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When a solid cylindrical rod is connected across a fixed potential difference, a current $I$ flows through the rod. What would be the current (in terms of $I$ ) if (a) the length were doubled, (b) the diameter were doubled, (c) both the length and the diameter were doubled?

$I_{2}=2 \mathrm{I}$

Physics 102 Electricity and Magnetism

Chapter 19

Current, Resistance, and Direct-Current Circuit

Electric Charge and Electric Field

Gauss's Law

Electric Potential

Capacitance and Dielectrics

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

Lectures

10:31

A capacitor is a passive two-terminal electrical component that stores electrical energy in an electric field. The effect of a capacitor is known as capacitance. The electrical charge stored in a capacitor is proportional to the potential difference between its terminals. A capacitor is defined as an electrical component that can store an electric charge. The effect of a capacitor is known as capacitance. The charge on a capacitor is directly proportional to the potential difference across its terminals. The unit of capacitance in the International System of Units (SI) is the farad (F), defined as one coulomb per volt. In electrical engineering, a common symbol for capacitance is the lowercase Greek letter "rho" (?). The capacitance of a capacitor is also expressed in farads.

18:38

In physics, electric flux is a measure of the quantity of electric charge passing through a surface. It is used in the study of electromagnetic radiation. The SI unit of electric flux is the weber (symbol: Wb). The electric flux through a surface is calculated by dividing the electric charge passing through the surface by the area of the surface, and multiplying by the permittivity of free space (the permittivity of vacuum is used in the case of a vacuum). The electric flux through a closed surface is zero, by Gauss's law.

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Here foster diagram shows the celendric rod when current i following through it let area cross section of the road is a length of the road is l. Resistance of the road is r, let potential applied across the road is v here. The second diagram shows the rot when its length is doubled here, length new length is l, prime, which is equal to twice of the original length here area procession is a potential applied, is the same as the first 1 we current flowing to the lord is a Prime resistance of the road is r prime, so we can use omdula of our first cassis veult. I r and here r, is equal to resist tut times length of the rod upon area of conductor. Now, for the second case, we can write the law as r prime equal to resisting times length upon area here. Length is twice of the original length upon area procession or we can write this as 2 times act times length upon area, so we can write it as r prime equals to 2 times here. The drum within the breaker, which in the breakers is the original resistance of the wire, so we have r prime equals to 2 r. So it is clear that if we increase the length by 2 times, then resistance will also increase by 2 times now, while using oma. For this case, we have v is equal to i, prime r prime, or we can write this as hi prime is equal to v upon r. Prime from here we have. I prime equal to v upon r prime is equal to 2 r. In the first case, we have earned equal to v upon or now in the second case we have, i prime is equal to 1 upon 2 into v upon or here the time within the breakers is the original current. So we can write it as new. Current is equal to initial current upon 2, so it is now clear that if we increase the length of the wire by 2 times, then the current flowing through the wire will decrease by 2 times now. In the second case, we have double the diameter of the wire. Now we want to check the effect on current here d is the original diameter of the wire and de prime is the new diameter which is equal to 2 times of the initial diameter. Here we have same length that is l in both cases. Now we can write the relations that are equal to rise to t times length upon area crosssection. We also write this expression as r equal to resistit length upon area. Crosssection is equal to pi into diameter square upon 4. The same relation can hold good for the second case. So here we have r prime equal to resistito resuited times length upon a prime which is equal to resist len upon here, a prime is pi into d. Prime square upon 4, which is equal to resistant upon pi here d prime, is equal to 2 g whole square upon 4, or we can write it as our prime is equal to 1 upon 4 into resistivity length upon pi d square upon 4 point. So the drums between the breakers is original volume of the wire, so we have r prime is equal to 1. Fourth of the original resistance. So it is clear that when we have double the diameter of the wire, then its resistance will be decreases by 4 times. Now, while using arms- for our first case, we have v is equal to, i r from here, we have is equal to v upon r now for second law. For second case we have osla equal to v, is equal to i, prime prime, or we can write it as i prime is equal to v upon r prime now by plugging the value of r prime. Here we have i prime equal to v prime upon 1. Upon 4 r- or we can write it as i prime- is equal to 4 into we upon or hear the drums within the breakers is original current, so we have new. Current is equal to 4 times of the initial current. So it is clear now that when we increase the diameter of the wire by 2 times, then the current flowing through it will be 4 times of the initial current. Now, in the third case, we have increased both length and diameter of the wire here. Initial length of wire is l, and new length is twice up. L, that is new length is prime here initial diametry is t and the new diameter, prime, which is twice the first 1. Let i primes new current now we can write. The relation for r is r is equal to resistivity times length upon pi times d square upon 4 pi. We can write the same relation for the second case as r prime equal to resist into l, prime upon pi into d prime square. Upon 4 point now by putting the value of l prime and d prime here we hve rowel prime, is 12 upon pi into here d. Prime is 2 d whole square. Upon 4 from here, we have r prime equal to 2 by 4 resistive t l upon pi d square upon 4 point. So here we have 1 upon 2 and the terms within the breakers is ordinal resistance of the wire that is are so. If we have increased diameter and length of the wire by twice, then the resistance will be decreased by 2 times now we're using the omsula for first case that v is equal to i, r from here. We have i equal to v upon r now by using ormslafor second case we have v is equal to i, prime prime, or we have i prime, is equal to v upon r prime, which is equal to v. U, here, prime, is half of an. We can write it, as i prime, is equal to 2 times v upon r here upon here. V upon r is orignal current, so we have. I prime, is equal to 2 into i, so the current increases by 2 times there's the end of the problem. Thanks for watching.

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