Define electric energy and power. Give their S.I. units and define them. Give relation between them.
Electric Energy :
The work done by a source to maintain a current in an electrical circuit is known as electric energy.
Consider an electric device or circuit element (e.g., an electric lamp, heater etc.) of resistance $R$ through which current $I$ flows from the end $A$ to the end $B$ for time $t$. Let $q$ be the charge flowing from $A$ to $B$ in time $t$, then
$q = It \quad \left( I = \frac{q}{t} \right)$
If $V$ be the potential difference between $A$ and $B$, then work done to carry the charge $q$ from point $B$ to $A$ is equal to the change in potential energy of the charge $q$ and is given by
$W = \Delta U = Vq = VIt$
This work done is equal to the electric energy $E$ consumed in the circuit and is given by
E = VIt
We know, $V = IR$ (from Ohm's law)
$E = (IR) It = I^2Rt = \left( \frac{V}{R} \right)^2 Rt = \frac{V^2}{R} t$
This electric energy appears as heat energy in the resistor or electric device. In other words, eqn. (iii) represents the amount of energy dissipated as heat in the resistor or conductor during time him intervall.
Eqn. (iii) is known as Joule's law of heating.
S.I. unit of electrical energy is joule (J)
$1 \text{ joule } = 1 \text{ volt } \times 1 \text{ ampere } \times 1 \text{ second } = 1 \text{ VAs}$
Electric Power :
Electric power can be defined in different ways :
(i) Electric power can be defined as the rate of doing electrical work.
(ii) Electric power (P) is defined as the heat energy produced (or energy dissipated) per unit time in a conductor of resistance $R$, when current $I$ is passing through it.
That is, Electric power, $P = \frac{\text{energy produced}}{\text{time}} $
$P= \frac{E}{t}$ ...(i)
Now $E = VIt$
$P = \frac{VIt}{t} = VI$ ...(ii)
Thus, electric power is simply defined as the product of the applied voltage and current flowing through the circuit.
Using ohm's law, $V = IR$, eqn. (ii) can be written as
$P = I^2R = \left( \frac{V}{R} \right)^2 R = \frac{V^2}{R}$
Eqn. (iii) represents power loss (or ohmic loss) in a conductor of resistance $R$ through which current $I$ passes due to potential difference $V$. This power heats up the devices like electric bulbs, electric heaters, electric toasters etc.
S.I. Units of Electric Power is watt (W).
We know,$P = IV$
$\therefore$ $1 \text{ watt } = 1 \text{ ampere } \times 1 \text{ volt}$
$1 \text{ W } = 1 \text{ ampere volt}$
Definition of 1 watt :
Electric power is said to be 1 watt if 1 ampere current flows through an electrical circuit, when a potential difference of 1 volt is applied across it.
Bigger unit of electric power is kilowatt (kW) and still bigger unit is megawatt (MW).
$1 kW = 10^3 W$
$1 MW = 10^6 W$
Practical unit of power is horse-power (h.p.).
$1 h.p. = 746 W$
Relation between electric energy and electric power :
Electric energy, E = VIt ...(i)
and Electric power, P = VI ...(ii)
where $V$ is applied potential difference, $I$ is the current flowing in the circuit and $t$ is the time for which current is flowing in the circuit. From eqns. (i) and (ii), we get
$E = Pt$
or
$P = \frac{E}{t}$
or
$Power = \frac{Energy}{Time}$
Commercial Unit or Board of Trade (BOT) Unit of electric energy is kilowatt hour (kW h).
$1 unit = 1 kW h= 1000 W h$
Important Conversion :
$1 kW h = 1000 watt h$
$\therefore 1W = 1 J s^{-1})$
$1 kW h= 1000 \frac{J}{s} \times 3600s$
$1 kW h= 3600,000 J$
$1 kW h= 3.6 \times 10^6J$
$1 kW h = 3.6 \times 10^6 J $
Electric Power Transmission :
Consider a device (say a bulb or heater) of resistance $R$. Let the power $P$ is delivered to the device through conductors or cables from the power station.
Let $V$ is potential difference across the device of resistance $R$ through which current $I$ passes, then the power delivered to the device is given by
$P= VI \tag{i}$
If $r$ is the resistance of the conductors or cables, then the power dissipated in the conductors or cables through which current $I$ passes given by
$P^{'}= I^2 r \tag{ii}$
From eqn. (i), we have
$I= \frac{P}{V} \tag{iii}$
Using eqn. (iii) in eqn. (ii), we get
$P^{'}= \frac{P^2r}{V^2} \tag{iv}$
which is the expression for the power dissipated in the transmission conductors or cables.
Thus,
$P^{'} \propto \frac{1}{V^2}$
($\because$ $P$ and $r$ are constants)
The power dissipated in the transmission cables or conductors will be less if the cables carry current at very high voltage(V).