IGCSE Physics 0625: Transformer Calculations and High-Voltage Transmission

A power station generator produces electrical power at a primary voltage of $V_p = 400\text{ V}$ . This is fed into an ideal (100% efficient) step-up transformer with $N_p = 200$ turns on the primary coil and $N_s = 20\,000$ turns on the secondary coil, before being sent along transmission cables. The generator supplies a constant power output of $P = 40\text{ kW}$ .

(a) Calculate the secondary (transmission) voltage, $V_s$ . [2]

(b) Assuming the transformer is 100% efficient, calculate the current flowing in the transmission cables (the secondary current, $I_s$ ). [2]

(c) Suppose instead the electricity were transmitted directly at the primary voltage of $400\text{ V}$ , with no step-up transformer, while still delivering the same power of $40\text{ kW}$ . Calculate the current that would flow in the cables in this case, and hence explain why transmitting power at high voltage instead greatly reduces the power wasted as heat in the cables. [3]

Part (a): Secondary voltage

For an ideal transformer, the voltages and numbers of turns are related by:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

Rearranging to make $V_s$ the subject:

$V_s = V_p \times \frac{N_s}{N_p}$

Substituting the values $V_p = 400\text{ V}$ , $N_s = 20\,000$ and $N_p = 200$ :

$V_s = 400 \times \frac{20\,000}{200} = 400 \times 100$

$V_s = \boxed{40\,000\text{ V}}$

Part (b): Current in the transmission cables

For a 100% efficient transformer, no power is lost in the transformer itself, so the power delivered on the secondary side equals the power supplied on the primary side, $P = 40\text{ kW} = 40\,000\text{ W}$ .

Power, voltage and current are related by $P = IV$ , so on the secondary side:

$I_s = \frac{P}{V_s} = \frac{40\,000\text{ W}}{40\,000\text{ V}}$

$I_s = \boxed{1\text{ A}}$

Part (c): Transmitting at the primary voltage instead

If the same power were sent at the (lower) primary voltage of $400\text{ V}$ with no step-up, the current needed would be:

$I_p = \frac{P}{V_p} = \frac{40\,000\text{ W}}{400\text{ V}}$

$I_p = \boxed{100\text{ A}}$

This current is $100$ times larger than the $1\text{ A}$ found in part (b) for the high-voltage cables.

The power wasted as heat in a cable of resistance $R$ is given by $P_{\text{loss}} = I^2R$ . Since power loss depends on the square of the current, a current that is $100$ times smaller (as in the high-voltage case) gives a power loss that is smaller by a factor of $100^2 = 10\,000$ , not just $100$ .

So stepping up to a high transmission voltage keeps the current in the cables very small for the same power delivered, and because power loss scales with $I^2$ , this dramatically reduces the amount of electrical energy wasted as heat along the transmission cables, which is why the National Grid transmits electricity at very high voltages.

Final answers

(a) Secondary voltage $V_s =$ $40\,000\text{ V}$ (40 kV)
(b) Secondary current $I_s =$ $1\text{ A}$
(c) Primary-voltage current $I_p =$ $100\text{ A}$ ; since $P_{\text{loss}} = I^2R$ , the $100\times$ smaller current at high voltage gives $10\,000\times$ less power wasted as heat.