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Obtain the formula for the 'power loss' (i.e., power dissipated) in a conductor of resistance R, carrying a current I. |
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Answer» <P> Solution : Consider a conductor with end points A and B in which acurrent I is flowing from A to B. Let electric potential at pointsA and B be V(A) and V(B) respectively. As current is flowing from A to B, obviously V(A) > V(B). Let V(A) – V(B) = V. In a time interval `Delta t`, a charge `Delta q = I. Delta t ` travels from A to B. Thus, potential energy of charge at A and B is` U(A) = Delta q.V(A)` and U(B)` = DeltaqV(B)` respectively. ` therefore ` Change in potential energy `DeltaU = U(B) - U(A) = Deltaq[V(B) - V(A)]=- Delta_q.V=-l Delta t. V` If charges were moving freely through the conductor under the action of an applied electric field then in accordance with law of conservation of energy, change in potential energy leads to a change in KINETIC energy too such that, ` Delta U = Delta K = 0` or ` Delta K = - Delta U = I.V Delta t ` In practice, charge carriers in a conductor move with a steady drift velocity. This is because of their collisions with ions and atoms during transit. During collisions, the energy gained by the charges is shared with the atoms. The atoms vibrate more vigourously and the conductor heats up. Thus, in a conductor energy DISSIPATED as heat during a time interval `Delta t` is ` I.V. Delta t`. The energy dissipated per unit time is called the “power loss”. Thus, Power loss `P = (IV Delta t)/(Delta t) = VI` As per Ohm.s law V =IR , hence `P = VI = (V^2)/( R) = I^2 R` |
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