Application of complex numbers to series a.c.

circuits 347

Figure 24.7 (a) Voltage triangle (b) Impedance triangle

(c) Argand diagram
Figure 24.8 (a) Circuit diagram (b) Phasor diagram
it may be seen that in complex form the supply voltage (c) Argand diagram
V is given by:
in Figure 24.8(c), where it may be seen that in complex
V = VR − jVC form the supply voltage V is given by:
Figure 24.7(a) shows the voltage triangle that is V = VR + j(VL − VC )
derived from the phasor diagram of Figure 24.6(b). From the voltage triangle the impedance triangle is
If each side of the voltage triangle is divided by cur- derived and superimposing this on the Argand diagram
rent I, the impedance triangle is derived as shown gives, in complex form,
in Figure 24.7(b). The impedance triangle may be
superimposed on the Argand diagram as shown in impedance Z = R + j (XL − XC ) or Z = | Z | ∠ φ

Part 3
Figure 24.7(c), where it may be seen that in complex where,
form the impedance Z is given by √
|Z |= [R 2 + (X L − X C )2 ] and
Z = R − jXC
φ = tan−1 (X L − X C )/R
Thus, for example, an impedance expressed as
(9 − j 14)  means that the resistance is 9  and the When V L = VC , X L = X C and the applied voltage V
capacitive reactance X C is 14  and the current I are in phase. This effect is called series
In polar form, Z = |Z |∠ φ where, from the impedance resonance and is discussed separately in Chapter 28.
√
triangle, |Z |= (R 2 + X C2 ) and φ = tan−1 (X C /R)
leading (g) General series circuit
In an a.c. circuit containing several impedances con-
(f) R–L–C series circuit nected in series, say, Z 1 , Z 2, Z 3, . . ., Z n , then the total
In an a.c. circuit containing resistance R, inductance L equivalent impedance Z T is given by
and capacitance C in series (see Figure 24.8(a)), the ZT = Z1 + Z2 + Z3 + · · ·+ Zn
applied voltage V is the phasor sum of V R , V L and VC
as shown in the phasor diagram of Figure 24.8(b) (where
Problem 1. Determine the values of the
the condition V L > VC is shown). The phasor diagram
resistance and the series-connected inductance or
may be superimposed on the Argand diagram as shown
 348 Electrical Circuit Theory and Technology

Thus Z = 2.20 × 106∠−30◦ 

capacitance for each of the following impedances:
(a) (12 + j 5)  (b) − j 40  (c) 30∠60◦  = (1.905 × 106 − j 1.10 ×106 ) 
(d) 2.20 × 106 ∠−30◦ . Assume for each a
represents a resistance of 1.905 ×106  (i.e.
frequency of 50 Hz.
1.905 M) and a capacitive reactance of
1.10 × 106  in series (from Section 24.2(e)).
(a) From Section 24.2(d), for an R–L series circuit,
Since capacitive reactance X C = 1/(2π f C),
impedance Z = R + j X L .
Thus Z = (12 + j 5)  represents a resistance of 1
12  and an inductive reactance of 5  in series. capacitance C =
2π f X C
Since inductive reactance X L = 2π f L,
1
= F
XL 5 2π(50)(1.10 × 106)
inductance L = = = 0.0159 H
2π f 2π(50)
= 2.894 × 10−9 F or 2.894 nF
i.e. the inductance is 15.9 mH.
Thus an impedance (12 + j5)  represents a Thus an impedance 2.2 × 106∠ −30 ◦  repre-
resistance of 12  in series with an inductance sents a resistance of 1.905 M in series with a
of 15.9 mH. 2.894 nF capacitor.

(b) From Section 24.2(c), for a purely capacitive

circuit, impedance Z = − j X C Problem 2. Determine, in polar and rectangular
Thus Z = − j 40  represents zero resistance and a forms, the current flowing in an inductor of
capacitive reactance of 40 . negligible resistance and inductance 159.2 mH
when it is connected to a 250 V, 50 Hz supply.
Since capacitive reactance X C = 1/(2π f C),
1 1
capacitance C = = F Inductive reactance
2π f X C 2π(50)(40)
X L = 2π f L = 2π(50)(159.2 × 10−3) = 50 
106
= µF = 79.6 µF Thus circuit impedance Z = (0 + j 50)  =50∠90◦ 
2π(50)(40)
Supply voltage, V = 250∠0◦ V (or (250 + j 0)V)
Thus an impedance −j 40  represents a pure
capacitor of capacitance 79.6 µF. (Note that since the voltage is given as 250 V, this is
assumed to mean 250∠0◦ V or (250 + j 0)V)
Part 3

