Example 10.5: The step-up transformer

Question: An electric power plant produces $P=1$GW of AC electric power at a peak voltage of $V_1=500$V. If it is desired to transmit this power at a peak voltage of $V_2= 50$kV, what is the appropriate turns-ratio of the step-up transformer? What peak current $I_1$ would be sent over the transmission line if the peak voltage were $V_1=500$V? What peak current $I_2$ would be sent over the transmission line if the peak voltage were $V_2= 50$kV? What is the ratio of the ohmic powers losses in the line in the two cases?
 
Answer: The appropriate turns-ratio is

$$\frac{N_2}{N_1} = \frac{(5\times 10^4)}{(500)} = 100.$$

Since the peak power is given by $P=I_1\,V_1$, it follows that

$$I_1 = \frac{P}{V_1} = \frac{(1\times 10^9)}{(500)} = 2\,{\rm MA}.$$

Since the peak power remains unchanged after the signal passes through the transformer (assuming that there are no power losses in the transformer), we have

$$I_2 = \frac{P}{V_2} = \frac{(1\times 10^9)}{(5\times 10^4)} = 20\,{\rm kA}.$$

The ratio of the power lost to ohmic heating in the two cases is

$$\frac{P_1}{P_2} = \frac{I_1^{~2}\,R}{I_2^{~2}\,R} = \left(\frac{2\times 10^6} {2\times 10^4}\right)^2 = 10000,$$

where $R$ is the resistance of the transmission line. Note that the ohmic power loss is much greater at low peak voltage than at high peak voltage.