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Keith_Richards [23]
3 years ago
5

Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long

er one has length 8L, diameter 4D, and resistance R2. How do the resistances of these two resistors compare
Physics
1 answer:
nordsb [41]3 years ago
7 0

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

R = \dfrac{\rho L }{A}

where;

A = πr²

R = \dfrac{\rho L }{\pi r ^2}

For the shorter cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{ 4 \rho L }{\pi \ D   ^2}

For the longer cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{32\rho L }{\pi \ (4 D)   ^2}

R = \dfrac{2\rho L }{\pi \ (D)   ^2}

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D   ^2}}{ \dfrac{2\rho L }{\pi \ (D)   ^2}}

\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D   ^2}}* { \dfrac  {\pi \ (D)   ^2} {2\rho L}}

\dfrac{R_s}{R_L} =2

{R_s}=2{R_L}

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

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After adding them, you will get the reciprocal again and then compute for the value. The problem says that there are 4 resistors in the circuit that have a resistance of 75.

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