Losses & Efficiency in a Transformer
Losses in Transformer
Loss in any machine is when your output is lesser than the input you gave. In transformer this is loss is in terms of power. As it is a static device, there are only electrical losses in a transformer, unlike a motor which also has mechanical losses. All types of losses are:
(a) Core Losses or Iron Losses
Losses occuring in the core of the transformer. These are:
Hysteresis loss in transformer : Hysteresis loss occurs when there is reversal in the process of magnetization in the transformer core. This loss depends upon the volume and grade of the iron, value of flux density and frequency of magnetic reversals.
Eddy Current Loss in transformer :
In transformer, AC current is supplied to the primary winding which sets up alternating magnetizing flux. When this flux links with secondary winding, it produces induced emf in it. But some part of this flux also gets linked with other conducting parts like steel core or iron body, which results in induced emf in those parts, resulting in small circulating currents in them. This current is called as Eddy current. Due to these eddy currents, some energy is dissipated in the form of heat causing Eddy current loss in transformer.
(b) Copper Loss
Copper loss occurs due to the resistance of the windings of the transformer. For primary winding, it is \(I_1^2R_1 \) and for secondary winding, it is \(I_2^2R_2\) where \(I_1 \) is primary current, \(R_1\) is primary winding resistance, \(I_2\) is secondary current and \(R_2\) is secondary winding resistance. As you can see, Copper loss is directly proportional to square of current and current depends on load, it implies that copper loss varies with the load connected to the transformer.
Efficiency of Transformer
Efficiency is given by ratio of output power to the input power.
Transformers are the most highly efficient electrical machines. Most of the transformers have full load efficiency between 95% to 98.5% .
As due to high efficiency, output is nearly equal to input power, so it becomes an impractical to measure efficiency using ratio of output and input. So a better way is to calculate the losses, subtract them from input and then calculate the ratio,
\[ \text{Efficieny } = \frac{\text{Input - Losses}}{\text{Input}} \]
Condition for Maximum Efficiency
Input Power \( = V_1I_1\cos{\phi_1} \)
Let,
Copper Loss \( = I_1^2R_1 \)
Iron Loss \( = W_i\)
\[ \text{Efficiency } = 1 - \frac{\text{Losses}}{\text{Input}} \]
\[ \eta = 1 - \frac{I_1^2R_1 + + W_i}{V_1I_1\cos{\phi_1}} \]
\[ \eta = 1 - \frac{I_1^2R_1}{V_1I_1\cos{\phi_1}} - \frac{W_i}{V_1I_1\cos{\phi_1}} \]
Differentiating w.r.t. \(I_1\)
\[ \frac{d\eta}{dI_1} = 0 - \frac{R_1}{V_1\cos{\phi_1}} + \frac{W_i}{V_1I_1^2\cos{\phi_1}} \]
\(\eta\) will be maximum at \(\large\frac{d\eta}{dI_1} = 0 \)
So, \[ \frac{R_1}{V_1\cos{\phi_1}} = \frac{W_i}{V_1I_1^2\cos{\phi_1}} \]
\[ \frac{I_1^2R_1}{V_1I_1^2\cos{\phi_1}} = \frac{W_i}{V_1I_1^2\cos{\phi_1}} \]
\[ \boxed{I_1^2R_1 = W_i} \]
Hence the efficiency of the transformer will be maximum when copper loss is equal to the iron loss.
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