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				This section includes 18 Mcqs, each offering curated multiple-choice questions to sharpen your Network Theory knowledge and support exam preparation. Choose a topic below to get started.
| 1. | Determine the power (kW) drawn by the load. | 
| A. | 21 | 
| B. | 22 | 
| C. | 23 | 
| D. | 24 | 
| Answer» E. | |
| 2. | A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IB. | 
| A. | 44.74∠303.4⁰A | 
| B. | 44.74∠-303.4⁰A | 
| C. | 45.74∠303.4⁰A | 
| D. | 45.74∠-303.4⁰A | 
| Answer» C. 45.74∠303.4⁰A | |
| 3. | A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IY. | 
| A. | 44.74∠183.4⁰A | 
| B. | 45.74∠183.4⁰A | 
| C. | 44.74∠183.4⁰A | 
| D. | 45.74∠-183.4⁰A | 
| Answer» D. 45.74∠-183.4⁰A | |
| 4. | A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IR. Assume the phase sequence to be RYB. | 
| A. | 44.74∠-63.4⁰A | 
| B. | 44.74∠63.4⁰A | 
| C. | 45.74∠-63.4⁰A | 
| D. | 45.74∠63.4⁰A | 
| Answer» B. 44.74∠63.4⁰A | |
| 5. | If the load impedance is Z∠Ø, the expression obtained for current (IB) is? | 
| A. | (V/Z)∠-240+Ø | 
| B. | (V/Z)∠-240-Ø | 
| C. | (V/Z)∠240-Ø | 
| D. | (V/Z)∠240+Ø | 
| Answer» C. (V/Z)∠240-Ø | |
| 6. | If the load impedance is Z∠Ø, the expression obtained for current (IY) is? | 
| A. | (V/Z)∠-120+Ø | 
| B. | (V/Z)∠120-Ø | 
| C. | (V/Z)∠120+Ø | 
| D. | (V/Z)∠-120-Ø | 
| Answer» E. | |
| 7. | If the load impedance is Z∠Ø, the current (IR) is? | 
| A. | (V/Z)∠-Ø | 
| B. | (V/Z)∠Ø | 
| C. | (V/Z)∠90-Ø | 
| D. | (V/Z)∠-90+Ø | 
| Answer» B. (V/Z)∠Ø | |
| 8. | In a balanced three-phase system-delta load, if we assume the line voltage is VRY = V∠0⁰ as a reference phasor. Then the source voltage VBR is? | 
| A. | V∠120⁰ | 
| B. | V∠240⁰ | 
| C. | V∠-240⁰ | 
| D. | V∠-120⁰ | 
| Answer» D. V∠-120⁰ | |
| 9. | In a balanced three-phase system-delta load, if we assume the line voltage is VRY = V∠0⁰ as a reference phasor. Then the source voltage VYB is? | 
| A. | V∠0⁰ | 
| B. | V∠-120⁰ | 
| C. | V∠120⁰ | 
| D. | V∠240⁰ | 
| Answer» C. V∠120⁰ | |
| 10. | IN_THE_QUESTION_7,_DETERMINE_THE_PHASE_CURRENT_IB.?$ | 
| A. | 44.74∠303.4⁰A | 
| B. | 44.74∠-303.4⁰A | 
| C. | 45.74∠303.4⁰A | 
| D. | 45.74∠-303.4⁰A | 
| Answer» C. 45.74‚Äö√Ñ√∂‚àö‚Ć‚Äö√тĆ303.4‚Äö√Ñ√∂‚àö√ñ‚Äö√†√ªA | |
| 11. | Determine_the_power_(kW)_drawn_by_the_load.$ | 
| A. | 21 | 
| B. | 22 | 
| C. | 23 | 
| D. | 24 | 
| Answer» E. | |
| 12. | In the question 7, determine the phase current IY? | 
| A. | 44.74∠183.4⁰A | 
| B. | 45.74∠183.4⁰A | 
| C. | 44.74∠183.4⁰A | 
| D. | 45.74∠-183.4⁰A | 
| Answer» D. 45.74‚Äö√Ñ√∂‚àö‚Ć‚Äö√тĆ-183.4‚Äö√Ñ√∂‚àö√ñ‚Äö√†√ªA | |
| 13. | A three phase, balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IR . Assume the phase sequence to be RYB.$ | 
| A. | 44.74∠-63.4⁰A | 
| B. | 44.74∠63.4⁰A | 
| C. | 45.74∠-63.4⁰A | 
| D. | 45.74∠63.4⁰A | 
| Answer» B. 44.74‚Äö√Ñ√∂‚àö‚Ć‚Äö√тĆ63.4‚Äö√Ñ√∂‚àö√ñ‚Äö√†√ªA | |
| 14. | In the question 4, the expression obtained for current (IY) is? | 
| A. | (V/Z)∠-120+Ø | 
| B. | (V/Z)∠120-Ø | 
| C. | (V/Z)∠120+Ø | 
| D. | (V/Z)∠-120-Ø | 
| Answer» E. | |
| 15. | If the load impedance is Z∠Ø, the current (IR ) is?$ | 
| A. | (V/Z)∠-Ø | 
| B. | (V/Z)∠Ø | 
| C. | (V/Z)∠90-Ø | 
| D. | (V/Z)∠-90+Ø | 
| Answer» B. (V/Z)‚Äö√Ñ√∂‚àö‚Ć‚Äö√тĆ‚Äö√†√∂‚àö‚â§ | |
| 16. | In a delta-connected load, the relation between line voltage and the phase voltage is? | 
| A. | line voltage > phase voltage | 
| B. | line voltage < phase voltage | 
| C. | line voltage = phase voltage | 
| D. | line voltage >= phase voltage | 
| Answer» D. line voltage >= phase voltage | |
| 17. | In the question 1, the source voltage VBR is? | 
| A. | V∠120⁰ | 
| B. | V∠240⁰ | 
| C. | V∠-240⁰ | 
| D. | V∠-120⁰ | 
| Answer» D. V‚Äö√Ñ√∂‚àö‚Ć‚Äö√тĆ-120‚Äö√Ñ√∂‚àö√ñ‚Äö√†√ª | |
| 18. | In a balanced three-phase system-delta load, if we assume the line voltage is VRY = V∠0⁰ as a reference phasor. Then the source voltage VYB is? | 
| A. | V∠0⁰ | 
| B. | V∠-120⁰ | 
| C. | V∠120⁰ | 
| D. | V∠240⁰ | 
| Answer» C. V‚Äö√Ñ√∂‚àö‚Ć‚Äö√тĆ120‚Äö√Ñ√∂‚àö√ñ‚Äö√†√ª | |