Question

As we know that in ohm's law, potential difference (V) is directly proportional to current (I) and when we remove sign of proportionality, why do we apply Resistance (R) as constant value?
Can we apply any another constant? Why not?​

Answer 1

Answer:

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the Concept of Ohm's Law has been used. We are asked that if we can use any other constant value. In order to support our answer we have to use the method of Dimensional Formula which can prove that our answer is correct.

Let's do it !!

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★ Solution :-

Given,

» Potential Difference = V = Voltage

» Current = I = Ampere

» Resistance = R = Ohm

These are the given things. Now let's find the solution step - by - step.

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~ Why do we apply Resistance as constant value in Ohm's Law ::

First of all we need to understand that what does 'Resistance' mean in terms of electricity.

✒ Resistance : It is the property of a conductor to resist the flow of charges through it. It is represented by Ω .

Now, we know that according to Ohm's Law,

$\\\;\bf{\rightarrow\;\;\;\red{V\;\;\propto\;\;I}}$

This means that Potential Difference is directly proportional to the Electric Current.

So if the Potential Difference increases in a conductor, then the Current also increases to overcome that difference.

Also, Ohm's Law states that "the electric current flowing through a metallic wire is directly proportional to the potential difference V, across its ends provided its temperature remains constant" .

We know that, according to the definition of resistance, that every material has that whether its less or more. And even Potential Difference also provides that resistance in material.

That is why, the ratio of Potential Difference by Current is given as a fixed constant that is Resistance.

This proves the Ohm's Law.

$\\\;\bf{\rightarrow\;\;\;\blue{\dfrac{V}{I}\;\;=\;\;Constant}}$

This because two proportional quantity will provide a constant value which will remain same by the change of Potential Difference and Current.

So this gives us,

$\\\;\bf{\rightarrow\;\;\;\green{\dfrac{V}{I}\;\;=\;\;R}}$

This is the reason why we apply Resistance as the Constant Value.

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~ Can we apply any other constant ? Why not ?

No, we can't apply any other constant.

This is because, we know that

$\\\;\bf{\rightarrow\;\;\;\orange{\dfrac{V}{I}\;\;=\;\;R}}$

From this, we get

$\\\;\sf{\Longrightarrow\;\;\;R\;=\;\dfrac{Potential\;Difference}{Current}\;=\;Volt\:Ampere^{-1}}$

• Since SI unit of Potential Difference is Volt and that of current is Ampere.

Also we know that,

$\\\;\sf{\Longrightarrow\;\;\;Potential\;Difference\;=\;\dfrac{Workdone}{Charge}\;=\;J\:Coulomb^{-1}}$

• Since SI unit of Workdone is Joule and charge is Coulomb.

$\\\;\sf{\Longrightarrow\;\;\;Current\;=\;\dfrac{Charge}{Time}\;=\;Coulomb\:sec^{-1}}$

• Since SI unit of Time is sec.

Now if we take out dimensional formula of the Ohm's Law, we get

$\\\;\sf{\mapsto\;\;[R]\;=\;\bigg[\dfrac{V}{I}\bigg]}$

$\\\;\sf{\mapsto\;\;[R]\;=\;\dfrac{[V]}{[I]}}$

Now we need to prove that dimensions of LHS should be equal to RHS. If this proves then equation will be correct .

Now we know that,

• [R] = [M L² T⁻³ A⁻²]

• [V] = [M L² T⁻³ A⁻¹]

• [I] = [A]

By applying these values, we get

$\\\;\sf{\mapsto\;\;\gray{[M\:L^{2}\:T^{-3}\:A^{-2}]\;=\;\dfrac{[M\:L^{2}\:T^{-3}\:A^{-1}]}{[A]}}}$

$\\\;\sf{\mapsto\;\;\purple{[M\:L^{2}\:T^{-3}\:A^{-2}]\;=\;[M\:L^{2}\:T^{-3}\:A^{-2}]}}$

This gives us LHS = RHS.

So Resistance is the correct constant value applied.

Now if we apply different Constant, then surely Ohm's Law won't be satisfied since here LHS ≠ RHS then because every constant has different values. Even a slight change in value can cause change in equation.

So, Resistance is the correct and only Constant Value that can be applied in Ohm's Law.

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★ More to know :-

$\\\;\tt{\leadsto\;\;Specific\;Resistivity\;=\;\rho\:\dfrac{l}{A}}$

$\\\;\tt{\leadsto\;\;Heat\;=\;V\:I\:t}$

$\\\;\tt{\leadsto\;\;Heat\;=\;I^{2}\:R\:t}$

$\\\;\tt{\leadsto\;\;Heat\;=\;\dfrac{V^{2}}{R}\:\times\:t}$

$\\\;\tt{\leadsto\;\;Heat\;=\;P\:\times\:t}$

$\\\;\tt{\leadsto\;\;Power,\;P\;=\;V\:\times\:I}$

$\\\;\tt{\leadsto\;\;Power,\;P\;=\;I^{2}\:\times\:R}$

$\\\;\tt{\leadsto\;\;Power\;=\;\dfrac{V^{2}}{R}}$