2. CONTENTS
2.STEADY STATE ANALYSIS OF AC CIRCUITS
2.1 Response to Sinusoidal Excitation-Pure Resistance
2.2 Response to Sinusoidal Excitation-Pure Inductance
2.3 Response to Sinusoidal Excitation-Pure Capacitance
2.4 Impedance concept and phase angle
2.5 Series Circuits-RL,RC,RLC
2.6 Steady state AC Mesh Analysis
2.7 Steady state AC Nodal Analysis
2.8 Star-Delta &Delta-Star Transformation
3. 2.STEADY STATE ANALYSIS OFAC CIRCUITS
2.1 Response to Sinusoidal Excitation-Pure Resistance :
β’ The circuit which contains only a resistance of R ohms in
the AC circuit is known as Pure Resistive AC Circuit.
β’ Let a sinusoidal alternating voltage is
applied across a pure resistance as shown in the Fig.2.1(a).
π£(π‘) = ππsinππ‘
β’ The current flowing through the resistance R is
π(π‘) =
)π£(π‘
π
=
ππ
π
sinππ‘
4. where
β’ The current flowing through the resistance is also
sinusoidal and it is in phase with the applied voltage.
β’ The phase angle between voltage and current is zero.
β’ In pure resistance, current and voltage are in phase.
i. Instantaneous power
π(π‘) = πΌ πsinππ‘
πΌ π =
ππ
π
)π(π‘) = π£(π‘) Γ π(π‘
π(π‘) = ππ πΌ πsin2
ππ‘
The average power is given by
5. πππ£ =
1
π
0
π
π(π‘)ππ‘
=
1
π
0
π
ππ πΌ π
2
β
ππ πΌ π
2
cos2ππ‘ π π‘
=
ππ πΌ π
2
=
ππ
2
πΌ π
2
πππ£ = ππ.π.π πΌπ.π.π
Where Pav is average power
Vr.m.s is root mean square value of supply voltage
Ir.m.s is root mean square value of the current
6. Example 2.1 A sinusoidal voltage is applied to the resistive circuit shown in
Fig.2.1(d).Determine the following values
(a) (b) (c) (d)πΌπππ πΌ ππ£ πΌ π πΌ ππ
Solution The function given to the circuit shown is
The current passing through the resistor is
π£(π‘) = ππsinππ‘ = 20sinππ‘
π(π‘) =
)π£(π‘
π
=
20
2 Γ 103
sinππ‘ = 10 Γ 10β3sinππ‘
7. Peak value
Peak to peak value
Rms value = 0.707Γ10 mA=7.07 mA
πΌ π=10 mA
πΌ ππ=20 mA
πΌπππ =
πΌ π
2
Average Value =
1
π
0
π
πΌ πsinππ‘ πππ‘πΌ ππ£
πΌ ππ£ =
1
π
βπΌ πcosππ‘
0
π
= 0.637 πΌ π
= 0.637 Γ 10ππ΄ = 6.37ππ΄
8. 2.2 Response to Sinusoidal Excitation-Pure Inductance :
β’ The circuit which contains only inductance (L) in the
Circuit is called a Pure inductive circuit.
β’ Let a sinusoidal alternating voltage is
applied across a pure inductance as shown in the
Fig.2.2(a).
π£(π‘) = ππsinππ‘
β’ As a result, an alternating current i(t) flows through the inductance which induces an
emf in it.
π = βπΏ
ππ
ππ‘
β’ The emf which is induced in the circuit is equal and opposite to the applied voltage
11. ii. The energy stored in a pure inductor is obtained by integrating power expression
over a positive half cycle of power variation.
Energy Stored
=
ππ πΌ π
2π
=
1
2
πΏπΌ π
2
Energy stored in a pure inductor =
1
2
πΏπΌ π
2 π½ππ’πππ
= β
ππ πΌ π
2
βcos2ππ‘
2π π
2
π
π =
π
2
π
π π‘ ππ‘ = β
ππ πΌ π
2
π
2
π
sin2ππ‘ ππ‘
12. Example 2.2 Determine the rms current in the circuit shown in Fig 2.2(d)
Solution Inductive reactance π πΏ = 2πππΏ
= 2π Γ 10 Γ 103
Γ 50 Γ 10β3
π πΏ = 3.141ππΊ
πΌπππ =
ππππ
π πΏ
πΌπππ =
10
3.141 Γ 103
= 3.18 mA
13. 2.3 Response to Sinusoidal Excitation-Pure Capacitance :
β’ The circuit which contains only a pure capacitor of
capacitance C farads is known as a Pure Capacitor
Circuit.