(c) 30∠60◦ = 30(cos 60◦ + j sin 60◦ ) = 15 + j 25.98

V 250∠0◦ 250
Thus Z = 30∠60◦  = (15 + j 25.98)  represents Hence current I = = = ∠(0◦ − 90◦ )
Z 50∠90◦ 50
a resistance of 15  and an inductive reactance of
25.98  in series (from Section 24.2(d)). ∠−90◦ A
= 5∠

Since X L = 2π f L , V (250 + j 0) 250(− j 50)

Alternatively, I = = =
XL 25.98 Z (0 + j 50) j 50(− j 50)
inductance L = =
2π f 2π(50) − j (50)(250)
= = −j 5 A
= 0.0827 H or 82.7 mH 502
Thus an impedance 30∠ ∠60◦  represents a
which is the same as 5∠−90◦ A
resistance of 15  in series with an inductance
of 82.7 mH.
Problem 3. A 3 µF capacitor is connected to a
(d) 2.20 × 106 ∠−30◦ supply of frequency 1 kHz and a current of
= 2.20 × 106 [cos(−30◦ ) + j sin(−30◦)] 2.83∠90◦ A flows. Determine the value of the
supply p.d.
= 1.905 × 106 − j 1.10 × 106
 Application of complex numbers to series a.c. circuits 349

1 Problem 5. A 200 V, 50 Hz supply is connected

Capacitive reactance X C =
2π f C across a coil of negligible resistance and inductance
1 0.15 H connected in series with a 32  resistor.
=
2π(1000)(3 × 10−6) Determine (a) the impedance of the circuit, (b) the
current and circuit phase angle, (c) the p.d. across
= 53.05  the 32  resistor, and (d) the p.d. across the coil.
Hence circuit impedance
(a) Inductive reactance X L = 2π f L = 2π(50)(0.15)
◦
Z = (0 − j 53.05) = 53.05∠−90  = 47.1 
Impedance Z = R + j X L
Current I = 2.83∠90◦ A (or (0 + j2.83)A) ∠ 55.81◦ 
= (32 + j 47.1) or 57.0∠

Supply p.d., V = IZ = (2.83∠90◦ )(53.05∠−90◦ ) The circuit diagram is shown in Figure 24.9

i.e. ∠ 0◦ V
p.d. = 150∠

Alternatively, V = IZ = (0 + j 2.83)(0 − j 53.05)

= − j 2(2.83)(53.05) = 150 V

Problem 4. The impedance of an electrical circuit

is (30 − j 50) 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 Figure 24.9
240 V, 50 Hz supply.
V 200∠0◦
(b) Current I = =
Z 57.0∠55.81◦
(a) Since impedance Z = (30 − j 50) , the resis-
∠−55.81◦ A
= 3.51∠
tance is 30 ohms and the capacitive reactance is
50  i.e. the current is 3.51 A lagging the voltage by
55.81◦
(b) Since X C = 1/(2π f C), capacitance,

Part 3
(c) P.d. across the 32 resistor,
C=
1
=
1
= 63.66 µF V R = IR = (3.51∠−55.81◦ )(32∠0◦ )
2π f X c 2π(50)(50)
∠−55.81◦ V
i.e. VR = 112.3∠
(c) The modulus of impedance, (d) P.d. across the coil,
√ √
|Z | = (R 2 + X C2 ) = (302 + 502 ) V L = IX L = (3.51∠−55.81◦)(47.1∠90◦ )
∠34.19◦ V
i.e. VL = 165.3∠
= 58.31 
The phasor sum of V R and V L is the supply voltage V
(d) Impedance Z = (30 − j 50)
as shown in the phasor diagram of Figure 24.10.
XC
= 58.31∠ tan−1 V R = 112.3∠−55.81◦ = (63.11 − j 92.89) V
R
= 58.31∠−59.04◦  V L = 165.3∠34.19◦ V = (136.73 + j 92.89) V
Hence
V 240∠0◦
Hence current I = =
Z 58.31∠−59.04◦ V = V R + V L = (63.11 − j 92.89) + (136.73 + j 92.89)
∠59.04 A
= 4.12∠ ◦ = (200 + j 0)V or 200∠0◦ V, correct to three
significant figures.
 350 Electrical Circuit Theory and Technology

