Home
Class 12
PHYSICS
Obtain the formula for the 'power loss' ...

Obtain the formula for the 'power loss' (i.e., power dissipated) in a conductor of resistance R, carrying a current I.

Text Solution

Verified by Experts


Consider a conductor with end points A and B in which a current I is flowing from A to B. Let electric potential at points A 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`
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    U-LIKE SERIES|Exercise LONG ANSWER QUESTIONS - I|47 Videos
  • CURRENT ELECTRICITY

    U-LIKE SERIES|Exercise LONG ANSWER QUESTIONS - II|8 Videos
  • CURRENT ELECTRICITY

    U-LIKE SERIES|Exercise VERY SHORT ANSWER QUESTIONS|39 Videos
  • CBSE SAMPLE QUESTION PAPER 2019-20 (SOLVED)

    U-LIKE SERIES|Exercise SECTION D|6 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    U-LIKE SERIES|Exercise SELF ASSESSMENT TEST SECTION-C (VERY SHORT ANSWER QUESTIONS)|3 Videos

Similar Questions

Explore conceptually related problems

A moving coil voltmeter is generally used to measure the potential difference across a conductor of resistance 'r' carrying a current i . The resistance of voltmeter is R . For more correct measurement of potential difference

The average power dissipated in a pure inductor L carrying an alternating current of rms value I is .

The average power dissipated in an AC circuit containing a resistance along is

The power dissipated (in watt) in 3Omega resistance in the following circuit is

Find the power dissipated in 3 Omega resistance of the network of resistors.

On giving 220 V to a resistor the power dissipated is 40 W then value of resistance is

A coil and an inductance - free resistance R=25 Omega are connected in parallel to the ac1 mains. Find the heat power generated in the coil provided a current I=0.90 A is drawn from the mains. The coil and the resistance R carry currents I_(1)=0.50 A and I_(2)=0.60 A respectively.

U-LIKE SERIES-CURRENT ELECTRICITY -SHORT ANSWER QUESTIONS
  1. A battery of emf epsi and internal resistance r, when connected acros...

    Text Solution

    |

  2. The reading on a high resistance voltmeter, when a cell is connected a...

    Text Solution

    |

  3. Obtain the formula for the 'power loss' (i.e., power dissipated) in a ...

    Text Solution

    |

  4. The adjoining graph shows the variation of terminal potential differe...

    Text Solution

    |

  5. Two cells of emf 2 epsi and epsi and internal resistance 2r and r resp...

    Text Solution

    |

  6. A 10 V cell of negligible internal resistance is connected in parallel...

    Text Solution

    |

  7. Two bulbs are rated (P1, V)" and "(P2, V) . If they are connected (i...

    Text Solution

    |

  8. A cell of emf epsi and internal resistance r is connected to two exte...

    Text Solution

    |

  9. Two cells epsi1 and epsi2 in the given circuit diagram have an emf of ...

    Text Solution

    |

  10. Draw the diagram of Wheatstone bridge. Why does no current flow throug...

    Text Solution

    |

  11. Explain the principle on which the working of a potentiometer is based...

    Text Solution

    |

  12. Two students X and Y perform an experiment on potentiometer separately...

    Text Solution

    |

  13. In a potentiometer arrangement for determining the emf of a cell, the ...

    Text Solution

    |

  14. In the given circuit, assuming point A to be at zero potential, use Ki...

    Text Solution

    |

  15. Use Kirchhoff's rules to determine the potential difference between th...

    Text Solution

    |

  16. A wire of 15 Omega resistance is gradually stretched to double its ori...

    Text Solution

    |

  17. Calculate the current drawn from the battery in the given network .

    Text Solution

    |

  18. Using Kirchhoff's rules in the circuit shown determine (a) the voltag...

    Text Solution

    |

  19. The circuit in Fig. shows two cells connected in opposition to each o...

    Text Solution

    |

  20. Figure shows two circuits each having a galvanometer and a battery of ...

    Text Solution

    |