Kapasitor bank functions primarily to improve power factor, especially for large AC current applications in industry. It also provides reactive power to maximize utilization of complex power (kVA), reduces voltage drop, prevents transformer overload, controls temperature rise in cables, improves efficiency by lowering total KVA usage, minimizes electrical network losses, and prevents penalties from electricity suppliers due to reactive power. Examples show how to calculate the required capacitor size in kVAR and Farads to improve the power factor of motors and alternators from given values using simple formulas and tables. Key formulas used in power factor improvement and these calculations are also outlined.

1. KAPASITOR BANK

2. FUNGSI UTAMA KAPASITOR BANK
•Memperbaiki faktor kuasa elektrik
khususnya pada pemasangan arus AC
dengan nilai kuasa besar seperti
penggunaan di industri.

3. FUNGSI KAPASITOR BANK SELAIN DARI
MEMPERBAIKI POWER FACTOR (PF);
• Membekalkan kuasa reaktif dengan tujuan untuk memaksimakan penggunaan kuasa
komplek (kva).
• Mengurangi terjadinya Voltage drop atau voltan menurun.
• Mencegah overload atau beban lebih pada transformer kerana kapasitor bank boleh
berfungsi juga sebagai kuasa tambahan.
• Mencegah kenaikan suhu kabel (temperature).
• Mengawal Efisiensi /kuasa dengan menurunkan KVA secara total kerana penggunaan
KVA lebih efisien dengan nilai Kw yang digunakan.
• Meminimakan kerugian pada jaringan elektrik.
• Mencegah denda daripada Pembekal Tenaga Seperti Tenaga Nasional
Berhad(Semenanjung Malaysia), Sabah Elektricity Sdn. Bhd.(SESB),Sarawak Energy
Berhad(SEB) kerana adanya kuasa reaktif.

4. EXAMPLE: 1
A 3 PHASE, 5 KW INDUCTION MOTOR HAS A P.F (POWER
FACTOR) OF 0.75 LAGGING. WHAT SIZE OF CAPACITOR IN
KVAR IS REQUIRED TO IMPROVE THE P.F (POWER
FACTOR) TO 0.90?
Solution #1 (By Simple Table Method)
• Motor Input = 5kW
• From Table, Multiplier to improve PF from 0.75 to 0.90 is .398
• Required Capacitor kVAR to improve P.F from 0.75 to 0.90
• Required Capacitor kVAR = kW x Table 1 Multiplier of 0.75 and 0.90 = 5kW x .398
= 1.99 kVAR
• And Rating of Capacitors connected in each Phase, 1.99/3 = 0.663 kVAR

5. EXAMPLE: 1
A 3 PHASE, 5 KW INDUCTION MOTOR HAS A P.F (POWER
FACTOR) OF 0.75 LAGGING. WHAT SIZE OF CAPACITOR IN
KVAR IS REQUIRED TO IMPROVE THE P.F (POWER
FACTOR) TO 0.90?
Solution # 2 (Classical Calculation Method)
• Motor input = P = 5 kW
• Original P.F = Cosθ1 = 0.75
• Final P.F = Cosθ2 = 0.90
• θ1 = Cos-1 = (0.75) = 41°.41; Tan θ1 = Tan (41°.41) = 0.8819
• θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843
• Required Capacitor kVAR to improve P.F from 0.75 to 0.90
• Required Capacitor kVAR = P (Tan θ1 – Tan θ2) = 5kW (0.8819 – 0.4843) = 1.99 kVAR
• And Rating of Capacitors connected in each Phase, 1.99/3 = 0.663 kVAR

6. EXAMPLE 2:
AN ALTERNATOR IS SUPPLYING A LOAD OF 650 KW AT A P.F (POWER FACT OR) OF
0.65. WHAT SIZE OF CAPACITOR IN KVAR IS REQUIRED TO RAISE THE P.F (POWER
FACTOR) TO UNITY (1)? AND HOW MANY MORE KW CAN THE ALTERNATOR
SUPPLY FOR THE SAME KVA LOADING WHEN P.F IMPROVED.
Solution #1 (By Simple Table Method)
• Supplying kW = 650 kW
• From Table 1, Multiplier to improve PF from 0.65 to unity (1) is 1.169
• Required Capacitor kVAR to improve P.F from 0.65 to unity (1)
• Required Capacitor kVAR = kW x Table 1 Multiplier of 65 and 100 = 650kW x 1.169 = 759.85
kVAR
• We know that P.F = Cosθ = kW/kVA . . .or
• kVA = kW / Cosθ = 650/0.65 = 1000 kVA
• When Power Factor is raised to unity (1)
• No of kW = kVA x Cosθ = 1000 x 1 = 1000kW
• Hence increased Power supplied by Alternator, 1000kW – 650kW = 350kW

