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Why can birds perch on the bare wire of a power line without being electrocuted?

Short Answer

Expert verified

A bird doesn't get electrocuted while sitting on a live wire since the point where its feet touch the wire are at a same potential and hence no current passes through its body.

Step by step solution

01

Given data

Birds can sit on live wires without getting electrocuted.

02

Flow of current

Current flows through a path when there is a potential difference between the end points of the path. Also in the presence of multiple paths between two points having a potential difference, the current prefers the path of least resistance.

03

Determination of the reason why birds can sit on live wires

There is very little potential drop when current passes through overhead wires. Thus when a bird sits on a live wire, the points where its feet touch the wire are practically at the same potential. Hence no current passes through the bird's body. If a slight potential difference is present, the current prefers the path of the wire to the bird's body since the wire is much more conducting.

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Most popular questions from this chapter

Question: in circuit 1 (Figure 19.72), an uncharged capacitor is connected in series with two batteries and one light bulb. Circuit 2 (Figure 19.72) contains two light bulbs identical to the bulb in the circuit; in all other respects, it is identical to circuit 1. In circuit 1, the light bulb stays lit for 25 s. The following questions refer to these circuits. You should draw diagrams representing the fields and charges in each circuit at the times mentioned, in order to answer the questions.

(a)One microsecond after connecting both circuits, which of the following are true? Chose all that apply: (1) the net electric field at location B in circuit 1 is larger than the net electric field at location B in circuit 2. (2) At location A in 1, electrons flow to the left. (3) At location A in circuit 1, the electric fields due to charges on the surface of the wires and batteries points to the right. (4) in circuit 1 the potential difference across the capacitor plates is equal to the emf of the batteries. (5) The current in circuit 1 is larger than the current in circuit 2.

(b)Two seconds after connecting both circuits, which of the following are true? Choose all that apply: (1) there is more charge on the plates of capacitor 1 than there is on the plates of capacitor 2. (2) there is negative charge on the right plate of the capacitor in circuit 1. (3) At location B in circuit 2 the net electric field points to the right. (4) At location B in circuit 2 the fringe field of the capacitor points to the right. (5) At location A in circuit 1 the fringe field of the capacitor points to the left.

(c)Which of the graphs in Figure 19.73 represents the amount of charge on the positive plate of the capacitor in circuit 1 as a function of time?

(d)Which of the graphs in Figure 19.73 represents the current in circuit 1 as a function of time?

The conductivity of tungsten at room temperature,1.8×107A/m2/V/m , is significantly smaller than that of copper. At the very high temperature of a glowing light-bulb filament (nearly 3000 kelvins), the conductivity of tungsten is 18 times smaller than it is at room temperature. The tungsten filament of a thick-filament bulb has a radius of about 0.015 mm. Calculate the electric field required to drive 0.20 A of current through the glowing bulb and show that it is very large compared to the field in the connecting copper wires.

A certain 6 V battery delivers 12 A when short circuited. How much current does battery deliver when 1Ω resistor is connected to it?

How is the initial current through a bulb affected by putting a capacitor in series in the circuit? Explain briefly.

The two circuits shown in Figure 19.59 have different capacitors but the same batteries and thin-filament bulbs. The capacitors in circuit 1and circuit 2areidentical exceptthat the capacitor in circuit 2was constructed with its plates closer together. Both capacitors have air between their plates. The capacitors are initially uncharged. In each circuit the batteries are connected for a short time compared to the time required to reach equilibrium, and then they are disconnected. In which circuit (1or 2) does the capacitor now have more charge? Explain your reasoning in detail.

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