EXAMPLES ON AC CIRCUITS (USE OF COMPLEX NOTATION).

1. For the circuit shown below, determine the circuit current I flowing and the p.d
across each impedance.

SOLUTION.
Total impedance ZT =Z 1+ Z 2 +Z 3
This gives ZT =( 15 ∠ 30 0) + ( 10 ∠−600 ) + ( 12 ∠ 500 )
This yields ZT =( 12.99+ j7.5 )+ (5− j8.66 )+ (7.713+ j 9.193 )
This simplifies ZT =( 25.703+ j 8.033 )=26.929 ∠ 17.360 .
0
V 100 ∠ 0 . 0
The current I = = =3.713 ∠−17.36 A .
Z 26.929 ∠ 210 .

P.d across Z1 is given by

V 1=I Z 1=( 3.713 ∠−17.360 ) × ( 15∠ 30 0 )=55.695 ∠ 12.64 0 V
P.d across Z 2 is given by
V 2=I Z 2=( 3.713 ∠−17.360 ) × ( 10∠ −600 )=37.13 ∠−77.36 0 V
P.d across Z3 is given by
V 3=I Z 3=( 3.713∠−17.36 0 ) × ( 12∠ 50 0 )=44.56 ∠32.640 V

2. Two Impedances Z1 =5∠ 30 0 Ω, Z 2=10∠ 450 Ω draw a current of 3.36 A when

connected in series to a certain ac supply. Determine the:
(a) Supply voltage.
(b) Phase angle between the voltage and current
(c) P.d across Z1 .
(d) P.d across Z 2 .

SOLUTION:

(a) Total impedance ZT =Z 1+ Z 2 .

This gives ZT =( 5 ∠ 30 0) + ( 10 ∠ 450 ) =( 4.33+ j 2.5 ) + ( 7.071+ j 7.071 ) .
This simplifies ZT =( 11.401+ j 9.571 )=14.886 ∠ 400 .

Now, V =IZ =3.36 ×14.886 ∠ 400=50 ∠ 400 V .

(b) Phase angle is 40 0 lagging.

(c) P.d across Z1 is given by V 1=I Z 1=3.36 ×5 ∠ 30 0=16.8 ∠ 300 V
(d) P.d across Z 2 is given by V 2=I Z 2=3.36 × 10∠ 450=33.6 ∠ 450 V

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 3. For the circuit diagram below, determine the supply current and its phase relative to
the applied voltage.

SOLUTION:

Let Z1 =( 10+ j 20 ) Ω and Z 2=( 20− j15 )

1 1 1
The total impedance ZT , is given by = + .
Z T Z1 Z 2

1 1 1 1 ( 10− j 20 ) ( 20+ j15 )

This gives = + . We get = +
Z T ( 10+ j20 ) ( 20− j15 ) Z T 102 +202 202+ 152

1 ( 10− j 20 ) ( 20+ j15 )

This yields to = + =0.052− j 0.016
ZT 500 625

1 0
This simplifies to =0.0544 ∠−17.1 .
ZT

V 1
=V × =( 30 ) × ( 0.0544 ∠−17.1 ) =1.632∠−17.1 A .
0 0
The supply current I =
Z Z
4. For the circuit diagram below, determine the supply current and its phase relative to
the applied voltage.

SOLUTION:
1 1
XC= = =26.526 Ω and X L =2 πfL=2 π ×50 × 0.12=37.7 Ω
2 πfc 2 π ×50 ×120 ×10−6

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 Let Z1 =( 50− j26.526 ) Ω , Z 2=40 Ω and Z3 =( 30+ j 37.7 ) Ω
1 1 1 1
The total impedance ZT , is given by = + +
Z T Z1 Z 2 Z 3
1 1 1 1
This gives = + +
Z T ( 50− j 26.526 ) 40 ( 30+ j37.7 )

1 ( 50+ j26.526 ) 1 30− j37.7

This yields to = + +
ZT 3203.63 40 2321.29

1 0
This simplifies to =0.05353− j0.00796=0.054119∠−8.46 .
ZT

V 1
=V × =( 100 ) × ( 0.054119∠−8.46 )
0
The supply current I =
Z Z
This gives the supply current I =5.4119 ∠−8.46 0 A

5. A coil of resistance 25 Ω and inductance 0.044 H is connected in parallel with a

branch made up of a 50 μF capacitor in series with a 40 Ω resistor and the whole is
connected to a 200 V ,50 Hz supply. Calculate the total current taken from the
supply and its phase angle and draw the complete phasor diagram.
SOLUTION:
1 1
XC= = =63.66 2 Ω , X L =2 πfL=2 π ×50 × 0.044=13.823 Ω
2 πfc 2 π ×50 ×50 ×10−6

Let Z1 =( 25+ j 13.823 ) Ω , and Z 2=( 40− j 63.66 2 ) Ω

1 1 1
The total impedance ZT , is given by = +
Z T Z1 Z 2
1 1 1
This gives = +
Z T ( 25+ j13.82 3 ) ( 40− j63.66 2 )

1 ( 25− j 13.823 ) 40+ j63.66 2

This yields to = +
ZT 816.08 5652.85

1 0
This simplifies to =0.03771− j 0.0056764=0.038135 ∠−8.56 .
ZT

V 1
=V × =( 200 ) × ( 0.038135 ∠−8.56 )
0
The supply current I =
Z Z
This gives the supply current I =7. 627 ∠−8.560 A .

