PDF chapter test TRY NOW

Have you ever watched a thunderstorm with

**mighty lightning bolts from the sky?***Lightning bolts*

\(\)

A lightning bolt is a sudden, massive surge of electricity between the clouds or sometimes between the sky and the ground. The electrical energy produced by a single lightning bolt is enough to light \(100\ -watt\) lamp.

Electric power is defined as the rate of doing work or spending energy. In other words, it is the rate of consumption of electrical energy.

The formula is,

$\begin{array}{l}\mathit{Power}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\mathit{Work}}{\mathit{Time}}\\ \\ P\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{W}{t}\end{array}$ ---- (eq. 1)

Let '\(I\)' be the current flowing through a conductor for time interval '\(t\)'. Let '\(V\)' be the potential difference across the two ends of the conductor of resistance \(R\).

Then, '\(W\)' is the work done to move the charge across the ends of the conductor. So, the equation is written as,

$W=\phantom{\rule{0.147em}{0ex}}VIt$ ---- (eq. 2)

Substituting the value of \(W\) in the equation of power,

$P\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\mathit{VIt}}{t}=\mathit{VI}$

Hence, the electric power is the product of the electric current and the potential difference due to which the current passes in a circuit.

According to Ohm's law, \(V = IR\). Now, by substituting this value in (eq. 2), we get

$W\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}{I}^{2}R\phantom{\rule{0.147em}{0ex}}t$ ---- (eq. 3)

On substituting the above value in (eq. 1), we get

$\begin{array}{l}P\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{{I}^{2}R\phantom{\rule{0.147em}{0ex}}t}{t}\\ \\ ={I}^{2}R\end{array}$

From Ohm's law, it is deduced that $I\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{V}{R}$. Hence, (eq. 3) becomes

$W\phantom{\rule{0.147em}{0ex}}=\frac{{V}^{2}}{R}\phantom{\rule{0.147em}{0ex}}t$

Now, by substituting the above value in (eq. 1), we get

$\begin{array}{l}P\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\frac{{V}^{2}}{R}\phantom{\rule{0.147em}{0ex}}t}{t}\\ \\ =\phantom{\rule{0.147em}{0ex}}\frac{{V}^{2}}{R}\end{array}$

**Unit of electric power:**

The SI unit of electric power is the watt (\(W\)). If \(1\ ampere\) of a current passes across the conductor at a potential difference of \(1\ volt\), then the electric power is given as,

$P=1\phantom{\rule{0.147em}{0ex}}\mathit{volt}\phantom{\rule{0.147em}{0ex}}\times 1\phantom{\rule{0.147em}{0ex}}\mathit{ampere}=1\phantom{\rule{0.147em}{0ex}}\mathit{watt}\hspace{0.17em}=1\phantom{\rule{0.147em}{0ex}}\mathit{VA}$

1 watt is defined as the power consumed when an electric device is operated at a potential difference of 1 volt, carrying a current of 1 ampere.

For larger unit of power, the term 'kilowatt' (\(kW\)) is used.

\(1\ kilowatt\) \(=\) \(1000\ watts\)