β’ Let a sinusoidal alternating voltage is
applied across a pure capacitance as shown in the
Fig.2.3(a).
π£ π‘ = ππsinππ‘
β’ Current flowing through the circuit is given by the equation
π(π‘) =
ππ
ππ‘
=
)π(πΆπ
ππ‘
14. π(π‘) = ππΆππcosππ‘
= πΌ πsin ππ‘ +
π
2
πΌ π = ππΆππ
ππ
πΌ π
=
1
ππΆ
=
1
2πππΆ
= π πΆ
β’ In the pure Capacitor circuit, the current flowing through the
capacitor leads the voltage by an angle of 90 degrees.
i. Instantaneous power )π(π‘) = π£(π‘) Γ π(π‘
= ππsinππ‘ Γ πΌ πsin Οπ‘ +
π
2
= ππ πΌ πsinππ‘cosππ‘
15. π(π‘) =
ππ πΌ π
2
sin2ππ‘
The average power is given by
πππ£ =
1
π 0
T
ππ πΌ π
2
sin 2ππ‘ π ππ‘ = 0
ii. The energy stored in a pure capacitor is obtained by integrating power
expression over a positive half cycle of power variation.
Energy Stored = π =
0
π 2
)π(π‘ ππ‘ =
ππ πΌ π
2
0
π 2
sin2ππ‘ ππ‘
=
ππ πΌ π
2π
=
1
2
πΆππ
2
Energy stored in a pure capacitor=
1
2
πΆππ
2 Joules
16. Example 2.3 Determine the rms current in the circuit shown in Fig 2.3(d)
Solution Capacitive reactance π πΆ =
1
2πππΆ
=
1
2π Γ 5 Γ 103 Γ 0.01 Γ 10β6
π πΆ = 3.18πΎπΊ
πΌπππ =
ππππ
π πΆ
πΌπππ =
5
3.18πΎ
= 1.57 mA
17. 2.4 Impedance and Phase angle :
β’ Impedance is defined as the opposition offered by the circuit elements to the flow
of alternating current.
β’ It can also be defined as the ratio of voltage function to current function and it is
denoted with Z.
β’ If voltage and current are both sinusoidal functions of time, the phase difference
between voltage and current is called phase angle.
Impedance=Z=
ππππ‘πππ ππ’πππ‘πππ
πΆπ’πππππ‘ ππ’πππ‘πππ
π =
π£
π
=
π π
πΌ π
=
π π ππ
πΌ π ππ
ohms
19. 2.5.1 Series RL Circuit :
β’ Consider a circuit consisting of pure resistance
connected in series with pure inductance.
β’ Let a sinusoidal alternating voltage is applied across a
series RL circuit as shown in the Fig.2.5(a).
By applying Kirchhoffβs voltage law to the circuit
shown in Fig.2.5(a)
We get,
π = ππ + ππΏ
π = πΌπ + πΌπ πΏ
β’ Generally, for series a.c. circuit, current is taken as the reference phasor and the
phasor diagram is shown in the Fig.2.5(b).
20. Steps to draw Phasor diagram:
1. Take current as a reference phasor.
2. In case of resistance, voltage and current are in phase, so
VR will be along current phasor.
3. In case of inductance, current lags voltage by 90 degrees.
4. Supply voltage is obtained by the vector sum of VL and
VR .
