transformer and electric-power transmission

A transformer works because of electromagnetic induction: the alternating current in the primary coil produces an alternating magnetic flux in the core, and this flux induces an emf in the secondary coil. Both coils are threaded by practically the same flux, so the emfs are proportional to the numbers of turns:

$$\frac{U_2}{U_1}=\frac{N_2}{N_1}.$$

For an ideal transformer the power is conserved:

$$U_1I_1=U_2I_2,$$

so raising the voltage by a factor $k$ lowers the current by the same factor. When energy travels along a line of resistance $R$, the loss is $P_{loss}=I^2R$, so a smaller current sharply reduces the loss. That is why the voltage is stepped up before the line and stepped down before the consumer. A transformer cannot work on direct current: without a changing magnetic flux there is no emf in the secondary coil.

An ideal transformer is connected to a $U_1 = 240$ V supply. The primary coil has $N_1 = 1000$ turns and the secondary coil has $N_2 = 50$ turns. Find the secondary voltage $U_2$ and state the type of the transformer.

A power station delivers $P = 10$ MW through a transmission line of resistance $R = 2$ Ω. Compare the power lost in the line when the transmission voltage is (a) $U_A = 10$ kV and (b) $U_B = 400$ kV. Express each loss as a percentage of $P$.

A generator produces $P = 100$ kW at 2 kV. A village is connected to it by a line of resistance $R = 5$ Ω.

• Option A: transmit directly at 2 kV.
• Option B: step up to 40 kV with an ideal transformer, transmit, then step down with an ideal transformer of turns ratio $100:1$.

For each option find the line loss and the voltage arriving at the village. Which option should be chosen?

Self-check in a simulation (safe, no mains).

1. Open an interactive AC-circuit model with an ideal transformer (adjustable $N_1$, $N_2$ and a resistive load). Do not connect real transformers to the mains: 220 V work is hazardous.
2. Set $U_1 = 100$ V and $N_1 = 1000$ turns. Vary $N_2$ from 100 to 2000 in steps of 100, recording $U_2$ and $I_2$ with an unchanged load resistance.
3. Plot $U_2$ against $N_2$ and check $U_1I_1 \approx U_2I_2$ at every point. Mark where the transformer steps up and where it steps down.
4. In a spreadsheet set $P = 1$ MW, $R = 4$ Ω and compute $P_{loss}\% = \frac{(P/U)^2 R}{P} \cdot 100\%$ for $U$ from 1 kV to 100 kV. Plot the loss fraction against the transmission voltage (logarithmic $U$ axis) and state the conclusion in one sentence.

Show the whole power-transmission chain and the quantitative link between the transmission voltage and the line loss.

A two-part diagram. On the left, the circuit: generator $\rightarrow$ step-up transformer ($N_1:N_2$) $\rightarrow$ line of resistance $R$ $\rightarrow$ step-down transformer $\rightarrow$ consumer; each element is labelled with $U$ and $I$, and arrows show the direction of power flow. On the right, a small graph: the horizontal axis is the transmission voltage $U$ (logarithmic scale), the vertical axis is the loss fraction $I^2R/P$ in percent; the curve falls monotonically, with marked points for 10 kV and 400 kV.

A power station sends $P = 1000$ MW to a city $200$ km away. The transmission line has a total resistance $R = 10\ \Omega$. A step-up transformer raises the generator voltage from $20$ kV to $400$ kV before the line. At $400$ kV the line current is $I = P/V = 2500$ A, so the heating loss in the line is $P_{\text{loss}} = I^2 R = 62.5$ MW, about $6.3\%$ of the transmitted power. If the same power were sent at $20$ kV, the current would be $50\,000$ A and the loss would be $25\,000$ MW, which no practical line can carry. Near the city, substations step the voltage down to $10$ kV and then to $400$ V for homes.

This uses the ideal-transformer relations $V_1/V_2 = N_1/N_2$ and $P_1 = P_2$, together with the current ratio $I_1/I_2 = N_2/N_1$. It also exercises the reason for high-voltage transmission: for fixed power, $I = P/V$, so raising $V$ lowers $I$ and the resistive loss $I^2 R$ falls as $1/V^2$.

A phone charger is plugged into a $230$ V socket and delivers $5$ V at $2$ A to the phone, so the output power is $10$ W. Inside the charger there is a small transformer with a turns ratio of about $46:1$, followed by a rectifier that changes the alternating current to direct current. The current drawn from the socket is about $10/230 \approx 0.043$ A.

This applies the step-down transformer equation $V_2 = V_1 N_2/N_1$: to get $5$ V from $230$ V the secondary must have about $1/46$ as many turns as the primary. It also shows power conservation $P_1 = P_2$ in the ideal transformer: the small primary current $0.043$ A at $230$ V supplies the same $10$ W as $2$ A at $5$ V.

An electric locomotive collects power from a $25$ kV overhead line and uses an on-board transformer to step the voltage down to $1.5$ kV for its traction motors. For a motor input power of $5$ MW, the secondary current is $I = P/V = 5\,000\,000/1500 \approx 3333$ A, while the primary current is only $5\,000\,000/25\,000 = 200$ A. The overhead wire therefore carries a much smaller current and wastes less energy in its resistance.

This exercises the same step-down transformer relations as the charger, now at high power: $V_1/V_2 = N_1/N_2$ and $I_1/I_2 = N_2/N_1$. It shows why transport systems use high voltage on the catenary: for the same power, lower current in the long overhead wire means lower $I^2 R$ loss.