# Transformer Nameplate Data

A typical iron-core power transformer rating, included in the nameplate data for the transformer, might be the following:
$$5 \mathrm{kVA} \quad 2000 / 100 \mathrm{~V} \quad 60 \mathrm{~Hz}$$
The $2000 \mathrm{~V}$ or the $100 \mathrm{~V}$ can be either the primary or the secondary voltage; that is, if $2000 \mathrm{~V}$ is the primary voltage, then $100 \mathrm{~V}$ is the secondary voltage, and vice versa. The $5 \mathrm{kVA}$ is the apparent power $( S=V I)$ rating of the transformer. If the secondary voltage is $100 \mathrm{~V}$, then the maximum load current is
$$I_{L}=\frac{S}{V_{L}}=\frac{5000 \mathrm{VA}}{100 \mathrm{~V}}=50 \mathrm{~A}$$
and if the secondary voltage is $2000 \mathrm{~V}$, then the maximum load current is
$$I_{L}=\frac{S}{V_{L}}=\frac{5000 \mathrm{VA}}{2000 \mathrm{~V}}=2.5 \mathrm{~A}$$
The transformer is rated in terms of the apparent power rather than the average, or real, power for the reason demonstrated by the circuit of Fig. 1. Since the current through the load is greater than that determined by the apparent power rating, the transformer may be permanently damaged. Note, however, that since the load is purely capacitive, the average power to the load is zero. The wattage rating would therefore be meaningless regarding the ability of this load to damage the transformer.
Fig. 1: Demonstrating why transformers are rated in kVA rather than kW.
The transformation ratio of the transformer under discussion can be either of two values. If the secondary voltage is $2000 \mathrm{~V}$, the transformation ratio is
$$a=N_{p} / N_{s}=V_{g} / V_{L}=100 \mathrm{~V} / 2000 \mathrm{~V}=1 / 20$$
and the transformer is a step-up transformer. If the secondary voltage is $100 \mathrm{~V}$, the transformation ratio is
$$a=N_{p} / N_{s}=V_{g} / V_{L}=2000 \mathrm{~V} / 100 \mathrm{~V}=20$$
and the transformer is a step-down transformer. The rated primary current can be determined simply by applying
$$I_{p}=\frac{I_{s}}{a}$$
which is equal to $$[2.5 \mathrm{~A} /(1 / 20)]=50 \mathrm{~A}$$ if the secondary voltage is $2000 \mathrm{~V}$, and $(50 \mathrm{~A} / 20)=2.5 \mathrm{~A}$ if the secondary voltage is $100 \mathrm{~V}$.
To explain the necessity for including the frequency in the nameplate data, consider equation:
$$E_{p}=4.44 f_{p} N_{p} \Phi_{m}$$
Fig. 2: Demonstrating why the frequency of application is important for transformers.
and the $B-H$ curve for the iron core of the transformer (Fig. 2).The point of operation on the $B-H$ curve for most transformers is at the knee of the curve. If the frequency of the applied signal should drop, and $Np$ and $E_p$ remain the same, then $\Phi_{m}$ must increase in magnitude, as determined by:
$$\Phi_{m} = { E_p \over 4.44 f_p N_p}$$
The result is that B will increase, as shown in Fig. 2, causing H to increase also. The resulting $\Delta I$ could cause a very high current in the primary, resulting in possible damage to the transformer.