MCQOPTIONS
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MCQOPTIONS
This section includes 27 Mcqs, each offering curated multiple-choice questions to sharpen your Electrical Engineering knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In the two-wattmeter method of measuring 3-phase power, the wattmeters indicate equal and opposite readings when load power factor is |
| A. | 90 leading |
| B. | 60 lagging |
| C. | 30 leading |
| D. | 30 lagging |
| Answer» B. 60 lagging | |
| 2. |
In case of power measurements by two wattmeter method in a balanced 3 – phase system with pure inductive load |
| A. | Both wattmeters will indicates the same value but with opposite signs |
| B. | Both wattmeters will indicate zero |
| C. | Both the wattemeters will indicate the same value same sign |
| D. | One wattmeter will indicate zero and the other will indicate same non zero value. |
| Answer» B. Both wattmeters will indicate zero | |
| 3. |
An electric iron designed for 110 V AC supply was rated at 500 W. It was put across a 220 V supply. Assuming that at 110 V it supplied 500 W output (i.e. no losses) at the new voltage it will supply |
| A. | 2500 W |
| B. | 250 W |
| C. | 500 W |
| D. | 2000 W |
| Answer» E. | |
| 4. |
A wattmeter is connected as shown in the figure. The wattmeter reads |
| A. | Zero always |
| B. | Total power consumed by Z1 and Z2 |
| C. | Power consumed by Z1 |
| D. | Power consumed by Z2 |
| Answer» E. | |
| 5. |
An iron cored choke coil has an equivalent resistance of 4 Ω. It draws 10A from a single phase ac source of voltage 200 V, 50 Hz. Then, power consumed by the coil and its power factor respectively are: |
| A. | 200 W, 0.2 lag |
| B. | 400 W, 0.2 lag |
| C. | 200 W, 0.2 lead |
| D. | 400 W, 0.2 lead |
| Answer» C. 200 W, 0.2 lead | |
| 6. |
Calculate the percentage error for a wattmeter which is so connected that the current coil is on the load side, The wattmeter has a current coil of 0.03Ω resistance and a pressure coil of 6000Ω resistance. It is also known that the load takes 20 A at a voltage of 220 V and 0.6 power factor. |
| A. | 0.45% |
| B. | 45% |
| C. | 5.5% |
| D. | 6.5% |
| Answer» B. 45% | |
| 7. |
In a balanced 3-phase 400V circuit, the line current is 115.5 A. When the power is measured by two wattmeter method, one meter reads 40 kW and the other zero. What is the power factor of the load |
| A. | 1 |
| B. | 0.5 |
| C. | 0.6 |
| D. | 0.8 |
| Answer» C. 0.6 | |
| 8. |
In a 3-phase system, two-wattmeter method is used to measure the power. If one of the wattmeters shows a negative reading and the other shows a positive reading, and the magnitude of the readings are not the same, then what will be the power factor (p.f.) of the load? |
| A. | 0.0 < p.f. < 0.5 |
| B. | 0.5 < p.f. < 1.0 |
| C. | 1 |
| D. | 0.5 |
| Answer» B. 0.5 < p.f. < 1.0 | |
| 9. |
A three-phase 500 V motor load has a power factor of 0.4. Two wattmeters connected to measure the input read 20 kW and 10 kW. Find the reactive power (Q) |
| A. | 51.96 kvar |
| B. | 10 kvar |
| C. | 17.32 kvar |
| D. | 30 kvar |
| Answer» D. 30 kvar | |
| 10. |
Power consumed by a balanced 3 – phase, 3 – wire load is measured by the two wattmeter method. The first wattmeter reads twice that of the second, then the load impedance angle in radians is – |
| A. | \(\frac{\pi }{{12}}\) |
| B. | \(\frac{\pi }{{8}}\) |
| C. | \(\frac{\pi }{{6}}\) |
| D. | \(\frac{\pi }{{3}}\) |
| Answer» D. \(\frac{\pi }{{3}}\) | |
| 11. |
If 3-phase power is measured with the help of two-wattmeter method in a balanced load with the application of 3-phase balanced voltage, variation in readings of wattmeters will depend on |
| A. | load only |
| B. | power factor only |
| C. | load and power factor |
| D. | neither load nor power factor |
| Answer» D. neither load nor power factor | |
| 12. |
A – 3 phase balanced load which has a power factor of 0.707 is connected to balanced supply. The power consumed by the load is 5 kW. The power is measured by the two-wattmeter method. The readings of the two watt meters are. |
| A. | 3.94 kW and 1.06 kW |
| B. | 2.50 kW and 2.50 kW |
| C. | 5.00 kW and 0.00 kW |
| D. | 2.96 kW and 2.04 kW |
| Answer» B. 2.50 kW and 2.50 kW | |
| 13. |
In case of two wattmeter method of measurement of 3-phase power measurement of balanced load, the power factor angle is |
| A. | \({\tan ^{ - 1}}\left( {\frac{{{w_1} + {w_2}}}{{{w_1} - {w_2}}}} \right)\) |
| B. | \({\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{w_1} - {w_2}} \right)}}{{{w_1} + {w_2}}}} \right)\) |
| C. | \({\tan ^{ - 1}}\left( {\frac{{{w_1} - {w_2}}}{{{w_1} + {w_2}}}} \right)\) |
| D. | \({\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{w_1} + {w_2}} \right)}}{{{w_1} - {w_2}}}} \right)\) |
