- If the maximum flux density of core of a transformer increases, the secondary output voltage of the transformer
The secondary induced voltage equation of a transformer is E
_{2}= 4.44 × AB_{m}fT_{2}Volt. Where, A is the cross – sectional area of the magnetic core, B_{m}is the maximum flux density in the core, f is the supply frequency, T_{2}is the number of turns in secondary winding of the transformer. Hence from the above equation, it is clear that the induced voltage in secondary is directly proportional to maximum flux density of core. Therefore, with increase in maximum flux density of core, the secondary output voltage of the transformer increases. This is to be remembered, over flux in transformer is occurred during over voltage or/and under frequency conditions. - If secondary numbers of turns in a transformer increases, the secondary output voltage of the transformer
The secondary induced voltage equation of a transformer is:

E_{2}= 4.44AB_{m}fT_{2}Volt.

Where, A is the cross – sectional area of the magnetic core, B_{m}is the maximum flux density in the core, f is the supply frequency, T_{2}is the number of turns in secondary winding of the transformer. Hence from the above equation, it is clear that the induced voltage in secondary is directly proportional to secondary numbers of turns. Therefore, with increase in secondary numbers of turns, the secondary output voltage of the transformer increases. - In an electrical power transformer
In an electrical power transformer and in any other type of transformers the relation between induced voltage and number of turns is E
_{1}/ E_{2}= N_{1}/ N_{2}

⇒ E_{1}/ N_{1}= E_{2}/ N_{2}= 4.44φf Volt i.e. emf per turn in primary = emf per turn in secondary. - For a transformer, if the full load copper loss and iron loss are A and B respectively, then the load at which these two losses would be equal is given by
Let us consider full load current in the transformer is I
_{L}then full load copper loss A is I_{L}^{2}R watts. Let us again consider I is the load current at which the copper loss and core loss are equal, hence I^{2}R = B

⇒ I = √( B / R )

⇒ I = I_{L}× √( B / I_{L}^{2}R )

⇒ I = I_{L}× √( B / A ). - If W is the weight of copper in 315 MVA, 400 / 220 KV two winding transformer then what will be the weight of copper in an auto-transformer of same rating?
Weight of copper or conductor in auto transformer = ( 1 – k ) × ( Weight of copper or conductor in two winding transformer ).

Where k is the voltage ratio or turns ratio of LV winding to HV winding in transformer. Here k = 220 / 400 = 0.55, hence correct answer will be ( 1 – 0.55 ) × k or 0.45 × k. - If a two winding step down transformer of voltage ratio LV / HV = k is connected as auto transformer, then what will be the MVA rating of this auto transformer ?
Let’s consider V
_{1}and V_{2}are the primary and secondary voltages of two winding transformer. I_{1}is the primary current drawn from the source. Hence input or the MVA rating of the transformer if losses are neglected, will be V_{1}I_{1}.

If now it is connected as auto transformer, that is both winding of the former transformer are connected in series, then input taken from the source will be ( V_{1}+ V_{2})I_{1}= V_{1}I_{1}( 1 + V_{2}/ V_{1})

= ( 1 + k ) × MVA rating of two winding transformer. - When a 400 Hz transformer is operated at 50 Hz, its kVA rating is
The voltage equation of a transformer is given by E = 4.44 × φ
_{m}fT Volt.

Where, φ_{m}is the maximum flux in the core, f is the supply frequency, T is the number of turns in winding of the transformer. Hence from the above equation, it is clear that the voltage is directly proportional to frequency. Therefore, with increase in frequency, the output voltage of the transformer increases proportionally. When 400 Hz transformer operated in 50 Hz the output voltage also becomes 50 / 400 or 1 / 8 times hence VI rating of the transformer will become 1 / 8 times. - For constant load current at which power factor the efficiency of a transformer will be maximum?
The efficiency of transformer can be expressed as

η = V_{2}I_{2}cosθ_{2}/ ( V_{2}I_{2}cosθ_{2}+ constant losses ).

