PW-C3 · Transmission and per-unit

Voltage drop and loss: raise the voltage, cut the loss

When current flows through the line impedance the receiving voltage drops, and the line resistance soaks power into heat. Yet to send the same power, raising the voltage lowers the current in inverse proportion, so the voltage drop falls as 1/V and the loss as 1/V². Drag the voltage and see why the loss falls so steeply.

Drag the transmission voltage and watch the loss

The same power down the same line. Drag the handle to raise the transmission voltage. The current bar falls as 1/V, but the loss bar, being 1/V², shrinks far more steeply. Double the voltage and the current halves, the loss drops to a quarter.

Drag the handle left and right to set the transmission voltage.

Current and loss (vs reference voltage = 1)

P_loss ≈ 10.0%   η ≈ 90.9%

V × 1.0   I × 1.00   (P_loss × 1/V²)

Current flows, voltage drops

When current flows through the line Z = R + jX, the receiving voltage falls below the sending voltage. The approximate drop is e ≈ I(R cosφ + X sinφ), depending not only on current but on the load power factor. Since X is large on HV lines, a lagging power factor lets the X sinφ term swell the drop. If the receiving voltage sags too far, equipment misbehaves, so voltage drop is always watched in transmission.

Loss goes as the current squared

The power loss the line resistance soaks up is, in three phases, P_loss = 3 I²R. The key is the square of the current. A small drop in current cuts the loss by its square, and a rise in current makes the loss climb steeply. So when sending power far, the way to tame loss is, above all, to reduce the current itself.

ObserveP_loss = 3 I² R

Three-phase line loss is current squared times resistance.

ChooseP_loss → 1 / ?

For the same power, loss is inverse to voltage squared.

Fill ine ≈ I(R cosφ + X ?)

The reactive term of the drop is X times sinφ.

On your ownV × 2 → P_loss × ?

Double the voltage, a quarter of the loss.

Raise the voltage, lower the current

To send the same power P = √3 V_L I_L cosφ, the current is I = P/(√3 V cosφ), inversely proportional to voltage. So raising the voltage cuts the drop as 1/V and the loss, being the current squared, as 1/V². Double the voltage and the loss is a quarter. This is exactly why power is stepped up to extra-high voltage to be sent far. Power-factor correction (B2) also lowers the current for the same power and so cuts the loss too.

Back to the first screen

As you dragged the voltage up the current bar fell as 1/V, but the loss bar collapsed far more steeply as 1/V². Sending the same power, cutting only the current makes the loss follow by its square. The voltage drop, proportional to current, shrank along with it. Lifting transmission to hundreds of kilovolts, and improving the power factor to shed reactive current, are in the end the same one thing — reducing the current in the line to tame the loss.

The voltage drop and power loss — current through the line Z = R + jX lowers the receiving voltage. The approximate drop e ≈ I(R cosφ + X sinφ), the line loss P_loss = 3 I²R. Since I = P/(√3 V cosφ) for the same power, raising the voltage cuts the drop as 1/V and the loss as 1/V² (double the voltage → a quarter of the loss). Extra-high-voltage transmission and power-factor correction both reduce the current to tame the loss.

The next step

The voltage drop swings as the load changes. At a light midnight load the receiving voltage is high, at a heavy midday load it sags. The next unit (PW-C4) looks at voltage regulation: changing transformer taps or adjusting reactive power with capacitors and reactors so that, however the load changes, the receiving voltage is held within its allowed band.