Question

what is the dimensional formula of expression RLC circuit​

Answer 1

Answer:

Dimensions of L, MT^(-2)L^2A^(-2)

Dimensions of R,

ML^2T^(-3)A^(-2)

Answer 2

Answer:

The dimension formula of R and L is $[ML^2T^{-3}A^{-2}]$ and $[MT^{-2}L^{2}A^{-2}]$

Explanation:

Let consider resistance

Using Ohm’s law

$V= IR$

$R=\dfrac{V}{I}$

Now, V has units

$V= electric\ field\times distance$

But electric field has units  

$E=\dfrac{F}{Q}$

The charge has dimensions  

$Q=I\times t$

$F= \dfrac{M\times L}{T^2}$

Thus, dimensions of V is,

$V=\dfrac{[LMLT^{-2}]}{[AT]}$

$V=[ML^2T^{-3}A^{-1}]$

Current I has dimensions  

$[I]=A$

Thus, the dimensions of resistance,

$R=\dfrac{V}{I}$

$R=[ML^2T^{-3}A^{-2}]$

For inductance,  

The defining equation is,

$\phi=L\ I$

But $\phi$ has units  

$\phi=B\times L^2$  

Magnetic field from Lorentz force law has units,

$B=\dfrac{F}{v\times q}$

Therefore, dimensions of magnetic field

$B=\dfrac{[MLT^{-2}]}{[LT^{-1}AT]}$

$B=[MT^{-2}A^{-1}]$

Therefore, The dimensions of magnetic flux,

$\phi=[MT^{-2}L^{2}A^{-1}]$

Thus finally, dimensions of inductance,

$L=\phi\times I$

$L=[MT^{-2}L^{2}A^{-2}]$

Hence, The dimension formula of R and L is $[ML^2T^{-3}A^{-2}]$ and $[MT^{-2}L^{2}A^{-2}]$

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