(c) R = 0, L =31.8 mH
(d) R = 0, C = 1.061µF
(e) R = 7.5 , L = 41.3 mH
(f) R = 4.243 M, C = 0.750 nF]
2. A 0.4 µF capacitor is connected to a 250 V,
2 kHz supply. Determine the current flowing.
[1.257∠90◦ A or j 1.257 A]
3. Two voltages in a circuit are represented
Figure 24.10
by (15 + j 10)V and (12 − j 4)V. Determine
the magnitude of the resultant voltage when
Problem 6. Determine the value of impedance if these voltages are added. [27.66 V]
a current of (7 + j 16) A flows in a circuit when the
supply voltage is (120 + j 200)V. If the frequency of 4. A current of 2.5∠−90◦ A flows in a coil of
the supply is 5 MHz, determine the value of the inductance 314.2 mH and negligible resis-
components forming the series circuit. tance when connected across a 50 Hz supply.
Determine the value of the supply p.d.
V (120 + j 200) [246.8∠0◦ V]
Impedance Z = =
I (7 + j 16) 5. A voltage (75 + j 90) V is applied across
233.24∠59.04◦ an impedance and a current of (5 + j 12) A
= flows. Determine (a) the value of the circuit
17.464∠66.37◦
impedance, and (b) the values of the compo-
= 13.36∠−7.33  or (13.25 − j 1.705) nents comprising the circuit if the frequency
is 1 kHz. [(a) Z = (8.61 − j 2.66) or
The series circuit thus consists of a 13.25  resistor and 9.01∠−17.19◦ 
a capacitor of capacitive reactance 1.705  (b) R = 8.61 , C = 59.83 µF]
1
Since X C = 6. A 30 µF capacitor is connected in series with
2π f C
a resistance R at a frequency of 200 Hz. The
1 resulting current leads the voltage by 30◦.
capacitance C = Determine the magnitude of R. [45.95 ]
2π f X C
Part 3

1 7. A coil has a resistance of 40  and an induc-

=
2π(5 × 106)(1.705) tive reactance of 75 . The current in the coil
= 1.867 × 10−8 F = 18.67 nF is 1.70∠0◦ A. Determine the value of (a) the
supply voltage, (b) the p.d. across the 40 
resistance, (c) the p.d. across the inductive
Now try the following exercise part of the coil, and (d) the circuit phase
angle. Draw the phasor diagram.
Exercise 97 Further problems on series a.c. [(a) (68 + j 127.5) V or 144.5∠61.93◦ V
circuits (b) 68∠0◦V (c) 127.5∠90◦V (d) 61.93◦
lagging]
1. Determine the resistance R and series induc-
tance L (or capacitance C) for each of the fol- 8. An alternating voltage of 100 V, 50 Hz is
lowing impedances, assuming the frequency applied across an impedance of (20 − j 30).
to be 50 Hz. (a) (4 + j 7) (b) (3 − j 20) Calculate (a) the resistance, (b) the capaci-
(c) j 10  (d) − j 3 k (e) 15∠(π/3) tance, (c) the current, and (d) the phase angle
(f) 6∠−45◦ M between current and voltage
[(a) R = 4 , L =22.3 mH [(a) 20  (b) 106.1 µF (c) 2.774 A
(b) R = 3 , C = 159.2 µF (d) 56.31◦ leading]
 Application of complex numbers to series a.c. circuits 351

9. A capacitor C is connected in series with Problem 8. A circuit comprises a resistance of

a coil of resistance R and inductance 90  in series with an inductor of inductive
30 mH. The current flowing in the circuit is reactance 150 . If the supply current is 1.35 ∠0◦ A,
2.5∠−40◦ A when the supply p.d. is 200 V determine (a) the supply voltage, (b) the voltage
at 400 Hz. Determine the value of (a) resis- across the 90  resistance, (c) the voltage across the
tance R, (b) capacitance C, (c) the p.d. across inductance, and (d) the circuit phase angle. Draw
C, and (d) the p.d., across the coil. Draw the the phasor diagram.
phasor diagram.
[(a) 61.28  (b) 16.59 µF The circuit diagram is shown in Figure 24.12
(c) 59.95∠−130◦ V (d) 242.9∠10.90◦ V]
10. If the p.d. across a coil is (30 + j 20)V at
60 Hz and the coil consists of a 50 mH
inductance and 10  resistance, determine
the value of current flowing (in polar and
Cartesian forms).
[1.69∠−28.36◦ A; (1.49 − j 0.80) A]

