201. What is the use of higher flux density in the transformer design?
- For increasing the weight/kVA
- For decreasing the weight/kVA
- For reducing iron losses
- For improving insulation
202. The errors in current transformers can be reduced by designing them with
- Using large cross section for both primary and secondary winding conductors
- Using primary and secondary winding as close to each other as possible
- High permeability and low loss core materials, avoiding any joints in the core and also keeping the flux density to a low value.
- All of the given options
203. Leakage flux in a transformer occurs because
- Applied voltage is sinusoidal
- Iron core has high permeability
- Transformer is not an efficient device
- Air is not a good magnetic insulator
204. The no-load primary current I0 is about _____ of full load primary current of a transformer.
- 30-40%
- 3-5%
- 15-30%
- Above 40%
205. The primary and secondary windings of a transformer are wound on the top of each other in order to reduce
- Leakage reactance
- Iron losses
- Winding resistance
- Copper losses
206. The no load input power to a transformer is practically equal to _______ loss in the transformer.
- Windage
- Copper
- Iron
- Eddy current
207. If the fault current is 2000 A, the relay setting at 50% and CT ratio is 400/5, then the plug setting multiplier will be
- 25 A
- 15 A
- 50 A
- 10 A
208. A transformer has maximum efficiency at full load when iron losses are 800 watts. Copper losses at half-load will be
- 1600 W
- 800 W
- 400 W
- 200 W
209. A 2 kVA transformer has iron loss of 150 W and full load copper loss of 250 W. The maximum efficiency of the transformer will occur when the total loss is
- 500 W
- 400 W
- 300 W
- 275 W
210. A transformer is working at its full load and its efficiency is also maximum. The iron loss is 1000 watts. Then, its copper loss at half of the full load will be
- 250 watt
- 300 watt
- 400 watt
- 500 watt
211. High leakage transformers are of
- Small voltage ampere rating
- High voltage ampere rating
- High voltage rating
- Low voltage rating
212. Which of the following will improve the mutual coupling between primary and secondary circuit?
- Transformer oil of high breakdown voltage
- High reluctance magnetic core
- Winding material of high resistivity
- Low reluctance magnetic core
213. The efficiency of a 100 kVA transformer is 0.98 at full as well as half load. For this transformer at full load the copper loss
- is less than core loss
- is equal to core loss
- is more than core loss
- All of the above
214. The power factor at which the transformer operates
- is unity
- is 0.8 lag
- is 0.8 lead
- Depends upon the power factor of the load
215. If a 500 KVA, 200Hz transformer is operated at 50Hz, its KVA rating will be
- 2000 KVA
- 125 KVA
- 250 KVA
- 1000 KVA
216. Copper loss of transformer at $\frac{7}{8}th$ full load is 4900 W, then its full load copper loss would be
- 5600 W
- 6400 W
- 373 W
- 429 W
217. In case of a power transformer, the no-load current in terms of rated current is
- 10 to 20%
- 2 to 6%
- 15 to 30%
- 30 to 50% and motor
218. The purpose of the conservator in a transformer is
- To cool the winding
- To prevent moisture in the transformer
- To prevent a short circuit of the primary and secondary winding
- To take up contraction and expansion of oil
219. Location of lighting arrester should be near a
- Generator
- Transformer
- Bus-bar
- Circuit breaker
220. Power transformers are designed such that maximum efficiency occurs at
- Half of the full load
- Near full load
- 1/4 th full load
- 3/4 th of full load