The document discusses power in AC circuits, including impedance, phasor and power diagrams for RL and RC loads. It describes the concepts of active power (P), reactive power (Q), apparent power (S) and power factor. Low power factor is disadvantageous and can be improved by connecting a capacitor bank parallel to the load to provide the required reactive power. Illustrative examples are provided to calculate the capacitance and kvar rating required to improve the power factor of single-phase loads.

1. Basic Electrical Technology
[ELE 1051]
SINGLE PHASE AC CIRCUITS
L19 – Power in AC circuits
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2. Topics covered…
 Impedance, phasor & Power Diagram
 Concept of power factor and its significance
 Need for power factor improvement
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3. Power associated in RL load
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For RL load:
𝜃
𝑹
𝐗𝐋
𝒁
Impedance diagram
× 𝐼
𝜃
𝑷 = 𝑰𝟐𝑹
𝑸 = 𝑰𝟐𝐗𝐋
𝑺 = 𝑰𝟐𝒁
Power diagram
× 𝐼
𝜃
𝑽𝑹 = 𝑰𝑹
𝑽𝑳 = 𝑰 𝐗𝐋
𝑽 = 𝑰𝒁
𝐼
Phasor diagram
𝑺 = 𝑷 + 𝒋𝑸
𝑆 = 𝑉𝐼
𝑃 = 𝑉𝐼 cos ∅
𝑄 = 𝑉𝐼 sin ∅
Where,
S = Apparent Power (VA)
P = Active Power (W)
Q = Reactive Power (var)

4. Power associated in RL load
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For RC load:
× 𝐼 × 𝐼
𝑺 = 𝑷 − 𝒋𝑸
Where,
S = Apparent Power (VA)
P = Active Power (W)
Q = Reactive Power (var)
𝜃
𝑹
𝐗𝐂
𝒁
Impedance diagram
𝜃
𝑽𝑹 = 𝑰𝑹
𝑽𝑳 = 𝑰 𝐗𝐂
𝑽 = 𝑰𝒁
𝐼
Phasor diagram
𝑷 = 𝑰𝟐
𝑹
𝑸 = 𝑰𝟐𝐗𝐂
𝑺 = 𝑰𝟐𝒁
Power diagram
𝜃
𝑆 = 𝑉𝐼
𝑃 = 𝑉𝐼 cos ∅
𝑄 = 𝑉𝐼 sin ∅

5. Power Factor
 For an impedance Z,
cos 𝜃 =
𝐼𝑅
𝑉
=
𝐼𝑅
𝐼𝑍
=
𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒
 Power factor is lagging when the current lags the supply
voltage
 Power factor is leading when the current leads the supply
voltage
 For a resistive load, power factor is Unity
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𝑷𝒐𝒘𝒆𝒓 𝑭𝒂𝒄𝒕𝒐𝒓 =
𝑨𝒄𝒕𝒊𝒗𝒆 𝑷𝒐𝒘𝒆𝒓 𝑷 𝒊𝒏 𝒘𝒂𝒕𝒕𝒔
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝑷𝒐𝒘𝒆𝒓 𝑺 𝒊𝒏 𝒗𝒐𝒍𝒕𝒂𝒎𝒑𝒆𝒓𝒆𝒔
cos 𝜃 =
𝑃
𝑆
=
𝑃
𝑉𝐼

6. Disadvantages of Low Power Factor
 Under utilisation of power system network
 Increased transmission losses
 Hence bulk consumers are advised to maintain the power
factor close to unity by power utilities
Remedial Measures
o Most of the industrial loads are inductive in nature
o Reactive power demand of Inductive loads can be
compensated with capacitive loads
o Hence it is possible to localise reactive power requirement by
connecting parallel capacitors across the load
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7. Power Triangle
 Practically, loads are in connected parallel
 Majority of the loads are inductive in nature
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A.C.
Mains
Load
2
Load
1
𝑺𝒕𝒐𝒕𝒂𝒍
𝑷𝒕𝒐𝒕𝒂𝒍 = 𝑷𝟏 + 𝑷𝟐
𝑸𝒕𝒐𝒕𝒂𝒍 = 𝑸𝟏 + 𝑸𝟐
𝑺𝒕𝒐𝒕𝒂𝒍 = 𝑷𝒕𝒐𝒕𝒂𝒍 + 𝒋𝑸𝒕𝒐𝒕𝒂𝒍
Load 2
∅2
𝑺𝟐
𝑷𝟐
𝑸𝟐
Load 1
∅1
𝑷𝟏
𝑸𝟏
𝑺𝟏
∅𝑇𝑂𝑇𝐴𝐿

8. Power Factor Improvement
 Connect capacitor parallel to the load
 Energy stored by the capacitor provides the required reactive power
by the load
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𝑺𝒕
𝑷𝒕
𝑸𝒕
∅𝑇𝑂𝑇𝐴𝐿
Load
2
Load
1
A.C.
Mains
capacitor
𝑺𝒕
𝑷𝒕
𝑸𝒄
∅𝑇𝑂𝑇𝐴𝐿
∅𝑛𝑒𝑤
𝑸𝒏𝒆𝒘 = 𝑸𝒕 − 𝑸𝒄
𝑺𝒏𝒆𝒘
• Calculate 𝑄𝑐needed to improve power factor to 𝑐𝑜𝑠∅𝑛𝑒𝑤
• Calculate 𝑋𝑐 =
𝑉2
𝑄𝐶
& c =
𝑋𝑐
2𝜋𝑓
Calculation of capacitor value

9. Illustration 1
A single-phase motor takes 8.3 A at a power factor of 0.866
lagging when connected to a 230 V, 50 Hz supply. Capacitance
bank is now connected in parallel with the motor to raise the
power factor to unity. Determine the capacitance value
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Ans: 𝟓𝟕. 𝟒 𝝁𝑭

10. Illustration 2
A single-phase load of 5 kW operates at a power factor of 0.6
lagging. It is proposed to improve this power factor to 0.95
lagging by connecting a capacitor across the load. Calculate the
kvar rating of the capacitor
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Ans: 5.02 kvar

11. Summary
 Power in AC circuit is given by, 𝑺 = 𝑽𝑰
 For any load with impedance Z,
o Active Power (watt), 𝑷 = 𝑽𝑰 𝐜𝐨𝐬 𝜽 = 𝑰𝟐
𝒁 𝐜𝐨𝐬 𝜽 = 𝑰𝟐
𝑹
o Reactive Power (var), 𝑸 = 𝑽𝑰 𝐬𝐢𝐧 𝜽 = 𝑰𝟐
𝒁 𝐬𝒊𝒏 𝜽 = 𝑰𝟐
𝑿
o Apparent Power (VA), 𝑺 = 𝑽𝑰 = 𝑰𝟐
𝒁
 𝑷𝒐𝒘𝒆𝒓 𝑭𝒂𝒄𝒕𝒐𝒓 =
𝑨𝒄𝒕𝒊𝒗𝒆 𝑷𝒐𝒘𝒆𝒓
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝑷𝒐𝒘𝒆𝒓
=
𝑃
𝑆
=
𝑅
𝑍
= cos 𝜃
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Impedance diagram
RL Load RC Load
Power diagram
RL Load RC Load
∅
P - Active Power
Q
–
Reactive
Power
∅
P - Active Power
Q
–
Reactive
Power

12. Summary
 Low power factor loads must be avoided
 Capacitor bank connected parallel to the load serves as the
source of reactive power for the load
o Improves load power factor
o Reduces transmission and distribution losses
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