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2 years ago

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The resistance, R, to electricity of a cylindrical-shaped wire is given by the equation R=pL/(pi)d^2 where p represents the resi

stivity of the wire’s material, L represents the length of the wire, and d represents the diameter of the wire.
What happens to the resistance of the wire as the diameter approaches 0?
The resistance approaches 0.
The resistance approaches pi.
The resistance approaches L.
The resistance approaches infinity.

2 answers:

almond37 [142]2 years ago

7 0

Answer:

Option 4th is correct.

The resistance approaches infinity.

Step-by-step explanation:

The resistance, R, to electricity of a cylindrical-shaped wire is given by the equation:

$R = \frac{p L}{\pi d^2}$ ....[1]

where,

p is the the resistivity of the wire’s material

L is the length of the wire.

d represents the diameter of the wire.

We have to find resistance of the wire as the diameter approaches 0.

As diameter approaches to 0.

⇒ $d \rightarrow 0$

Substitute in [1] we have;

$R = \frac{p L}{\pi 0^2}$

⇒ $R \rightarrow \infty$

therefore. the resistance approaches infinity as the diameter approaches 0.

sasho [114]2 years ago

6 0

Answer:

D.) The resistance approaches infinity.

Step-by-step explanation:

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2 years ago