
Dimensions of electrical resistance are:
$\text{A}\text{. }\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-1}} \right]$
$\text{B}\text{. }\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right]$
$\text{C}\text{. }\left[ M{{L}^{3}}{{T}^{-3}}{{A}^{-2}} \right]$
$\text{D}\text{. }\left[ M{{L}^{2}}{{T}^{3}}{{A}^{2}} \right]$
Answer
582k+ views
Hint: According to Ohm’s law, V = iR, where V is potential difference, i is current and R is resistance. Hence, $R=\dfrac{V}{i}$. Use this formula to find the dimensional formula of R. V is work done per unit charge.
Formula used:
V = iR
$\left[ W \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$
[q] = [AT]
Complete step by step answer:
Electrical resistance of a given substance is the ability of the substance to resist the flow of electrons or charges when a potential difference is created across the substance.
The flow of charges per unit time is called current in the circuit.
Suppose a conductor with a resistance of R is connected across a cell of emf V. This cell will create a potential difference V across the conductor. Due to the potential difference there will be current in the circuit. Let the current in the circuit be i.
According to Ohm’s law, V = iR.
Hence, we get that
$R=\dfrac{V}{i}$.
Hence, the dimensional formula of resistance will be equal to the ratio of dimensional formulas of potential difference to current. $\left[ R \right]=\dfrac{\left[ V \right]}{\left[ i \right]}$ …. (i).
Therefore, let us find the dimensional formula of V and i.
Potential difference is equal to the work done per unit charge.
The dimensional formulas of work done is $\left[ W \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$.
The dimensional formula of charge is [q] = [AT].
Hence, the dimensional formula of potential difference is $\left[ V \right]=\dfrac{\left[ W \right]}{\left[ q \right]}=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ AT \right]}=\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-1}} \right]$.
The dimensional formula of current is [A].
Substitute the dimensional formulas of potential difference and current in equation (i).
$\left[ R \right]=\dfrac{\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-1}} \right]}{\left[ A \right]}=\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right]$.
Hence, the correct answer is option B.
Note:
We can also use the units of the quantities for finding the dimensional formula of resistances.
The unit of work is $kg{{m}^{2}}{{s}^{-2}}$.
The unit of charge is As.
Therefore, the unit of potential difference is $kg{{m}^{2}}{{s}^{-3}}{{A}^{-1}}$.
The unit of current is A.
Hence,the unit of resistance is $\dfrac{kg{{m}^{2}}{{s}^{-3}}{{A}^{-1}}}{A}=kg{{m}^{2}}{{s}^{-3}}{{A}^{-2}}$.
The dimensions of units kg, m, s and A are M, L, T and A.
Hence, the dimensions of resistance are $\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right]$.
Formula used:
V = iR
$\left[ W \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$
[q] = [AT]
Complete step by step answer:
Electrical resistance of a given substance is the ability of the substance to resist the flow of electrons or charges when a potential difference is created across the substance.
The flow of charges per unit time is called current in the circuit.
Suppose a conductor with a resistance of R is connected across a cell of emf V. This cell will create a potential difference V across the conductor. Due to the potential difference there will be current in the circuit. Let the current in the circuit be i.
According to Ohm’s law, V = iR.
Hence, we get that
$R=\dfrac{V}{i}$.
Hence, the dimensional formula of resistance will be equal to the ratio of dimensional formulas of potential difference to current. $\left[ R \right]=\dfrac{\left[ V \right]}{\left[ i \right]}$ …. (i).
Therefore, let us find the dimensional formula of V and i.
Potential difference is equal to the work done per unit charge.
The dimensional formulas of work done is $\left[ W \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$.
The dimensional formula of charge is [q] = [AT].
Hence, the dimensional formula of potential difference is $\left[ V \right]=\dfrac{\left[ W \right]}{\left[ q \right]}=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ AT \right]}=\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-1}} \right]$.
The dimensional formula of current is [A].
Substitute the dimensional formulas of potential difference and current in equation (i).
$\left[ R \right]=\dfrac{\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-1}} \right]}{\left[ A \right]}=\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right]$.
Hence, the correct answer is option B.
Note:
We can also use the units of the quantities for finding the dimensional formula of resistances.
The unit of work is $kg{{m}^{2}}{{s}^{-2}}$.
The unit of charge is As.
Therefore, the unit of potential difference is $kg{{m}^{2}}{{s}^{-3}}{{A}^{-1}}$.
The unit of current is A.
Hence,the unit of resistance is $\dfrac{kg{{m}^{2}}{{s}^{-3}}{{A}^{-1}}}{A}=kg{{m}^{2}}{{s}^{-3}}{{A}^{-2}}$.
The dimensions of units kg, m, s and A are M, L, T and A.
Hence, the dimensions of resistance are $\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-2}} \right]$.
Recently Updated Pages
Three resistances each of 4Omega are connected in the class 12 physics CBSE

Assertion Bell shaped age pyramid represents a stable class 12 biology CBSE

StatementI HCHO and HCOOH can be distinguished by the class 12 chemistry CBSE

Compound A is an aromatic amine which reacts with NaNO2 class 12 chemistry CBSE

The full form of the term EPI is aEngineered product class 12 biology CBSE

system of irrigation is more useful on uneven lands class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

