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3 years ago

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Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

L 2, how do L 1 and L 2 compare if wire 1 is made from copper and wire 2 is made from aluminum? The resistivity of copper is 1.7 × 10 –5 Ω·m and the resistivity of aluminum is 2.82 × 10 –5 Ω·m.

1 answer:

SVEN [57.7K]3 years ago

7 0

Answer with Explanation:

We are given that

Length of wire 1= $L_1$

Length of wire 2= $L_2$

Resistivity of copper wire= $\rho_1=1.7\times 10^{-5}\Omega-m$

Resistivity of aluminum wire= $\rho_2=2.82\times 10^{-5}\Omega-m$

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance = $\frac{\rho l}{A}$

When diameter of wires are same then their cross section area are also same .

$l=\frac{RA}{\rho}$

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity is small then the length of wire will be large.

$\rho_1$

Therefore, $L_1>L_2$

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

$\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66$

$L_1=1.66L_2$

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A 12-V battery is connected to an air-filled capacitor that consists of two parallel plates,

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- The area of each plate, A = 7.6 cm^2

- The separation between plates, d = 0.3 cm

- The charge of the proton. q = 1.6*10^-19 C

- The initial velocity of proton, vi = 0 m/s

Solution:-

- The electric field ( E ) between the parallel plates of the air-filled capacitor is determined from the applied potential difference by the battery on the two ends of the plates.

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E = 4,000 V/m ... Answer

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- The capacitance of the charged plates would be ( C ):

C = k*ε*A / d

Where,

k: the di-electric constant of material = 2.1

ε: permittivity of free space = 8.85 × 10^-12

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