| Answer» C. \({\tan ^{ - 1}}\left( {\frac{{{w_1} - {w_2}}}{{{w_1} + {w_2}}}} \right)\) | |
| 14. |
A 400 V, three-phase, rated frequency balanced source is supplying power to a balanced three-phase load carrying a line current of 5 A at an angle of 30° lagging. The readings of the two wattmeters W1 and W2, used for measuring the power drawn by the circuit, are respectively |
| A. | 2000 W and 1000 W |
| B. | 1500 W and 1500 W |
| C. | 2000 W and 1500 W |
| D. | 1500 W and 1000 W |
| Answer» B. 1500 W and 1500 W | |
| 15. |
In a three phase supply driving a load, what is the minimum number of single phase watt meters required to measure the power and power factor? |
| A. | One |
| B. | Two |
| C. | Three |
| D. | Four |
| Answer» C. Three | |
| 16. |
In the measurement of power of balanced load by two wattmeter method in a 3-phase circuit, The readings of the wattmeters are 4 kW and 2 kW respectively, the later is being taken after reversing the connections of current coil. the power factor and reactive power of the load is |
| A. | 0.2 & 6 kVAR |
| B. | 0.2 & 6 √3 kVAR |
| C. | 0.32 & 2 kVAR |
| D. | 0.32 & 2 √3 kVAR |
| Answer» C. 0.32 & 2 kVAR | |
| 17. |
A 400 V three phase 50 Hz balanced source is supplying power to a balanced three phase load. Line current flowing through the load is 5A at a power factor angle of 30 degrees lagging. Reading of two wattmeters used to measure the load power are: |
| A. | 1000 W, 2000 W |
| B. | 2000 W, 4000 W |
| C. | 2000 W, 3000 W |
| D. | 1000 W, 3000 W |
| Answer» B. 2000 W, 4000 W | |
| 18. |
In the circuit shown below, what would be the wattmeter reading? |
| A. | 640 W |
| B. | 320 W |
| C. | 800 W |
| D. | 400 W |
| Answer» B. 320 W | |
| 19. |
In a two-wattmeter method of measuring power in a balance 3-phase circuit, the ratio of the two wattmeter reading is 1 : 2. The circuit power factor is |
| A. | 0.707 |
| B. | 0.5 |
| C. | 0.866 |
| D. | indeterminate |
| Answer» D. indeterminate | |
| 20. |
In two wattmeter method of measuring 3 phase power, power factor is 0.5 , then one of the wattmeter will read |
| A. | W/2 |
| B. | 0 |
| C. | 0.577 W |
| D. | 1.414 W |
| Answer» C. 0.577 W | |
| 21. |
In a balanced 3-phase 200 V circuit, the line current is 115.5 A. When the power is measured by two wattmeter method, one of the wattmeter reads 20 kW and the other one reads zero. What is the power factor of the load? |
| A. | 0.5 |
| B. | 0.6 |
| C. | 0.7 |
| D. | 0.8 |
| Answer» B. 0.6 | |
| 22. |
In electrodynamometer type wattmeters, pressure coil inductance produce error which is: |
| A. | constant irrespective of power factor |
| B. | higher at low power factors of load |
| C. | zero at leading power factors of load |
| D. | lower at low power factors of load |
| Answer» C. zero at leading power factors of load | |
| 23. |
A wattmeter is measuring the power supplied to a circuit whose power factor is 0.7. The frequency of the supply is 50 cycles/second. The wattmeter has a potential coil circuit of resistance 1000Ω and inductance 0.5H. The error in the meter reading is |
| A. | 4% |
| B. | 8% |
| C. | 12% |
| D. | 16% |
| Answer» E. | |
| 24. |
In 3-phase power measurement by two-wattmeter method, the two wattmeters read as W1 = 300 W and W2 = 300 W. Then the load is said to be operating at _________. |
| A. | Unity power factor |
| B. | Lagging power factor |
| C. | Zero power factor |
| D. | Leading power factor |
| Answer» B. Lagging power factor | |
| 25. |
A wattmeter reads 10 kW, when its current coil is connected in R phase and the potential coil is connected across R and neutral of a balanced 400 V (RYB sequence) supply. The line current is 54 A. If the potential coil reconnected across B-Y phases with the current coil in R phase, the new reading of the wattmeter will be nearly |
| A. | 10 kW |
| B. | 13 kW |
| C. | 16 kW |
| D. | 19 kW |
| Answer» C. 16 kW | |
| 26. |
In the two-wattmeter method of measurement of three-phase power of a balanced load, if both wattmeters indicate the same reading, then the power factor of the load is: |
| A. | 0.5 lag |
| B. | < 0.5 lag |
| C. | Unity |
| D. | > 0.5 lag |
| Answer» D. > 0.5 lag | |
| 27. |
Measurement of power factor for balanced load by two wattmeters for lagging power factor is: |
| A. | \(\tan \phi = \sqrt 3 \left( {\frac{{{W_1} + {W_2}}}{{{W_1} - {W_2}}}} \right)\) |
| B. | \(\tan \phi = - \sqrt 3 \left( {\frac{{{W_1} - {W_2}}}{{{W_1} + {W_2}}}} \right)\) |
| C. | \(\tan \phi = \sqrt 3 \left( {\frac{{{W_1} - {W_2}}}{{{W_1} + {W_2}}}} \right)\) |
| D. | \(\tan \phi = - \sqrt 3 \left( {\frac{{{W_1} + {W_2}}}{{{W_1} - {W_2}}}} \right)\) |
| Answer» D. \(\tan \phi = - \sqrt 3 \left( {\frac{{{W_1} + {W_2}}}{{{W_1} - {W_2}}}} \right)\) | |