7. EXAMPLE 2:
AN ALTERNATOR IS SUPPLYING A LOAD OF 650 KW AT A P.F (POWER FACT OR) OF
0.65. WHAT SIZE OF CAPACITOR IN KVAR IS REQUIRED TO RAISE THE P.F (POWER
FACTOR) TO UNITY (1)? AND HOW MANY MORE KW CAN THE ALTERNATOR
SUPPLY FOR THE SAME KVA LOADING WHEN P.F IMPROVED.
Solution # 2 (Classical Calculation Method)
• Supplying kW = 650 kW
• Original P.F = Cosθ1 = 0.65
• Final P.F = Cosθ2 = 1
• θ1 = Cos-1 = (0.65) = 49°.45; Tan θ1 = Tan (41°.24) = 1.169
• θ2 = Cos-1 = (1) = 0°; Tan θ2 = Tan (0°) = 0
• Required Capacitor kVAR to improve P.F from 0.75 to 0.90
• Required Capacitor kVAR = P (Tan θ1 – Tan θ2) = 650kW (1.169– 0)
= 759.85 kVAR

8. HOW TO CALCULATE THE
REQUIRED CAPACITOR BANK
VALUE IN BOTH KVAR AND
FARADS?
(HOW TO CONVERT FARADS
INTO KVAR AND VICE VERSA)

9. EXAMPLE: 3
A single phase 400V, 50hz, motor takes a supply current of 50A at a
P.F (power factor) of 0.6. The motor power factor has to be improved
to 0.9 by connecting a capacitor in parallel with it. Calculate the
required capacity of capacitor in both kvar and farads.
Solution #1 (by simple table method)
• Motor input = p = v x i x cosθ = 400V x 50A x 0.6 = 12kw
• From table, multiplier to improve pf from 0.60 to 0.90 is 0.849
• Required capacitor kvar to improve P.F from 0.60 to 0.90
• Required capacitor kvar = kw x table multiplier of 0.60 and 0.90 = 12kw
x 0.849 = 10.188 kvar

10. EXAMPLE: 3
A SINGLE PHASE 400V, 50HZ, MOTOR TAKES A SUPPLY CURRENT OF 50A
AT A P.F (POWER FACTOR) OF 0.6. THE MOTOR POWER FACTOR HAS TO
BE IMPROVED TO 0.9 BY CONNECTING A CAPACITOR IN PARALLEL WITH
IT. CALCULATE THE REQUIRED CAPACITY OF CAPACITOR IN BOTH KVAR
AND FARADS.
Solution # 2 (Classical Calculation Method)
• Motor Input = P = V x I x Cosθ = 400V x 50A x 0.6 = 12kW
• Actual P.F = Cosθ1 = 0.6
• Required P.F = Cosθ2 = 0.90
• θ1 = Cos-1 = (0.60) = 53°.13; Tan θ1 = Tan (53°.13) = 1.3333
• θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843
• Required Capacitor kVAR to improve P.F from 0.60 to 0.90
= P (Tan θ1 – Tan θ2) = 5kW (1.3333– 0.4843) =10.188 kVAR

11. EXAMPLE: 3
A SINGLE PHASE 400V, 50HZ, MOTOR TAKES A SUPPLY CURRENT OF 50A
AT A P.F (POWER FACTOR) OF 0.6. THE MOTOR POWER FACTOR HAS TO
BE IMPROVED TO 0.9 BY CONNECTING A CAPACITOR IN PARALLEL WITH
IT. CALCULATE THE REQUIRED CAPACITY OF CAPACITOR IN BOTH KVAR
AND FARADS.
Solution #1 (Using a Simple Formula)
We have already calculated the required Capacity of Capacitor in kVAR,
we can easily convert it into Farads by using this simple formula
• Required Capacity of Capacitor in Farads/Microfarads = C = kVAR / (2
f V2) in microfarad
• Putting the Values in the above formula = (10.188kVAR) / (2 x π x 50 x
4002)
= 2.0268 x 10-4 = 202.7 x 10-6 = 202.7μF