6. In the diagram below, the p.d between points A and B is 50 ∠−68.130 V . Determine
the:
(a) Supply current I
(b) Equivalent input impedance
(c) Supply voltage V

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(d) Supply voltage angle.
(e) P.d across points B and C
(f) Value of the currents I 1 and I 2.

SOLUTION:

Let Z BCU =10+ j 6 and Z BCL=4 + j 15. The effective Impedance Z BC for both Z BCL
1 1 1 1 1 1
and Z BCU is given by = + . This gives = + .
Z BC Z BC Z BCL Z BC ( 10+ j 6 ) ( 4+ j 15 )

1 10− j 6 4− j 15 1
This yields to = + . And simplifies to =0.09− j 0. 1064
Z BC 100+36 16 +225 Z BC

1 0
Or simply =0.13936 ∠−49.8 . We get Z BC =7.176 ∠ 49.8 0 Ω
Z BC

We later write Z BC =4.632+ j5.481

Now Total impedance ZT =Z AB+ Z BC =( 3.36− j2.47 ) + ( 4.632+ j5.481 )

This gives ZT =7.992+ j3.011=8.54 ∠ 20.64 0 .

But Z AB =3.36− j 2.47=4.17 ∠−36.320

V AB 50 ∠−68.130 0
(a) Supply Current I = = =11.99 ∠−31.81 A .
Z AB 4.17 ∠−36.32 0
(b) Equivalent in put Impedance ZT =8.54 ∠ 20.64 0 Ω
(c) Supply Voltage V =I Z T =( 11.99∠−31.810 ) ( 8.54 ∠ 20.64 0 . ).
This gives V =I Z T =102 .4 ∠−11. 170 V
(d) ϕ=−11. 170 +31.810=20. 64 0 lagging.
(e) V BC =I Z BC =( 11.99∠−31.810 ) ( 7.176 ∠ 49.8 0 )=86.0 ∠ 18 0 V .
V B C 86.0 ∠180 86.0∠ 18
0
0
(f) I 1= = = 0
=7. 374 ∠−12. 96 A
Z B CU 10+ j6 11.662 ∠30.96
V B C 86.0 ∠ 18 0
86.0∠ 18
0
0
I 2= = = 0
=5.54 ∠−57.07 A
Z B CL 4+ j15 15. 524 ∠75.07

7. A current of ( 15+ j8 ) A flows in a circuit whose supply voltage is ( 120+ j200 ) V .

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 Determine the:
(i) Active power
(ii) Reactive power
SOLUTION:
Apparent Power P=I V . This gives
S= ( 15− j8 )( 120+ j200 )=3400+ j2040=P+ jQ
Therefore, Active Power¿ 3400 W and Reactive Power¿ 2040 var .

8. A current of ( 12+ j 5 ) A flows in a circuit when the supply voltage is ( 150+ j220 ) V
. Determine the:
(a) Active power
(b) Reactive Power
(c) Apparent Power
SOLUTION:
Apparent Power S=V I =( 150+ j220 )( 12− j 5 )=2900+ j1890=P+ jQ
Active Power¿ 2900 W , Reactive power¿ 1890 var lagging ,
Apparent Power¿ 3461.5 VA .
9. A capacitor of capacitive reactance 40 Ω and a resistance of 30 Ω are connected in
series to a supply of 200 ∠ 600 V . Determine the active power in the circuit.
SOLUTION:
Z=( 30− j 40 ) Ω . This gives ϕ=tan ( )
−1 40
30
=¿53.13 0 ¿ and |Z|=50 Ω

We get Z=50 ∠−53.13 . 0

V ( 200 ∠ 600 ) 0
From I = , we get I = =4 ∠ 113.3 A .
Z ( 50 ∠−53.13 ) 0

Active Power¿ IV cos ϕ=200 ×4 cos (−53.13 0 )=480W .

Method 2:
Active Power¿ I R2 R=42 ×30=480 W .

10. Two impedances Z1 =6 ∠ 400 Ω , Z 2=10 ∠30 0 Ω are connected in series and have
a total reactive power of 1650 var lagging. Determine the:
(a) Average Power
(b) Apparent Power
(c) Power factor
SOLUTION:
Total impedance ZT =Z 1+ Z 2 .
This gives ZT =( 6 ∠ 400 ) + ( 10 ∠ 30 0 )=( 4.596+ j 3.8587 ) + ( 8.66 + j 5 ) .
This simplifies ZT =( 13.2565+ j 8.8567 )=15.9429 ∠ 33.750 .
1650
Now IV sin 33.750=1650. This gives apparent Power¿ 0
=2969.92VA
sin 33.75
Apparent Power¿ 2969.9 VA
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 Average Power P=IV cos 33.750=2969.92 cos 33.75 0=2469.4 W .
Power factor ¿ cos ϕ=cos 33.75 0=0.8315 lagging.