π = ππ
2 + ππΏ
2 = πΌπ 2 + πΌ Γ π πΏ
2
= πΌ π 2 + π πΏ
2
ππ =
Consider the right angle triangle OAB,
21. π = πΌπ
π = π 2 + π πΏ
2Impedance,
From impedance triangle,
tanπ =
π πΏ
π
In polar form, impedance can be represented as
π = |π|β π
π = π + ππ πΏ
In rectangular form, impedance can be represented as
|π| = π 2 + π πΏ
2 π = tanβ1
π πΏ
π and
π = πcosπ, π πΏ = πsinπ
22. Instantaneous power )π(π‘) = π£(π‘) Γ π(π‘
Where andπ£(π‘) = ππsinππ‘ )π(π‘) = πΌ πsin(ππ‘ β π
π(π‘) = ππsinππ‘ Γ )πΌ πsin(ππ‘ β π
π(π‘) =
ππ πΌ π
2
2sin ππ‘ β π sinππ‘
π(π‘) =
ππ
2
πΌ π
2
)cosπ β cos(2ππ‘ β π
π(π‘) =
ππ
2
πΌ π
2
cosπ β
ππ
2
πΌ π
2
cos 2ππ‘ β π
The average power consumed in the circuit over one complete cycle is given by
23. πππ£ =
1
π
0
π
π(π‘)π π‘ =
ππ πΌ π
2π
0
π
cosπ β cos 2ππ‘ β π π π‘
= ππ.π.π πΌπ.π.π cosπ = ππΌcosππππ£ =
ππ
2
πΌ π
2
cosπ
Example 2.4 To the circuit shown in the Fig.2.5(e),consisting a 1KW resistor connected
in series with a 50mH coil, a 10Vrms,10KHZ signal is applied. Find impedance Z,current
I, phase angle ,voltage across the resistance and the voltage across the inductance .ππ ππΏ
Solution Inductive reactance
In rectangular form,
Total impedance
π πΏ = ππΏ = 2πππΏ = 6.28 104 50 Γ 10β3 = 3140πΊ
π = 1000 + π3140 πΊ
= π 2 + π πΏ
2 = 1000 2 + 3140 2 = 3295.4πΊ
π
24. Current
Phase angle
Therefore, in polar form in total impedance
Voltage across the resistance
Voltage across the inductance
πΌ =
ππ
π
=
10
3295.4
= 3.03ππ΄
π = tanβ1
π πΏ
π
= tanβ1
3140
1000
= 72.330
π = 3295.4β 72.330
ππ = πΌπ = 3.03 Γ 10β3
Γ 1000 = 3.03π
ππΏ = πΌπ πΏ = 3.03 Γ 10β3 Γ 3140 = 9.51π
Example 2.5 Determine the source voltage and the phase angle, if voltage across the
resistance is 70V and the voltage across the inductance is 20V as shown in Fig.
Solution Source voltage is given by ππ = ππ
2
+ ππΏ
2
= 70 2 + 20 2 = 72.8π
25. The angle between the current and source voltage is
π = tanβ1
20
70
= 15.940
2.5.2 Series RC Circuit :
β’ Consider a circuit consisting of pure resistance R ohms
connected in series with a pure capacitor of capacitance C
farads.
β’ Let a sinusoidal alternating voltage is applied across a series
RC circuit as shown in the Fig.2.6(a).
By applying Kirchhoffβs voltage law to the circuit shown in Fig.2.6(a)
π = ππ + ππΆ
26. π = πΌπ + πΌπ πΆ
β’ Generally, for series a.c. circuit, current is taken as the reference phasor and the
phasor diagram is shown in the Fig.2.6(b).
Steps to draw Phasor diagram:
1. Take current as a reference phasor.
2. In case of resistance, voltage and current are in phase, so VR
will be along current phasor.
3. In case of pure capacitance, current leads the voltage by 90
degrees.
4. Supply voltage is attained by the vector sum of VC and VR .
28. In polar form, impedance can be represented as
π = |π|β π
|π| = π 2 + π πΆ
2 and π = tanβ1
π πΆ
π
Instantaneous power )π(π‘) = π£(π‘) Γ π(π‘
Where andπ£(π‘) = ππsinππ‘ )π(π‘) = πΌ πsin(ππ‘ + π
π(π‘) = ππsinππ‘ Γ )πΌ πsin(ππ‘ + π
π(π‘) =
ππ πΌ π
2
2sin ππ‘ + π sinππ‘
π(π‘) =
ππ
2
πΌ π
2
)cosπ β cos(2ππ‘ + π
The average power consumed in the circuit over one complete cycle is given by
29. πππ£ =
1
π
0
π
π(π‘)π ππ‘ =
ππ πΌ π
2π
0
π
cosπ β cos 2ππ‘ + π π ππ‘
= ππ.π.π πΌπ.π.π cosπ = ππΌcosππππ£ =
ππ
2
πΌ π
2
cosπ
Example 2.6 Determine the source voltage and phase angle when the voltage across the
Resistor is 20V and the capacitor is 30V as shown in Fig.
