1.

Deduce an expression to measure electrc power.

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.


Discussion

No Comment Found