A cylindrical resistor of length 2 cm and diameter of 0.5 cm has a voltage of 1 V applied between the two ends. This voltage results in a cu

Question

A cylindrical resistor of length 2 cm and diameter of 0.5 cm has a voltage of 1 V applied between the two ends. This voltage results in a current of 2 mA.
a. Find the conductivity of the material from which the resistor is made.
b. Next, the resistor is reshaped so that one half of its length remains the same while the other half has its diameter reduced to 0.2 cm. Find the total resistance.

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Kim Cúc 3 months 2021-08-09T12:33:18+00:00 1 Answers 2 views 0

Answers ( )

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    2021-08-09T12:35:04+00:00

    Answer:

    a) 2.037 ohm/meter

    b) 1487.9 Ω

    Explanation:

    a)

    We all know that:

    R = \rho \frac{l}{A}

    where:

    \rho = resistivity

    and which can be written as:

    \rho = \frac{1}{ \sigma}

    where \sigma = conductivity

    l = length = 2 cm = 2*10^{-2}m

    A = area = \pi r ^2

    where

    r = radius of the cylinder

    r = \frac{D}{2}

    = \frac{0.5}{2}

    = 0.25 cm

    Also by ohm’s law

    V = I R

    R = \frac{V}{I}

    = \frac{1 V }{2*10^{-3} A}

    = 500 Ω

    R = \rho \frac{l}{A}

    R = \frac{l}{\sigma A}

    \sigma = \frac{l}{R A}

    \sigma = \frac{2*10^{-2}m}{500* \pi * (0.25)^2 * 10^{-4}}

    \sigma = 2.037 ohm/ meter

    b)

    Since  R = \frac{\rho * l}{A *l}

    R = \frac{\rho *l}{V}

    where ; V = \pi  r^2 l

    \pi r^2 l = \pi r_1^2 l_1+  \pi r_2^2 l_2

    r = 0.25 cm, l = 2cm , r_1 = 0.25cm , r_2 = 0.1 cm

    So ; we are tasked to determine l_1 and l_2

    l' = l_1 + l_2

    l_1 = \frac{l'}{2} =l_2

    l_1 = l_2

    l_1 = \frac{\pi r^2 l}{\pi r^2_1+ \pi r_2^2}

    l_1 = \frac{\pi 0.25^2 *2}{\pi 0.25^2_1+ \pi 0.1_2^2}

    = 1.724 cm

    l' = l_1 + l_2

    l' = 1.724+1.724

    l' ≅ 3.45 cm

    Total Resistance R’ is calculated as follows:

    R' = \frac{\rho l'^2}{V}

    = \frac{1}{2.037} *\frac{3.45^2*10^{-4}}{\pi * (0.25)^2*2*10^{-6}}

    R’ = 1487.9 Ω

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