Electrical Engineering ⇒ Topic : Current and kVA at Maximum Efficiency

Harrison
 
Let us assume the iron losses = P_{i} The total copper losses referred to the primary side = From Equation , ........... (1) In a given transformer, V_{1} is approximately constant and hence for a given value of /_{1} and cos Φ_{1} the efficiency will be maximum when the denominator of Eq. (1) is minimum. Thus, differentiating both the sides with respect to /_{1}, we get
will be maximum when d/d/_{1} = O
........... (2) Hence, the efficiency of a transformer is maximum when the variable copper loss is equal to the constant iron loss. From Eq. (2), it follows that at maximum efficiency, the primary current Similarly, at maximum efficiency, the load current .............. (3) where R_{e2} is the equivalent resistance of the transformer referred to the secondary side given by R_{e2} = R_{2} + (V_{2}/V_{1})^{2}R_{1} The efficiency versus load current curves are shown in Figure for different power factors
FIGURE (a) Efficiency of a transformer as a function of the load current for different power factors For a given transformer, the efficiency decreases with a decrease in power factor. Distribution transformers are generally designed to give maximum efficiency at 75% of fullload  
 
Sonali
 
CURRENT AND kVA AT MAXIMUM EFFICIENCY Let S_{fl} = V_{2}I_{2} is full load in V_{A}, S_{m} = V_{2}I_{m} in VA at maximum efficiency, I_{2m} secondary current at maximum efficiency, I_{2fI} is fullload secondary current, and P_{cufl} cup is copper loss at full load ............. (1)  
 
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