Solution Source voltage is given by
ππ = ππ
2
+ ππΆ
2
= 20 2 + 30 2 = 36π
The angle between the current and source voltage is
π = tanβ1
30
20
= 56.30
31. Current πΌ =
ππ
π
=
10
3760.6
= 2.66ππ΄
Capacitive Voltage ππΆ = πΌπ πΆ = 2.66 Γ 10β3 Γ 3184.7 = 8.47π
Resistive Voltage ππ = πΌπ = 2.66 Γ 10β3 Γ 2000 = 5.32π
Total applied voltage in rectangular form, ππ = 5.32 β π8.47π
Total applied voltage in polar form, ππ = 10β β 57.870 π
2.5.3 Series RLC Circuit :
β’ Consider a circuit consisting of a pure resistance R
ohms, a pure inductance L Henry and a pure
capacitor of capacitance C farads are connected in
series.
32. β’ Let a sinusoidal alternating voltage is applied across a series RLC circuit as shown in the
Fig.2.7(a)
By applying Kirchhoffβs voltage law to the circuit shown in Fig.2.7(a)
π = ππ + ππΏ + ππΆ
β’ Generally, for series a.c. circuit, current is taken as the reference phasor and the phasor
diagram is shown in the Fig.2.7(b).
Steps to draw Phasor diagram:
1.Take current as reference.
2. is in phase with I.
3. leads current I by
ππ
ππΏ 900
33. 4. Lags current I by
5. Obtain the resultant of and .Both and are in phase opposition ( out of phase).
6.Add that with by law of parallelogram to get the supply voltage.
ππΆ 900
ππΏ ππΆ ππΏ ππΆ 1800
ππ
i) :πΏ π³ > πΏ πͺ
π = ππ
2 + ππΏ β ππΆ
2 = πΌπ 2 + πΌπ πΏ β πΌπ πΆ
2
= πΌ π 2 + π πΏ β π πΆ
2
From the Voltage triangle,
43. 2.6 Steady State AC Mesh Analysis:
A mesh is defined as a loop which does not contain any other loops within it.
Number of equations=branches-(nodes-1)
M=B-(N-1)
By applying Kirchhoffβs voltage law around the first mesh
π1 = πΌ1 π1 + πΌ1 β πΌ2 π2
By applying Kirchhoffβs voltage law around the second mesh
π2 πΌ2 β πΌ1 + π3 πΌ2 = 0
47. 2.7 Steady State AC Nodal Analysis:
β’ In general, in a N node circuit, one of the nodes is choosen as reference or datum node,
then it is possible to write N-1 nodal equations by assuming N-1 node voltages.
β’ The node voltage is the voltage of a given node with respect to one particular node,
called the reference node (which is assumed at zero potential).
ππ β π1
π1
+
ππ
π2
+
ππ β ππ
π3
= 0
βπ1
π1
+ ππ
1
π1
+
1
π2
+
1
π3
β
ππ
π3
= 0 β¦ β¦ β¦ (1)
ππ β ππ
π3
+
ππ
π4
+
ππ
π5 + π6
= 0
β
ππ
π3
+ ππ
1
π3
+
1
π4
+
1
π5 + π6
= 0 β¦ β¦ β¦ (2)
48.
49.
50.
51.
52.
53. 2.8 Delta-Star transformation:
Three resistances may be connected in star (or Y) and delta(or Ξ) connection as shown
in figure
In the star connection,
π ππ = π π + π π β¦ β¦ β¦ (1)
π ππ = π π + π π β¦ β¦ β¦ (2)
π ππ = π π + π π β¦ β¦ β¦ (3)
Similarly in delta connection, the resistance seen from
ab,bc and ca are given by
π ππ = π 1|| π 2 + π 3 β¦ β¦ β¦ (4)
π ππ = π 2|| π 1 + π 3 β¦ β¦ β¦ (5)
π ππ = π 3|| π 1 + π 2 β¦ β¦ β¦ (6)