[ ( )]
11. A current i=7.5 sin ωt −
π
4
A flows in a circuit which has an applied voltage

v=180 sin ⁡[ωt +(π /12)]V . Determine the:

(a) Circuit Impedance (b) Active Power (c) Reactive Power (d) Apparent Power.
SOLUTION:

( )
0
180 ∠ 15 0
Impedance Z= 0
Ω=24 ∠60 Ω.
7.5 ∠−45

Active power ¿ IV cos ϕ= ( )( )

7.5
√ 2
×
180
√ 2
cos 600 =337.5W

Reactive Power¿ IV sin ϕ=( ) ( √2 )

7.5 180 0
× sin 60 =584.56 var lagging
√2

Apparent Power¿ IV =( ) ( √2 )
7.5 180
× =675 VA
√2

12. The diagram blow shows an R-L series Circuit.

(a) Draw its phasor diagram.
(b) Derive an expression for determining its circuit impedance in complex notation.

SOLUTION:
(a) The Phasor diagram is shown below

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(b) Figure (a) below shows a voltage triangle that is derived from the phasor diagram.
Now if each side of the voltage triangle is divided by the current I , we get the
impedance triangle figure(b) below.
The impedance triangle may be superimposed on the Argand diagram figure (c)
below.

From the Argand diagram, it can be seen that the Impedance Z=R+ j X L.

PARALLEL AC NETWORKS:

The diagram below shows three impedance Z1 , Z 2 and Z3 connected in parallel. The
potential difference across each is the same and equal to the supply voltage V .

V V V
Currents I 1 , I 2 and I 3 are given by I 1= , i = and I 3= .
Z1 2 Z2 Z3

V
The ZT total equivalent impedance of the circuit, then I = .
ZT

The supply current I =I 1+ I 2 + I 3 ( phasorially )

V V V V
Thus = + + .
Z T Z1 Z 2 Z 3

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 1 1 1 1
This gives = + +
Z T Z1 Z 2 Z 3

This can be written as Y T =Y 1 +Y 2 +Y 3 .

In general, for n Impedances connected in parallel,

Y T =Y 1 +Y 2 +Y 3 +…+Y n (Phasorially).

The use of admittance has greatest advantage in parallel circuits.

CURRENT DIVISION IN AC CIRCUITS:

Consider a special case of two impedances Z1 and Z 2 in parallel.

1 1 1
The equivalent impedance ZT is given by = + .
Z T Z1 Z 2

Z1 Z2
The total equivalent Impedance ZT is given by ZT = .
Z1+ Z2

The supply Voltage V is given by V =I z T =I ( Z1 Z2

Z1+ Z2
.)
Also V =I 1 Z 1 and ¿ I 2 Z 2

On substitution we get I ( Z1 Z2
)
Z 1 +Z 2
=I 1 Z 1 and I
(
Z1 Z2
Z 1 +Z 2 )
=I 2 Z 2

From which we can deduce that I 1=I ( Z2

Z 1+ Z 2 )
and I 2=I (
Z1
Z 1+ Z 2
.
)
DEFINITIONS:
Admittance is defined as current flowing in an a.c circuit divided by the supply
voltage. It is the reciprocal of impedance Z. The symbol for Admittance is Y .
I 1
Thus Y = = .
V Z
The Admittance may be resolved into two parts, the real G and the imaginary part
B.
G is called Conductance and B is called Susceptance.
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In Complex form, Y =G ± jB
The diagram below shows an Admittance triangle where
Y = Admittance , G=conductance∧B=Susceptance

Note that Y cos ϕ=G and Y sin ϕ=B.

TRIAL QUESTION:
1. In the diagram shown below, Determine;
(a) Equivalent circuit impedance
(b) Supply current I
(c) Circuit phase angle
(d) Currents I 1 and I 2 .

[ 10.33 ∠−6.310 Ω , 4.84 ∠6.310 A ,6.31 0 leading , I 1=0.953 ∠−73.78 0 A , ]

0
I 2=4.765 ∠ 17.66 A .

2. In the diagram below, determine the:

(a) Value of impedance Z 2 .
(b) Current flowing in Z 2 .
(c) Components comprising of Z 2 if the supply frequency is 2 kHz .

[ 6.25 ∠ 52.34 0 Ω, 16.0 ∠ 7.660 A , R=3.819 Ω , L=0.394 mH ]

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3. The p.d across and the current in a certain circuit are represented by
( 190+ j 40 ) V and ( 9− j 4 ) A respectively. Determine the active power and the
reactive power sating clearly whether the latter is leading or lagging.
[ 1550 W ,1120 var lagging . ]
4. A circuit consists of an impedance 5 ∠−450 Ω in parallel with a resistance of
10 Ω . The supply is 4 A . Determine for the circuit:
(a) Active power
(b) Reactive power
(c) Power factor
[ 49.34 W ,28.90 var leading ,0.863 leading ].

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