# Transmission wires loss (1 Viewer)

#### superstar12

##### Member
Since P=VI and V=IR, we get

Power loss = I^2R

and find that a low current results in less power loss. Thus, electricity is stepped up to high voltages since according to P=VI, a high voltage results in a low current. But, if I=V/R is substituted into the P=VI equation, giving us,

Power loss = (V^2)/R

This implies that a high voltage results in a greater power loss, as opposed to what we see in both the previous equation and in practice. Why is this?

#### anomalousdecay

No no no.

$\bg_white P _{loss} = \frac{V _{drop} ^2}{R}$

This is not implying that the voltage source is responsible for power losses. Simply, the voltage drop across the wires is dependent on the current through them, $\bg_white V_{drop} = I R$.

If you really want the overall power delivered by the circuit, we can use:

$\bg_white Power delivered = V _{source} I _{source} - \frac{V _{drop} ^2}{R _{Thevenin}} = V _{source} I _{source} - I_{delivered} ^ 2 R _{Thevenin}$

Thevenin resistance is just the equivalent resistance as measured from that point of the whole circuit.

So what can be deduced here? Well if the current increases, so does the voltage drop across the transmission lines. Hence, the power delivered can be maximised with using a lower current across the transmission lines.

#### Fizzy_Cyst

##### Well-Known Member
Exactly what anamolous said.

Less current = lower voltage drop across the wires = less power loss

From P = V^2/R; V is the voltage drop not the transmitting voltage