Chapter 9, Problem 1ICA

Complete the following table.

			Dimensions
	Quantity	SI Units	M	L	T	Θ	N	J	I
Example	Acoustic impedance	(Pa s)/m	1	–2	–1	0	0	0	0
(a)	Circuit resistance	V/A
(b)	Luminous efficacy	cd/W
(c)	Thermal conductivity	cal/(cm s oC)

To determine

Fill the appropriate dimensions for given quantity and complete the table.

Answer to Problem 1ICA

The completed table of appropriate dimension for given quantity is shown in Table 1.

Explanation of Solution

Given data:

The SI unit of circuit resistance is $\text{V}/\text{A}$.

The SI unit of luminous efficacy is $\text{cd}/\text{W}$.

The SI unit of thermal conductivity is $\text{cal}/\left(\text{cm}\cdot \text{s }°\text{C}\right)$.

Calculation:

Refer to Table 9-1 in the textbook for fundamental dimensions and base unit.

Refer to Table 8-1 in the textbook for common derived units in the SI system, to find the fundamental dimension of voltage.

$1\text{V}=1\frac{\text{kg}\cdot {\text{m}}^2}{{\text{s}}^3\cdot \text{A}}$

Substitute M for kg, L for $\text{m}$, I for A, and T for s to find dimension of voltage.

$\begin{array}{c}1\text{V}=1\frac{\text{M}\cdot {\text{L}}^2}{{\text{T}}^3\cdot \text{I}}\\ ={\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}\end{array}$

Find the fundamental dimension of resistance.

$\text{Circuit resistance}=\text{V}/\text{A}$

Substitute ${\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}$ for V and I for A to find the dimension of resistance.

$\begin{array}{c}\text{Circuit resistance}=\frac{{\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}}{\text{I}}\\ ={\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-2}\end{array}$

Luminous efficacy is the measure of visible light produced by the source and is the ratio of luminous flux to power.

$\text{luminous efficacy}=\frac{\text{luminous flux}}{\text{power}}$

$\text{luminous efficacy}=\text{cd}/\text{W}$ (1)

Fundamental dimension of luminous flux (cd) is J.

Find the fundamental dimension of power.

Refer to Table 8-14 in the textbook for summary of electrical properties.

$\text{W}=\text{VA}$

Substitute ${\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}$ for V and I for A to find the dimension of power.

$\begin{array}{c}\text{W}=\left({\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}\right)\left(\text{I}\right)\\ ={\text{ML}}^2{\text{T}}^{-3}\end{array}$

Substitute J for cd and ${\text{ML}}^2{\text{T}}^{-3}$ for W in equation (1) to find dimension of power.

$\begin{array}{c}\text{luminous efficacy}=\frac{\text{J}}{{\text{ML}}^2{\text{T}}^{-3}}\\ ={\text{M}}^{-1}{\text{L}}^{-2}{\text{T}}^3\text{J}\end{array}$

Calories can be measured in terms of Joules.

Find the fundamental dimension of thermal conductivity.

$\text{thermal conductivity}=\text{cal}/\left(\text{cm}\cdot \text{s }°\text{C}\right)$ (2)

Refer to Table 8-11 in the textbook for dimension of energy, to represent the dimensions of required parameters.

Substitute J for cal in equation (2) to find the thermal conductivity in terms of Joule.

$\text{thermal conductivity}=\text{J}/\left(\text{cm}\cdot \text{s }°\text{C}\right)$

Substitute ${\text{ML}}^2{\text{T}}^{-2}$ for J, L for cm, T for s and $\Theta$ for $°\text{C}$ to find the fundamental dimension of thermal conductivity.

$\begin{array}{c}\text{thermal conductivity}=\frac{{\text{ML}}^2{\text{T}}^{-2}}{\left(\text{L}\right)\left(\text{T}\right)\left(\Theta \right)}\\ ={\text{ML}}^{2-1}{\text{T}}^{-2-1}{\Theta }^{-1}\\ ={\text{ML}}^1{\text{T}}^{-3}{\Theta }^{-1}\end{array}$

Obtained dimensions for circuit resistance, luminous efficacy and thermal conductivity is tabulated in Table 1.

Table 1

			Dimensions
	Quality	SI Units	M	L	T	$\Theta$	N	J	I
Example	Acoustic impedance	$\left(\text{Pa}\cdot \text{s}\right)/\text{m}$	1	$-2$	$-1$	0	0	0	0
(a)	Circuit resistance	$\text{V}/\text{A}$	1	2	$-3$	0	0	0	$-2$
(b)	Luminous efficiency	$\text{cd}/\text{W}$	0	$-3$	0	0	1	0	0
(c)	Thermal conductivity	$\text{cal}/\left(\text{cm}\cdot \text{s }°\text{C}\right)$	1	1	$-3$	$-1$	0	0	0

Conclusion:

Thus, the completed table of appropriate dimension for given quantity is shown in Table 1.