Select the best answer for items 1 through 10 to make each...

Select the best answer for items 1 through 10 to make each statement true, and place the letter for the answer in the space provided. A conductor has a resistance of $10 \mathrm{ohms}$. A conductor of the ___________________ same material but twice the diameter will have a resistance of a. $1 / 5 \mathrm{ohm}$. d. 10 ohms. b. 2.5 ohms. e. 20 ohms. c. 5 ohms.

Select the best answer for items 1 through 10 to make each statement true, and place the letter for the answer in the space provided.
A conductor has a resistance of $10 \mathrm{ohms}$. A conductor of the ___________________ same material but twice the diameter will have a resistance of
a. $1 / 5 \mathrm{ohm}$.
d. 10 ohms.
b. 2.5 ohms.
e. 20 ohms.
c. 5 ohms.

Key Concepts

Electrical Resistance

Electrical resistance is the property of a material to oppose the flow of electric current. It is determined by the material’s resistivity, the length of the conductor, and its cross-sectional area, as described by the formula R = ?L/A.

Resistivity

Resistivity is an intrinsic property of a material that quantifies how strongly it opposes electric current. It remains constant for a given material regardless of the size or shape of the conductor.

Conductor Geometry

The geometry of a conductor, specifically its length and cross-sectional area, directly influences its resistance. An increase in cross-sectional area, for instance, leads to a decrease in resistance, while an increase in length results in higher resistance.

Scaling Laws in Geometry

Scaling laws involve understanding how changes in the dimensions of a conductor affect its electrical properties. When the diameter of a conductor is doubled, the cross-sectional area increases by a factor of four (since area is proportional to the square of the diameter), which consequently reduces the resistance by the same factor.

Transcript

00:01 Belongs to the current electricity in which we have a diameter d equals to 2 .50mm and a length l it is equal to 14 meter and resistance r equals to 0 .104 om.

00:15 So for the part a we have to determine the resistivity of the material.

00:19 So we know that resistance are it is equals to row l by area a.

00:24 So we can write that row l divided by area a, it will be pi by 4 multiplied by d squared.

00:30 So from here after rearranging this equation, we will get that resistivity row, it is equals to pi by 4 d squared by resistance r divided by length l.

00:42 So substituting values, so we get pi by 4 multiplied by d which is 2 .50mm that is 10 to the power minus 3 meter.

00:49 So whole square and resistance r which is 0 .104 0 .104 ome divided by this length l which is 14 meters.

01:01 So from here after solving, we get resistivity row, it is equals to 3 .65 multiplied by 10 to the power minus 8 ome meter.

01:09 So this become the answer for the part a of the problem.

01:13 Now moving to the next part b in which we have to determine the total current i when electric field e it is equals to 1 .28 volt per meter.

01:24 So current i, it is equals to v by r from the oms law and v, it can be written as e multiplied by length l, electric field multiplied by length l divided by resistance r.

01:34 So substituting values, so electric wheel is 1 .28 length l is this 14 meter divided by resistance r which is this 0 .104 om...

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