Chapter 1, Problem 1.1P

Interpretation Introduction

Interpretation:

Units for the electric power should be determined as the combination of fundamental S. I. units.

Concept Introduction:

SI stands for International System of units.

SI defines 7 units of different quantities as the basic units from which all other SI units are derived. These basic units are:

Quantity	Unit
Mass	Kilogarm (Kg)
Length	Metre (m)
Time	Second (s)
Electric current	Ampere (A)
Temperature	Kelvin (K)
Luminous intensity	Candela
Amount of substance	Mole

Answer to Problem 1.1P

Units for the electric power as the combination of fundamental S. I. units is - $\frac{Kg.m^2}{s^3}$.

Explanation of Solution

Given Information:

The given quantity is Electric Power.

• Electric power is defined as the rate at which electrical energy is transferred by a circuit.
• SI unit of Electric Power − Watt

But watt is not a fundamental unit, so it has to be converted into fundamental units as following-
Electric Power = $\frac{Energy}{Time}$

Therefore, unit of electric power =watt = $\frac{N.m}{s}$

= $\frac{Kg.m}{s^2}.\frac{m}{s}=\frac{Kg.m^2}{s^3}$

Interpretation Introduction

Interpretation:

The units for the electric charge as the combination of fundamental S. I. units should be determined.

Concept Introduction:

SI stands for International System of units.

SI defines 7 units of different quantities as the basic units from which all other SI units are derived. These basic units are:

Quantity	Unit
Mass	Kilogarm (Kg)
Length	Metre (m)
Time	Second (s)
Electric current	Ampere (A)
Temperature	Kelvin (K)
Luminous intensity	Candela
Amount of substance	Mole

Answer to Problem 1.1P

Units for the electric charge as the combination of fundamental S. I. units is - $A.s$.

Explanation of Solution

Given Information:

The given quantity is Electric Charge.

• It is a physical property which enables the substance to experience a force in the magnetic field. Its SI unit is Coulomb (C), but it is also not a fundamental unit. Thus it is need to be converted into fundamental units.

1 coulomb = 1 ampere-second.

Therefore, unit of electric charge = $A.s$.

Interpretation Introduction

Interpretation:

The units for the electric potential difference as the combination of fundamental S. I. units should be determined.

Concept Introduction:

SI stands for International System of units.

SI defines 7 units of different quantities as the basic units from which all other SI units are derived. These basic units are:

Quantity	Unit
Mass	Kilogarm (Kg)
Length	Metre (m)
Time	Second (s)
Electric current	Ampere (A)
Temperature	Kelvin (K)
Luminous intensity	Candela
Amount of substance	Mole

Answer to Problem 1.1P

Units for the electric potential difference as the combination of fundamental S. I. units is - $\frac{Kg.m^2}{A.s^3}$.

Explanation of Solution

Given Information:

The given quantity is Electric Potential Difference.

• It is the amount of work done in carrying a charge from one point to another in an electric field. Its SI unit is Volt (V).

Therefore, unit of electric potential difference = V = $\begin{array}{l}\frac{Joules}{coulomb}\\ \end{array}$

$\begin{array}{l}=\frac{N.m}{A.s}=\frac{Kg.m}{s^2}.\frac{m}{A.s}\\ =\frac{Kg.m^2}{A.s^3}\end{array}$

Interpretation Introduction

Interpretation:

The units for the electric resistance as the combination of fundamental S. I. units should be determined.

Concept Introduction:

SI stands for International System of units.

SI defines 7 units of different quantities as the basic units from which all other SI units are derived. These basic units are:

Quantity	Unit
Mass	Kilogarm (Kg)
Length	Metre (m)
Time	Second (s)
Electric current	Ampere (A)
Temperature	Kelvin (K)
Luminous intensity	Candela
Amount of substance	Mole

Answer to Problem 1.1P

Units for the electric resistance as the combination of fundamental S. I. units is - $\frac{Kg.m^2}{A^2.s^3}$.

Explanation of Solution

Given Information:

The given quantity is Electric Resistance.

• Electric Resistance is the measure of the difficultyto pass an electric current through the substance. It unit is Ohms (O).

In fundamental units, its unit = $\Omega =\frac{volt}{A}=\frac{watt}{A^2}$

$\begin{array}{l}watt=\frac{Kg.m^2}{s^3}\\ \therefore \Omega =\frac{\raisebox{1ex}{$Kg.m^2$}\!\left/ \!\raisebox{-1ex}{$s^3$}\right.}{A^2}=\frac{Kg.m^2}{A^2.s^3}\end{array}$

Interpretation Introduction

Interpretation:

The units for the electric capacitance as the combination of fundamental S. I. units should be determined.

Concept Introduction:

SI stands for International System of units.

SI defines 7 units of different quantities as the basic units from which all other SI units are derived. These basic units are:

Quantity	Unit
Mass	Kilogarm (Kg)
Length	Metre (m)
Time	Second (s)
Electric current	Ampere (A)
Temperature	Kelvin (K)
Luminous intensity	Candela
Amount of substance	Mole

Answer to Problem 1.1P

Units for the electric capacitance as the combination of fundamental S. I. units is - $\frac{A^2.s^4}{Kg.m^2}$.

Explanation of Solution

Given Information:

The given quantity is Electric Capacitance.

• Electric Capacitance is the ratio of the change in electric charge in the system to the change in voltage. Its SI unit is Farad which is not a fundamental SI unit.

In fundamental units, its unit $F=\frac{C}{V}=\frac{A.s}{Kg.m/A.s^3}$

Conclusion:

Unit for the Electric Capacitance as the combination of fundamental S. I. units is $\frac{A^2.s^4}{Kg.m^2}$.