Question

4.
Compute the dimensional formula of electrical
resistance​

Answer 1

Answer:

Potential Difference :

$\longrightarrow\:\sf V = \dfrac{W}{q}$

• W = Work done
• q = charge

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

First we will find the Dimensional formula of work:

$\longrightarrow\sf [W] = [F][s]$

$\longrightarrow\sf [W] = [MLT^{-2}][L]$

$\longrightarrow\bold{[W] = [M^{1} L^{2} T^{-2}]}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

Now, let's find the Dimensions of charge 'q':

$\longrightarrow\sf [q] = [I][t]$

$\longrightarrow\sf [q] = [A][T]$

$\longrightarrow\bf [q] = [A^{1} T ^{1} ]$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

Now, we can calculate the potential difference:

$\longrightarrow\:\sf [V] = \dfrac{ [M^{1} L^{2} T^{-2}]}{[A^{1} T ^{1} ]}$

$\longrightarrow\:\sf [V] = [M^{1} L^{2} T^{-2}][A^{ - 1} T ^{ - 1} ]$

$\longrightarrow\:\bf [V] = [M^{1} L^{2} T^{-3}A^{ - 1} ]$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

Electrical Resistance :

By applying ohm's law :

$\longrightarrow\sf [V] = [I][R]$

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

$\longrightarrow\sf [R] = \dfrac{ [M^{1} L^{2} T^{-3}A^{ - 1} ]}{ [A^{1} ]}$

$\longrightarrow\sf [R] = [M^{1} L^{2} T^{-3}A^{ - 1} ] [A^{ - 1} ]$

$\longrightarrow \underline{ \boxed{ \orange{\bf [R] = [M^{1} L^{2} T^{-3}A^{ - 2} ] }}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$