JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Electric Power

The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e. \[P=\frac{W}{t}=Vi={{i}^{\mathbf{2}}}R=\frac{{{V}^{\mathbf{2}}}}{R}\]

(1) Units : It's S.I. unit is Joule/sec or Watt Bigger S.I. units are KW, MW and HP, remember 1 HP = 746 Watt

(2) Rated values : On electrical appliances (Bulbs, Heater, Geyser ... etc.). Wattage, voltage, ... etc. are printed called rated values e.g. If suppose we have a bulb of 40 W, 220 V then rated power \[({{P}_{R}})=40\,W\] while rated voltage \[({{V}_{R}})=220\,V\]. 

(3) Resistance of electrical appliance : If variation of resistance with temperature is neglected then resistance of any electrical appliance can be calculated by rated power and rated voltage i.e. by using \[R=\frac{V_{R}^{\mathbf{2}}}{{{P}_{R}}}\].

(4) Power consumed (illumination) : An electrical appliance (Bulb, heater, .. etc.) consume rated power \[({{P}_{R}})\] only if applied voltage \[({{V}_{A}})\] is equal to rated voltage \[({{V}_{R}})\] i.e. If  \[{{V}_{A}}={{V}_{R}}\]

So \[{{P}_{consumed}}={{P}_{R}}\]. If \[{{V}_{A}}<{{V}_{R}}\] then \[{{P}_{consumed}}=\frac{V_{A}^{2}}{R}\] also we have \[R=\frac{V_{R}^{2}}{{{P}_{R}}}\] so \[{{P}_{Consumed}}(Brightness)=\left( \frac{V_{A}^{2}}{V_{R}^{2}} \right)\,.\,{{P}_{R}}\]

(5) Long distance power transmission : When power is transmitted through a power line of resistance R, power-loss will be \[{{i}^{2}}R\]

Now if the power P is transmitted at voltage V then \[P=Vi\] 

i.e.  \[i=(P/V)\]    So,    \[\text{Power loss}=\frac{{{P}^{2}}}{{{V}^{2}}}\times R\]

Now as for a given power and line, P and R are constant so \[\text{Power loss }\propto (1/{{V}^{2}})\]

So if power is transmitted at high voltage, power loss will be small and vice-versa. This is why long distance power transmission is carried out at high voltage.