The power factor at which maximum efficiency occurs can be obtained by equating dη/dΘ_{2}= 0 from which we will get cosθ_{2}= 1. Thus the maximum efficiency for constant load current occurs at unity power factor. - Distribution transformers are usually designed to have maximum efficiency
The load on a distribution transformer varies over a wide range during a 24 hours day. The primary of distribution transformer is always energized and therefore core losses take place continuously. Keeping this in mind, distribution transformers are designed to have very low value of core losses. Again at maximum efficiency core losses = copper losses. At full load of transformer copper loss is maximum. Again at maximum efficiency at full load, core losses will be same as maximum copper losses thus core loss at this condition would be high. Hence for minimizing core losses the distribution transformers are designed to have maximum efficiency at half of the full load.
- Power transformers are usually designed to have maximum efficiency
The power transformers are manipulated to operate almost always at or near their rated capacity. Power transformers are disconnected from system during light periods. Therefore power transformers are designed to have maximum efficiency at its full load condition.
- The energy efficiency is the term related to
Energy efficiency of a transformer is defined as the ratio of total energy output for a certain period to the total energy input for the same period. The load on a distribution transformer varies over a wide range during 24 hours a day. Therefore, choice of a distribution transformer is based on their energy efficiency.
- The all day efficiency is the term related to
Energy efficiency of a transformer is defined as the ratio of total energy output for a certain period to the total energy input for the same period. When energy efficiency is computed for a day of 24 hours, it is called all day efficiency. The load on a distribution transformer varies over a wide range during a 24 hours day. Therefore, all day efficiency of a distribution transformer is taken as an essential parameter for judging the performance of the transformer.
- The phase difference between primary and secondary voltages in an ideal transformer is generally
The phase difference between primary and secondary voltage in an ideal transformer is generally 180°. This can be explained in many ways but the simple way is to consider that the transformer primary voltage is cause and the secondary voltage is effect. Therefore, according to Lenz’s law they will oppose each other hence directed 180° apart.
- A transformer has a iron loss of 900 W and copper loss of 1600 W. At what percentage of the load the transformer will have maximum efficiency?
Let us consider full load current in the transformer is I
_{L}then full load copper loss is A = I_{L}^{2}R watts. Let us again consider I is the load current at which the copper loss and core loss are equal, hence I^{2}R = B, where B is the iron loss or core loss.

⇒ I = √( B / R).

⇒ I = I_{L}× √( B / I_{L}^{2}R).

⇒ I = I_{L}× √( B / A).

Hence, the answer will be √( 900 / 1600 ) = 0.75 or 75 %. - Two transformer A and B, with identical ratings, are to be designed with flux densities of 1.2 and 1.4 Wb / m
^{2}respectively. The weight of transformer A per kVA isThe voltage of a transformer can be expressed as V = 4.44AB_{m}fT Volt.

Where, A is the cross – sectional area of the magnetic core, B_{m}is the maximum flux density in the core, f is the supply frequency, T is the number of turns in winding of the transformer. Hence from the above equation, it is clear that for constant voltage cross - sectional(A) area of transformer core is inversely proportional to flux density ( B_{m}). Therefore, as the flux density in transformer A ( B_{m}= 1.2 Wb / m^{2}) is less than that of Transformer B ( B_{m}= 1.4 Wb / m^{2}) the cross - section of core in transformer A will be larger than that of transformer B. Hence weight of transformer A is more than that of transformer B. - In a single phase transformer which of the following relation is true?
In a transformer the emf per tern and mmf in both LV and HV windings are same. That means E
_{1}/N_{1}= E_{2}/N_{2}and I_{1}N_{1}= I_{2}N_{2}. - The flux involved in emf equation of a transformer is
The emf equation of a transformer is derived from the expression of sinusoidal magnetizing flux i.e. φ
_{m}sinωt where φ_{m}is maximum value of flux. Hence the flux involved in emf equation is maximum flux. - A transformer has N
_{1}and N_{2}turns in primary and secondary winding respectively. Its secondary winding reactance of x_{2}ohms, when referred to primary, isFor referring secondary reactance to primary side of a transformer, we have to multiply the reactance with the square of turns ratio of primary and secondary, hence the correct answer will be x_{2}( N_{1}/ N_{2})^{2}. - In 132 / 66 KV power transformer has its LV winding resistance 0.02 pu. The resistance when referred to HV side should be
Per unit or pu value of resistance remains same whether it is referred to HV side or LV side.
- Leakage flux in a transformer depends on
Whenever a transformer is put in load, the load currents start to flow in both HV and LV winding, which produce opposite flux in the transformer core. But still there will be some flux in both winding which will not pass through the core referred as leakage flux. These leakage flux is the cause of leakage reactance of the transformer, hence it can be said that leakage reactance of transformer depends upon load current.

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