The document presents a method for measuring three-phase power using the two wattmeter technique, aimed at enabling learners to determine power effectively. It includes objectives, a list of required equipment, circuit diagrams, theoretical equations, and a data table for recording measurements. The methodology involves calculations using wattmeter readings, leading to conclusions about power factor and overall power measurement.

1. Measurement of 3-Phase
Power Using 2 Wattmeter
Method
Presented By:
Md. Shaheen Hawlader
B.Sc in EEE
Green University of Bangladesh

2. Content of the presentation
• Objectives
• Learning Outcome
• List of Equipment
• Circuit Diagram
• Theory
• Data Table

3. Objective & Learning Outcome
The objective of this experiment is to use the Two
Wattmeter method to determine Three-Phase
power.
We will be able to measure Three-Phase power
using Two Wattmeter method.

4. List of Equipment
• 3 – Phase supply cord
• AC ammeter – 1pcs
• Multi-meter – 1pcs
• 3-phase Load
• Wattmeter – 2pcs
• Wires for Connection

5. Circuit Diagram

6. Theory
• Here,
VRY, VYB, VBR = VL [Line Voltage]
IR, IY, IB = IL [ Line Current]
VRN, VYN, VBN = VPh [ Phase Voltage]
IRN, IYN, IBN = IPh [ Phase Current]

7. Theory
• Wattmeter – 1
Current(I): IR
Voltage(V): VRB
Power:
W1 = VRBIRCos(VRBˆIR)
• Wattmeter – 2
Current(I): IY
Voltage(V): VYB
Power:
W2 = VYBIYCos(VYBˆIY)

8. VRB = VRN – VBN
VYB = VYN – VBN
Phasor Diagram

9. From Phasor Diagram we can Write,
W1 = VRBIRCos(VRBˆIR)
= VLILCos(300 + Φ)
W2 = VYBIYCos(VYBˆIY)
= VLILCos(300 - Φ)

10. W1 + W2
• 𝑊1 + 𝑊2 = 𝑉𝐿 𝐼𝐿[𝐶𝑜𝑠 30° + Φ + 𝐶𝑜𝑠 30° − Φ ]
= 𝑉𝐿 𝐼𝐿 [2𝐶𝑜𝑠
30°+Φ+30°−Φ
2
. 𝐶𝑜𝑠
30°+Φ−30°+Φ
2
]
= 2 𝑉𝐿 𝐼𝐿 𝐶𝑜𝑠30°𝐶𝑜𝑠Φ
= 2 𝑉𝐿 𝐼𝐿
3
2
𝐶𝑜𝑠Φ
= 𝑉𝐿 𝐼𝐿 3𝐶𝑜𝑠Φ ………………(1)

11. W1 - W2
𝑊1 − 𝑊2 = 𝑉𝐿 𝐼𝐿[𝑐𝑜𝑠 30° + Φ − 𝑐𝑜𝑠 30° − Φ ]
= 𝑉𝐿 𝐼𝐿[2𝑠𝑖𝑛
30° + Φ + 30° − Φ
2
. 𝑠𝑖𝑛
30° − Φ − 30° − Φ
2
]
= 2 𝑉𝐿 𝐼𝐿 𝑠𝑖𝑛30°sin(−Φ)
= −2 𝑉𝐿 𝐼𝐿
1
2
𝑠𝑖𝑛Φ
= − 𝑉𝐿 𝐼𝐿 𝑠𝑖𝑛Φ…………………(2)

12. 𝑾 𝟏− 𝑾 𝟐
𝑾 𝟏+𝑾 𝟐
• Deviding Eqn (2) By Eqn (1) We Get,
𝑊1 − 𝑊2
𝑊1 + 𝑊2
=
− 𝑉𝐿 𝐼𝐿 𝑠𝑖𝑛Φ
𝑉𝐿 𝐼𝐿 3𝐶𝑜𝑠Φ
𝑊1 − 𝑊2
𝑊1 + 𝑊2
= −
1
3
𝑡𝑎𝑛Φ
𝑡𝑎𝑛Φ = − 3(
𝑊1 − 𝑊2
𝑊1 + 𝑊2
)
Φ = 𝑡𝑎𝑛−1
{− 3(
𝑊1−𝑊2
𝑊1+𝑊2
)}
• Power Factor of Load = cosΦ

13. Data Table
Sl.
No.
For W1 For W2 W =
W1
+ W2
Voltage(V) Current(I)
Power
W1 = V*I
Voltage(V) Current(I)
Power
W2 = V*I
01.

14. Thank You