# Transformer for Low Voltage Compensation

At times during the year, peak demands from the power company can result in a reduced voltage down the line. In midsummer, for example, the line voltage may drop from 120 V to 100 V because of the heavy load often due primarily to air conditioners. However, air conditioners do not run as well under low-voltage conditions, so the following option using an autotransformer may be the solution.
Fig. 1: Maintaining a 120-V supply for an air conditioner: (a) using an autotransformer; (b) using a traditional step-up transformer.
In Fig. 1(a), an air conditioner drawing 10 A at 120 V is connected through an autotransformer to the available supply that has dropped to 100 V. Assuming 100% efficiency, the current drawn from the line would have to be 12 A to ensure that $Pi = Po = 1200 W$. Using the analysis, we will find that the current in the primary winding is 2 A with 10 A in the secondary. The 12 A will exist only in the line connecting the source to the primary. If the voltage level were increased using the traditional step-up transformer shown in Fig. 1(b), the same currents would result at the source and load. However, note that the current through the primary is now 12 A which is 6 times that in the autotransformer. The result is that the winding in the autotransformer can be much thinner due to the significantly lower current level. Let us now examine the turns ratio required and the number of turns involved for each setup (associating one turn with each volt of the primary and secondary). For the autotransformer:
$$\frac{N_{s}}{N_{p}}=\frac{V_{s}}{V_{p}}=\frac{10 \mathrm{~V}}{100 \mathrm{~V}} \Rightarrow \frac{10 \mathrm{t}}{100 \mathrm{t}}$$
$$\frac{N_{s}}{N_{p}}=\frac{V_{s}}{V_{p}}=\frac{120 \mathrm{~V}}{100 \mathrm{~V}} \Rightarrow \frac{120 \mathrm{t}}{100 \mathrm{t}}$$