Resistance Depends on Area of Cross Section


 
 
Concept Explanation
 

Resistance Depends on Area of Cross Section

Resistance Depends on the Area of Cross Section of the wire:

The resistance of ohmic circuit elements such as metal wires or carbon resistor depends on the area of cross section of the wire. If we fix the length of the copper wire and vary the thickness or cross-sectional area A then we find the resistance of the copper decreases as the cross-sectional area A of the wire increases. To find dependence,we will follow the following steps

1 Keep the length of the wire as fixed and just increase or decrease the area of cross section of the wire.

2. Apply a potential difference across the ends of the wire and calculate the value of R from the values of current and potential difference

3. Calculate the value of R for different values of the area of cross section of the wire.

4. Plot a graph as shown below by taking the reciprocal of area of cross section along x axis and resistance along y axis

From the first graph we observe that as the reciprocal of area of cross section increases the resistance also increases. So we find that there is a proportionality between the reciprocal of area of cross section A of the wire and the resistance of the wire R. The resistance of a conductor R is directly proportional to the reciprocal of area of cross section 

5. We can also plot a graph as shown below by taking the area of cross section along x axis and resistance along y axis

From this graph we observe that as the area of cross section increases the resistance decreases. So we find that there is a proportionality between the area of cross section A of the wire and the resistance of the wire R. The resistance of a conductor R is inversly proportional to the area of cross section 

In fact we get an inverse relationship between area of cross section and resistance and it can be stated as

large R;infty ;frac{1}{A}or,

large R=k_1;frac{1}{A}

We know that Resistance is directly proportional to the length of the wire. It can be stated that

R; alpha; l

Rightarrow R; =;k_1 l

Combining the expressions of relation of resistance with length and area of cross section of a wire we get

: large R=k_1k_2frac{l}{A}

To simplify this expression, we can replace the product of the two constants with a single new constant and write

large R=rho ;frac{l}{A}

This constant is known as as Resistivity and it depends on the material of the conductor.

Illustration:  In a circuit  when a current of 2 A flows through a  certain wire having resistance R a potential difference of 8 V is generated across the wire. What will be the new potential difference generated across the wire if the area of cross section of the wire is doubled.

Solution:  It is given that

V= 8 V

I = 2 A

R= ?

V= IR

Rightarrow R = frac{V}{I}=frac{8}{2}= 4Omega

As the area of cross section of the wire is doubled so the resistance will be halved because resistance is inversely proportional to area of cross section..

Rightarrow R_{new} =frac{1}{2}times 4= 2Omega

Rightarrow V_{new} =IR_{new}= 2times 2= 4 V

So new potential difference of 4V is generated across the wire if the area of cross section of the wire is doubled.

Illustration: Calculate the resistance of a conductor when its length is doubled

Solution: We are given that the length is doubled so

Original Length = l1

New Length = l2

We know that

large R= rho frac{l}{A}

Let Original Resistance be Rand New Resistance be R2. So

R_1= rho frac{l_1}{A}; and; R_2 = rhofrac{l_2}{A}

Substituting the value of New length we get

R_2 = rhofrac{l_2}{A}= rhofrac{2l_1}{A}= 2rhofrac{l_1}{A}= 2R_1

Hence when length of a conductor is doubled its resistance gets doubled.

Illustration: A certain wire has a resistance R. Find the resistance of another wire made of same material but its area of cross-section is doubled.

Solution: We are given that the area of cross section is doubled so

Original Area of Cross Section = A1

New Area of Cross Section  = A2

We know that

large R= rho frac{l}{A}

Let Original Resistance be Rand New Resistance be R2. So

R_1= rho frac{l}{A_1}; and; R_2 = rhofrac{l}{A_2}

Substituting the value of New length we get

large R_2= rhofrac{l}{A_2} = rho frac{l}{2A_1} =frac{1}{2} ;timesrhofrac{l}{A_1}=frac{1}{2}R_1

Hence we can say that as area of cross-section is doubled the resistance is halved

Sample Questions
(More Questions for each concept available in Login)
Question : 1

Resistivity of a conductor depends on ______________________.

Right Option : D
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Explanation
Question : 2

The resistance of a thin wire in comparison of thick wire of the same meterial -

Right Option : D
View Explanation
Explanation
Question : 3

Resistance of some materials decreases with increase of temperature. These materials are ____________________ .

Right Option : D
View Explanation
Explanation
 
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