to enroll in courses, follow best educators, interact with the community and track your progress.
Enroll
698
Previous years GATE problems based on Two Wattmeter Method (Part 3)
951 plays

More
In this lesson I've discussed previous years GATE problems based on Two Wattmeter Method.

## Pawan Chandani is teaching live on Unacademy Plus

Pawan Chandani
Faculty of GATE/ESE (Electrical Engineering) , Also a YouTuber and teaches Electrical Eng. subjects on YouTube | Jony Ive Award Winner

U
Thanks ma'am, Cut off kam btne ke liye... 😇🤗
4 part tk pdh lia h..... 5 part kha milega
Pawan Chandani
a year ago
it will be published tomorrow
sir next part upload kro plz
Pawan Chandani
a year ago
1. Previous Year GATE problems based on Two Wattmeter Method By- Pawan Chandani

2. Q 14. A wattmeter reads 400W when its current coil is connected in the R phase and its pressure coil is connected between this phase and the neutral of a symmetrical 3-0 system supplying a balanced star connected 0.8 pf inductive load. The phase sequence is RYEB What will be the reading of this wattmeter if its pressure coil alone is reconnected between the B and Y phases, all other connections remaining as before?

3. Q 14. A wattmeter reads 400W when its current coil is connected in the R phase and its pressure coil is connected between this phase and the neutral of a symmetrical 3-0 system supplying a balanced star connected 0.8 pf inductive load. The phase sequence is RYB. What will be the reading of this wattmeter if its pressure coil alone is reconnected between the B and Y phases, all other connections remaining as before? -cos-10-8 = 36.86 36. 86 DO = Ph Ph

4. h- = 3%.lensinp Ph Ph Ph Ph N3 = N3 x 500 x sin (36.86") 3 X 300 3

5. Q 15. The circuit in fig. is used to measure the power consumed by the load. The current coil and the voltage coil of the wattmeter have 0.02 and 1000 resistance respectively. The measured power compared to the load power will be 0.020 20A 200 VI LOAD 00052

6. Q 15. The circuit in fig. is used to measure the power consumed by the load. The current coil and the voltage coil of the wattmeter have 0.02 and 1000 resistance respectively. The measured power compared to the load power will be 0.02n 20A dol J-200U = 20A 200 V LOAD 00052 M C cos = 1 = cos = 200 x 20 x 1 = 4000W

7. MC 11 Rc Connections At = 202x0.02 8 h m: 400048 = 4008 W