24.3 Further worked problems on Figure 24.12

series a.c. circuits
(a) Circuit impedance Z = R + j X L = (90 + j 150)
Problem 7. For the circuit shown in Figure 24.11, or 174.93∠59.04◦
determine the value of impedance Z 2 Supply voltage,
V = IZ = (1.35∠0◦ )(174.93∠59.04◦)
∠59.04◦ V or (121.5 + j202.5) V
= 236.2∠

(b) Voltage across 90  resistor, V R = 121.5 V (since

V = V R + j VL )

Part 3
(c) Voltage across inductance, V L = 202.5 V leading
V R by 90◦

(d) Circuit phase angle is the angle between the supply

current and voltage, i.e. 59.04◦ lagging (i.e. cur-
Figure 24.11 rent lags voltage). The phasor diagram is shown in
Figure 24.13.
Total circuit impedance
V 70∠30◦
Z= =
I 3.5∠−20◦
= 20∠50◦  or (12.86 + j 15.32)
Total impedance Z = Z 1 + Z 2 (see Section 24.2(g)).
Hence (12.86 + j 15.32) =(4.36 − j 2.10) + Z 2
from which, impedance
Z 2 = (12.86 + j 15.32) −(4.36 − j 2.10) Figure 24.13
∠63.99◦
= (8.50 + j 17.42)  or 19.38∠
 352 Electrical Circuit Theory and Technology

Inductive reactance, X L = 2π f L =2π(50)(0.10)

Problem 9. A coil of resistance 25  and
inductance 20 mH has an alternating voltage given = 31.4 
by v = 282.8 sin(628.4t + (π/3)) volts applied Capacitive reactance,
across it. Determine (a) the r.m.s. value of voltage 1 1
XC = = = 26.5 
(in polar form), (b) the circuit impedance, (c) the 2π f C 2π(50)(120 × 10−6)
r.m.s. current flowing, and (d) the circuit phase
Impedance Z = R + j(X L − X C ) (see Section 24.2(f))
angle.
i.e. Z = 12 + j (31.4 −26.5)
(a) Voltage v = 282.8 sin(628.4t + (π/3)) volts means = (12 + j 4.9) or 13.0∠ 22.2◦ 
Vm = 282.8 V, hence r.m.s.
 voltage 
1 V 240∠0◦
V = 0.707 × 282.8 or √ × 282.8 Current flowing, I = =
2 Z 13.0∠22.2◦
i.e. V = 200 V ∠−22.2◦ A,
= 18.5∠
In complex form the r.m.s. voltage may be i.e. the current flowing is 18.5 A, lagging the voltage
expressed as 200∠∠π/3 V or 200∠ ∠60◦ V by 22.2◦.
(b) ω = 2π f = 628.4 rad/s, hence frequency The phasor diagram is shown on the Argand diagram in
Figure 24.15
f = 628.4/(2π) = 100 Hz
Inductive reactance
X L = 2π f L =2π(100)(20 × 10−3 ) = 12.57 

Hence circuit impedance

Z = R + j X L = (25 + j12.57) or
∠26.69◦ 
27.98∠
V 200∠60◦
(c) Rms current, I = =
Z 27.98∠26.69◦
∠33.31◦ A
= 7.148∠
(d) Circuit phase angle is the angle between current I
and voltage V , i.e. 60◦ − 33.31◦ = 26.69◦ lagging. Figure 24.15
Part 3

Problem 10. A 240 V, 50 Hz voltage is applied

across a series circuit comprising a coil of resistance Problem 11. A coil of resistance R ohms and
12  and inductance 0.10 H, and 120 µF capacitor. inductance L henrys is connected in series with a
Determine the current flowing in the circuit. 50 µF capacitor. If the supply voltage is 225 V at
50 Hz and the current flowing in the circuit is
The circuit diagram is shown in Figure 24.14. 1.5∠ −30◦ A, determine the values of R and L.
Determine also the voltage across the coil and the
voltage across the capacitor.

Circuit impedance,
V 225∠0◦
Z= =
Z 1.5∠−30◦
= 150∠30◦  or (129.9 + j 75.0)
Capacitive reactance,
1 1
XC = = = 63.66 
Figure 24.14 2π f C 2π(50)(50 × 10−6)
 Application of complex numbers to series a.c. circuits 353

Circuit impedance Z = R + j(X L − X C )

Problem 12. For the circuit shown in
i.e. 129.9 + j 75.0 = R + j(X L − 63.66) Figure 24.17, determine the values of voltages V1
and V2 if the supply frequency is 4 kHz. Determine
Equating the real parts gives: resistance R = 129.9. also the value of the supply voltage V and the
circuit phase angle. Draw the phasor diagram.
Equating the imaginary parts gives: 75.0 = X L − 63.66,

from which, X L = 75.0 + 63.66 =138.66 

XL 138.66
Since X L = 2π f L, inductance L = =
2π f 2π(50)

= 0.441 H

The circuit diagram is shown in Figure 24.16.

