"Derive"P=I^(2)R.

1.	"Derive"P=I^(2)R.
Answer» Solution :1. Consider that a charge (Q) Coulomb passes through a point A. MOVES to point B in time interval 't' SECONDS. 2. Let V be the potential difference between the points A and B. 3. The work done by electric field in time 't' is GIVEN by W = QV-(1) 4. The work is equal to the energy lost by the charge while passing through the conductor for time 't'. 5. Energy lost by the conductor in `sec = (W)/(t).` From (1), `(W)/(t)=(QV)/(t)-(2)` We know, `(Q)/(t)=Iand(W)/(t)=P("POWER")` Then `(2)rArrP=VI-(3)` This equation can be used to calculate power consumption by any electric devie that is connected in a circuit. From Ohm's law V = I R `therefore(3)rArrPI^(2)R. (or)(V^(2))/(R)-(4)` 6. To know the power that can be EXTRACTED from battery or any source can be calculated by `P=epsilonI.` Whose `epsilon` is the emf of the battery or source.

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