A transformer is a device that changes the voltage of an alternating current (AC) without changing the power (ignoring small losses). Think of it like a water‑pipe system: if you want to move a lot of water (power) over a long distance, you can use a larger pipe (higher voltage) so the water flows faster and with less friction (loss). The transformer does the same for electricity – it “steps up” the voltage so that the current can be lower, reducing the energy lost as heat in the wires.
Recall and use the equation \$P = I^2 R\$ to explain why power losses in cables are smaller when the voltage is greater.
The power lost as heat in a cable of resistance \$R\$ carrying a current \$I\$ is:
\$P_{\text{loss}} = I^2 R\$
Because the current \$I\$ is inversely proportional to the voltage \$V\$ for a given transmitted power \$P{\text{trans}}\$ (i.e. \$I = \dfrac{P{\text{trans}}}{V}\$), increasing \$V\$ reduces \$I\$ and therefore reduces \$P_{\text{loss}}\$.
Suppose we need to transmit \$P_{\text{trans}} = 1000\ \text{W}\$ through a cable with \$R = 10\ \Omega\$.
Case 1 – Low voltage (230 V):
\$I = \dfrac{1000}{230} \approx 4.35\ \text{A}\$
\$P_{\text{loss}} = (4.35)^2 \times 10 \approx 189\ \text{W}\$
Case 2 – High voltage (400 V):
\$I = \dfrac{1000}{400} = 2.5\ \text{A}\$
\$P_{\text{loss}} = (2.5)^2 \times 10 = 62.5\ \text{W}\$
The higher voltage case loses only about one third of the power in the cable. This is why power stations send electricity at very high voltages and step it down near homes with transformers.
| Voltage (V) | Current (A) | Power Loss (W) |
|---|---|---|
| 230 | 4.35 | 189 |
| 400 | 2.5 | 62.5 |