12. Example: 3
A single phase 400V, 50hz, motor takes A supply current of
50A at A P.F (power factor) of 0.6. The motor power factor
has to be improved to 0.9 by connecting A capacitor in
parallel with it. Calculate the required capacity of
capacitor in both kvar and farads.
Solution # 2 (Simple Calculation Method)
• kVAR = 10.188 … (i)
• We know that; IC = V/ XC , Whereas XC = 1 / 2 π F C
• IC = V / (1 / 2 π F C) = V 2 F C = (400) x 2π x (50) x C = 125663.7 x C
And,
• kVAR = (V x IC) / 1000 … [kVAR =( V x I)/ 1000 ] = 400 x 125663.7 x C
• IC = 50265.48 x C … (ii)

13. EXAMPLE: 3
A SINGLE PHASE 400V, 50HZ, MOTOR TAKES A SUPPLY CURRENT OF 50A
AT A P.F (POWER FACTOR) OF 0.6. THE MOTOR POWER FACTOR HAS TO
BE IMPROVED TO 0.9 BY CONNECTING A CAPACITOR IN PARALLEL WITH
IT. CALCULATE THE REQUIRED CAPACITY OF CAPACITOR IN BOTH KVAR
AND FARADS.
• Solution # 2 (Simple Calculation Method)
Equating Equation (i) & (ii), we get,
• 50265.48 x C = 10.188C = 10.188 / 50265.48 = 2.0268 x 10-4 = 202.7 x 10-6 = 202.7μF

14. EXAMPLE 4
WHAT VALUE OF CAPACITANCE MUST BE CONNECTED IN PARALLEL WITH
A LOAD DRAWING 1KW AT 70% LAGGING POWER FACTOR FROM A 208V,
60HZ SOURCE IN ORDER TO RAISE THE OVERALL POWER FACTOR TO
91%.
• Solution:
• You can use either Table method or Simple Calculation method to find the required value of Capacitance in Farads or
to improve Power factor from 0.71 to 0.97. So I used table method in this case.
• P = 1000W
• Actual Power factor = Cosθ1 = 0.71
• Desired Power factor = Cosθ2 = 0.97
• From Table, Multiplier to improve PF from 0.71 to 0.97 is 0.783
• Required Capacitor kVAR to improve P.F from 0.71 to 0.97
• Required Capacitor kVAR = kW x Table Multiplier of 0.71 and 0.97
• = 1kW x 0.783
• =783 VAR (required Capacitance Value in kVAR)

15. EXAMPLE 4
WHAT VALUE OF CAPACITANCE MUST BE CONNECTED IN PARALLEL WITH
A LOAD DRAWING 1KW AT 70% LAGGING POWER FACTOR FROM A 208V,
60HZ SOURCE IN ORDER TO RAISE THE OVERALL POWER FACTOR TO
91%.
Solution:
• Current in the Capacitor = IC = QC / V = 783 / 208 = 3.76A
• And
• XC = V / IC = 208 / 3.76 = 55.25Ω
• C = 1/ (2 π f XC) = 1 (2 π x 60 x 55.25) = 48 μF (required Capacitance Value in Farads)

16. IMPORTANT FORMULAS WHICH IS USED FOR
POWER FACTOR IMPROVEMENT CALCULATION AS
WELL AS USED IN THE ABOVE CALCULATION
Power in Watts
• kW = kVA x Cosθ
• kW = HP x 0.746 or (HP x 0.746) / Efficiency … (HP = Motor Power)
• kW = √ ( kVA2– kVAR2)
• kW = P = VI Cosθ … (Single Phase)
• kW = P =√3x V x I Cosθ … (Three Phase)

17. IMPORTANT FORMULAS WHICH IS USED FOR
POWER FACTOR IMPROVEMENT CALCULATION AS
WELL AS USED IN THE ABOVE CALCULATION
Apparent Power in VA
• kVA= √(kW2+ kVAR2)
• kVA = kW/ Cosθ
Reactive Power in VA
• kVAR= √(kVA2– kW2)
• kVAR = C x (2 π f V2)

18. IMPORTANT FORMULAS WHICH IS USED FOR
POWER FACTOR IMPROVEMENT CALCULATION AS
WELL AS USED IN THE ABOVE CALCULATION
Power factor (from 0.1 to 1)
• Power Factor = Cosθ = P / V I … (Single Phase)
• Power Factor = Cosθ = P / (√3x V x I) … (Three Phase)
• Power Factor = Cosθ = kW / kVA … (Both Single Phase & Three Phase)
• Power Factor = Cosθ = R/Z … (Resistance / Impedance)
XC = 1/ (2 π f C) … (XC = Capacitive reactance)
• IC = V/ XC … (I = V / R)
Required Capacity of Capacitor in Farads/Microfarads
• C = kVAR / (2 π f V2) in microfarad
Required Capacity of Capacitor in kVAR
• kVAR = C x (2 π f V2)