Concept of RL series circuit and Dimensions

Question

Impedance is expressed by !!Z^2= A^2+B^2!! in the given figure. Then dimension of !!AB!! is?

(a) !!M^2L^2T^(-4)A^(-4)!!

(b) !!M^2L^4T^(-6)A^(-4)!!

(c) !!M^2L^3T^(-4)A^(-4)!!

(d) !!M^1L^3T^(-4)A^(-4)!!

Solution

We know that,

!!Z^2= R^2+(omegaL)^2!! for !!RL!! series circuit.

So, comparing with given equation.

!!A = R!! and !!B= omegaL!!

Dimension of !!R!! and !!omegaL!! is equal.

Dimension of !!R!! = [ML^2T^(-1)Q^(-2)]!!

Ww can write !!Q = A^1T^1!!

Therefore,

Dimension of !!R= [ML^(2)T^(-1)(A^1T^1)^(-2)]!!

!!=> [ML^2T^(-3)A^(-2)]!!

So, dimension of !!AB = [ML^2T^(-3)A^(-2)] [ML^2T^(-3)A^(-2)]!!

!!=> [M^2L^4T^(-6)A^(-4)]!!

Get it on Google Play