Before you read this, I suggest you read posts 17.44 and 17.45

You often see birds perched on cables that are carrying electricity but aren’t insulated. In the UK electricity can be sent through overhead cables at 400 kV. Why aren’t the birds electrocuted?

As charge moves along the cable, the voltage drops. Why? Because work is done moving the charge against the resistance of the cable (post 17.44). So less energy is available to move the charge further (post 16.21). This is equivalent to saying that the voltage has dropped (post 17.44). An analogy is a falling object whose potential energy, that makes it fall, decreases as it drops (post 16.21). As a charge is moved along a conductor by a potential difference, the potential difference between its current position and its destination decreases in the same way.

Now let’s return to our bird. When it perches on the cable, its right foot (at R) is a few centimetres from its left foot (at L). As a result, a current will flow through the bird. But L and R are also connected by the cable. So how much current flows through the bird and how much flows through the cable? Both the bird and the cable have resistance. The left-hand side of the figure above shows a bird on a wire; the right-hand side shows how this can be represented in electrical currents and resistances. On the right-hand side, current I flows along the cable from the right (when you look at the picture – this is on the left-hand side of the bird). Some of this current, I_b, flows through the bird and some, I_c, flows through the cable. Since the charge is flowing, the sum of the currents flowing out of R must be equal to the current flowing in. It’s just like water flowing through a branch in a pipe – all the water that flows out of the main a pipe, in a given time, must flow out of the branches in the same time – none of the water suddenly stops flowing. When we apply this idea to electrical currents, it’s called Kirchhoff’s current rule. In this example we can write the result as

I = I_b + I_c.                    (1)

Let’s represent the potential difference between L and R by V. If the resistance of the bird is R_b and the resistance of the cable is R_c, V is given (from Ohm’s law, post 17.44) by

V = I_bR_b   and   V = I_cR_c.

This means that

I_bR_b = I_cR_c                    (2).

Combining equations 1 and 2 gives

I_b = IR_c/(R_b + R_c).                      (3)

Try deriving this result for yourself; but, if you can’t, the derivation is in the appendix.

Let’s put some numbers into equation 3. All the information I have used about cables comes from http://nl.prysmiangroup.com/nl/business_markets/markets/hv-and-submarine/downloads/datasheets/Prysmian-Delft-HVac.pdf and the information about the effects of electricity on the human body comes from www.ncbi.nlm.nih.gov/pmc/articles/PMC2763825/.

The current in a cable is no more than about 1.5 × 10³ A; we will use this value for I. (If you’re not sure about writing numbers in the form 1.5 × 10³, see post 16.7). Let’s suppose that our bird’s feet are about 2 cm apart, so this is the distance between L and R. The resistance of 1 km of cable is about 0.01 Ω, so the resistance of 2 cm of cable is

(2 × 10^-2/1 × 10³) × 0.01 = 2 × 10^-7 Ω;

We will use this value for R_c.

A dry human hand covered by thick skin has a resistance of 100 kΩ; this can’t be very different to the resistance of a bird. So, we’ll assume that R_b = 100 kΩ.

If we put these numbers into equation 3, we get the result that

I_b = (1.5 × 10³) × (2 × 10^-7)/(100 × 10³) = 3 × 10^-9 A or 3 × 10^-6 mA.

 

What effect will this current have on our bird? The table above shows this current would have no adverse effect on the human body. But a bird is much smaller than a person. I’ve assumed that the effect of a current depends on the mass of a living thing. Since a small bird has a mass of about 20 g, a person has a mass of about 10⁴ times the mass of a small bird. In the table above, I’ve used this ratio of human to bird mass to estimate a “bird equivalent” of the currents that effect the human body.

We can see, from the “bird equivalent” column in the table that the current flowing through the bird is so small that it has no adverse effect.

Now let’s suppose our bird had very long legs – so that it could touch the cable and the ground at the same time, as shown in the picture above. Now the potential difference between the bird and the ground could be 400 kV (If the cable transmits at 400 kV and the earth is assigned a potential of 0 V, see post 17.44). The current flowing through the bird (resistance 100 kΩ) would be

I_b’ = (400 ×1000)/(100 × 1000) = 4 A.

According to the numbers in the table, our bird would be electrocuted.

When our bird sits on the cable, there is much less current flowing through it – so it is safe from electrocution.

 

Related posts

17.45 Electrical energy
17.44 Amps, volts and ohms
17.24 Fields and vectors
16.25 Electrical charge

 

Appendix

Combining equations 1 and 2 gives

I_bR_b = (I – I_b)R_c = IR_c – I_bR_c.

Adding I_bR_c to both sides of this equation gives

I_bR_b + I_bR_c = IR_c   or   I_b(R_b + R_c) = IR_c.

Dividing both sides of this final equation by (R_b + R_c) gives equation 3.