Figure 24.17

For impedance Z 1 ,
1 1
XC = = = 15 
2π f C 2π(4000)(2.653 × 10−6)
Figure 24.16
Hence Z 1 = (8 − j 15) or 17∠−61.93◦ 

Voltage across coil, VCOIL = IZ COIL and voltage V 1 = IZ 1

Part 3
Z COIL = R + j X L = (6∠0◦ )(17∠−61.93◦)
= (129.9 + j 138.66) or 190∠46.87◦  ∠−61.93◦ V or (48 −j90)V
= 102∠
Hence VCOIL = (1.5∠−30◦ )(190∠46.87◦ ) For impedance Z 2 ,
∠16.87◦ V or (272.74 + j82.71) V
= 285∠ X L = 2π f L = 2π(4000)(0.477 × 10−3) = 12 
Voltage across capacitor, Hence Z 2 = (5 + j 12) or 13∠67.38◦ 
V C = I X C = (1.5∠−30◦ )(63.66∠−90◦)
and voltage V 2 = IZ 2 = (6∠0◦ )(13∠67.38◦ )
= ∠−120◦ V
∠−
95.49∠− or (−47.75− j 82.70) V
∠67.38◦ V or (30 +j72)V
= 78∠
[Check: Supply voltage,
Supply voltage, V = V1 + V2 = (48 − j 90)+(30 + j 72)
V = VCOIL + VC
= (272.74 + j 82.71) +(−47.75 − j 82.70) = (78 −j18)V or 80 ∠−13◦ V

= (225 + j 0) V or 225∠0◦ V] Circuit phase angle, φ = 13◦ leading. The phasor dia-
gram is shown in Figure 24.18.
 354 Electrical Circuit Theory and Technology

Figure 24.20

3. A series circuit consists of a 10  resistor,

a coil of inductance 0.09 H and negligible
resistance, and a 150 µF capacitor, and is
connected to a 100 V, 50 Hz supply. Calcu-
late the current flowing and its phase relative
Figure 24.18 to the supply voltage.
[8.17 A lagging V by 35.20◦ ]
4. A 150 mV, 5 kHz source supplies an a.c. cir-
cuit consisting of a coil of resistance 25  and
Now try the following exercise inductance 5 mH connected in series with a
capacitance of 177 nF. Determine the current
Exercise 98 Further problems on series a.c. flowing and its phase angle relative to the
circuits source voltage. [4.44∠42.31◦ mA]

1. Determine, in polar form, the complex 5. Two impedances, Z 1 = 5∠30◦  and Z 2 =

impedances for the circuits shown in 10∠45◦  draw a current of 3.36 A when
Figure 24.19 if the frequency in each case connected in series to a certain a.c. supply.
is 50 Hz. Determine (a) the supply voltage, (b) the
[(a) 44.53∠−63.31◦  (b) 19.77∠52.62◦  phase angle between the voltage and cur-
(c) 113.5 ∠−58.08◦] rent, (c) the p.d. across Z 1, and (d) the p.d.
across Z 2 .
[(a) 50 V (b) 40.01◦ lagging (c) 16.8∠30◦ V
(d) 33.6∠45◦ V]
Part 3

6. A 4500 pF capacitor is connected in series

with a 50  resistor across an alternating volt-
age v = 212.1 sin(π106 t + π/4) volts. Cal-
culate (a) the r.m.s. value of the voltage,
(b) the circuit impedance, (c) the r.m.s. cur-
rent flowing, (d) the circuit phase angle, (e)
the voltage across the resistor, and (f) the
voltage across the capacitor.
[(a) 150∠45◦ V (b) 86.63∠−54.75◦ 
(c) 1.73∠99.75◦ A (d) 54.75◦ leading
(e) 86.50∠99.75◦ V (f) 122.38∠9.75◦ V]

Figure 24.19
7. Three impedances are connected in
series across a 120 V, 10 kHz supply. The
impedances are:
2. For the circuit shown in Figure 24.20 deter-
mine the impedance Z in polar and rectangu- (i) Z 1 , a coil of inductance 200 µH and
lar forms. resistance 8 
[Z = (1.85 + j 6.20) or 6.47 ∠73.39◦ ] (ii) Z 2 , a resistance of 12