Ω in series.

Since capacitive reactance X_C=1/(2πfC),

Thus, an impedance 2.2×10⁶∠–30°Ωrepresents a resistance of 1.905 MΩ in series with a 2.894 nF capacitor.

Example 7.6

Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance 159.2 mH when it is connected to a 250 V, 50 Hz supply.

Solution

Inductive reactance

Thus, circuit impedance Z=(0+j50) Ω=50∠90°Ω

Supply voltage, 

(Note that since the voltage is given as 250 V, this is assumed to mean 250∠0°V or (250+j0)V.)

Hence, current 

which is the same as 5∠–90°A

Example 7.7

A 3-μF capacitor is connected to a supply of frequency 1 kHz and a current of 2.83∠90 A flows. Determine the value of the supply voltage.

Solution

Hence, circuit impedance

Current 

Supply voltage, 

i.e., voltage=150∠0°V

Example 7.8

The impedance of an electrical circuit is (30 –j50) ohms. Determine (a) the resistance, (b) the capacitance, (c) the modulus of the impedance, and (d) the current flowing and its phase angle, when the circuit is connected to a 240 V, 50 Hz supply.

Solution

.

(a) Since impedance Z=(30 –j50) Ω, the resistance is 30 ohms and the capacitive reactance is 50 Ω.

(b) Since X_C=1/(2πfC), capacitance,

(c) The modulus of impedance,

(d) 

Example 7.9

A 200 V, 50 Hz supply is connected across a coil of negligible resistance and inductance 0.15 H connected in series with a 32 Ω resistor. Determine (a) the impedance of the circuit, (b) the current and circuit phase angle, (c) the voltage across the 32 Ω resistor, and (d) the voltage across the coil.

Solution

(a) 

The circuit diagram is shown in Figure 7.12.

Figure 7.12 Circuit diagram for Example 7.9

(b) Current 
i.e., the current is 3.51A lagging the voltage by 55.81°

(c) Voltage across the 32 Ω resistor,

i.e., 

(d) Voltage across the coil,

i.e., 

The phasor sum of V_R and V_L is the supply voltage V as shown in the phasor diagram of Figure 7.13.

Figure 7.13 Phasor diagram for Example 7.9

Hence,

Example 7.10

Determine the value of impedance if a current of (7+j16)A flows in a circuit when the supply voltage is (120+j200)V. If the frequency of the supply is 5 MHz, determine the value of the components forming the series circuit.

Solution

The series circuit consists of a 13.25 Ωresistor and a capacitor of capacitive reactance 1.705 Ω.

Since