Home
Class 12
PHYSICS
A bulb with rating 250V, 100W is connect...

A bulb with rating `250V, 100W` is connected to a power supply of `220 V` situatd `10 m` away using a copper wire of area of cross section `5 mm^2`. How much power will be consumed by the connecting wire? Resistivity of copper `= 1.7 xx 10^(-8) Omegam`.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will calculate the resistance of the bulb, the resistance of the copper wire, the total resistance in the circuit, and finally the power consumed by the connecting wire. ### Step 1: Calculate the resistance of the bulb (Rb) The resistance of the bulb can be calculated using the formula: \[ R_b = \frac{V^2}{P} \] Where: - \(V = 250 \, \text{V}\) (voltage rating of the bulb) - \(P = 100 \, \text{W}\) (power rating of the bulb) Substituting the values: \[ R_b = \frac{(250)^2}{100} = \frac{62500}{100} = 625 \, \Omega \] ### Step 2: Calculate the resistance of the copper wire (Rw) The resistance of the wire can be calculated using the formula: \[ R_w = \frac{\rho \cdot L}{A} \] Where: - \(\rho = 1.7 \times 10^{-8} \, \Omega \cdot m\) (resistivity of copper) - \(L = 10 \, m\) (length of one wire) - \(A = 5 \, mm^2 = 5 \times 10^{-6} \, m^2\) (cross-sectional area) Substituting the values: \[ R_w = \frac{1.7 \times 10^{-8} \cdot 10}{5 \times 10^{-6}} = \frac{1.7 \times 10^{-7}}{5 \times 10^{-6}} = 0.034 \, \Omega \] Since there are two wires, the total resistance of the wires (R_total) will be: \[ R_{\text{total}} = 2 \cdot R_w = 2 \cdot 0.034 = 0.068 \, \Omega \] ### Step 3: Calculate the total resistance in the circuit (R_total_circuit) The total resistance in the circuit is the sum of the resistance of the bulb and the total resistance of the wires: \[ R_{\text{total circuit}} = R_b + R_{\text{total}} = 625 + 0.068 = 625.068 \, \Omega \] ### Step 4: Calculate the current (I) in the circuit Using Ohm's law, the current can be calculated as: \[ I = \frac{V}{R_{\text{total circuit}}} \] Where \(V = 220 \, V\) (voltage from the power supply). Substituting the values: \[ I = \frac{220}{625.068} \approx 0.352 \, A \] ### Step 5: Calculate the power consumed by the connecting wire (P_wire) The power consumed by the wire can be calculated using the formula: \[ P_{\text{wire}} = I^2 \cdot R_{\text{total}} \] Substituting the values: \[ P_{\text{wire}} = (0.352)^2 \cdot 0.068 \approx 0.0084 \, W = 8.4 \, mW \] ### Final Answer: The power consumed by the connecting wire is approximately **8.4 mW**. ---

To solve the problem step by step, we will calculate the resistance of the bulb, the resistance of the copper wire, the total resistance in the circuit, and finally the power consumed by the connecting wire. ### Step 1: Calculate the resistance of the bulb (Rb) The resistance of the bulb can be calculated using the formula: \[ R_b = \frac{V^2}{P} \] ...
Promotional Banner

Topper's Solved these Questions

  • THERMAL AND CHEMICAL EFFECT OF ELECTRIC CURRENT

    HC VERMA ENGLISH|Exercise Short Question|9 Videos
  • THERMAL AND CHEMICAL EFFECT OF ELECTRIC CURRENT

    HC VERMA ENGLISH|Exercise Objective 2|5 Videos
  • THE SPECIAL THEORY OF RELATIVITY

    HC VERMA ENGLISH|Exercise Short answer|2 Videos
  • X-Rays

    HC VERMA ENGLISH|Exercise Ques for Short Ans|10 Videos

Similar Questions

Explore conceptually related problems

What length of a copper wire of cross-sectional area 0.01 mm^2 will be needed to prepare a resistance of 1 kOmega ? Resistivity of copper = 1.7 xx 10^(-8) Omega m .

Calculate the electric field in a copper wire of cross-sectional area 2.0mm^(2) carrying a current of 1A .The resistivity of copper =1.7xx10^(-8)(Omega)m .

Calculate the resistance of 1 km long copper wire of radius 1 mm. (Resistivity of copper is 1.72 xx10^(-8) Omega m ).

A copper wire of cross - sectional area 3.4 m m^(2) and length of the wire 400m, specific resistivity of copper is 1.7 xx 10^(-8)Omega-m . Then the resistance of the wire is

An inductor coil of inductance 17 mH is constructed from a copper wire of length 100 m and cross - sectional are 1mm^2 . Calculate the time constant of the circuit if this inductor is joined across an ideal battery. The resistivity of copper = 1.7 X 10^(-8) Omega m.

In house wiring, copper wire 2.05 mm in diameter is often used. Find the resistane of 35.0 m long wire. Specific resistance of copper is 1.72xx10^-8Omega-m .

An electric bulb, when connected across a power supply of 220V , consumes a power of 60 W . If the supply drops supply is suddently increased to 240 V , what will be the power consumed?

Find the resistance of a copper coil of total wire-length 10m and area of cross section 1.0mm^(2). What would be the resistance of a similar coil of aluminium? The resistivity of copper =1.7xx10^(-8)(Omega)m and that of aluminium 2.6xx10^(-8)(Omega).

A small insect crawls in the direction of electron drift along bare copper wire that carries a current of 2.56A. It travels with the drift speed of the electron in the wire of uniform cross-section area 1mm^2 Number of free electrons for copper = 8 times 10^28// cc and resistivity of copper = 1.6 times 10^-8 Omega m How much time would the insect take to crawl 1.0 cm if it crawls at the drift speed of the electron in the wire?

Calculate the drift speed of the electrons when 1A of current exists in a copper wire of cross section 2 mm^(2) .The number of free electrons in 1cm^(3) of copper is 8.5xx10^(22) .

HC VERMA ENGLISH-THERMAL AND CHEMICAL EFFECT OF ELECTRIC CURRENT-Exercise
  1. The specification on a heater coil is 250 V, 500 W. Calculate the resi...

    Text Solution

    |

  2. A heater coil is to be constructed with a nichrome wire (rho = 1.0 xx ...

    Text Solution

    |

  3. A bulb with rating 250V, 100W is connected to a power supply of 220 V ...

    Text Solution

    |

  4. An electric bulb, when connected across a power supply of 220V, consum...

    Text Solution

    |

  5. A servo voltage stabiliser restricts the voltage output to 220 V +- 1%...

    Text Solution

    |

  6. An electric bulb marked 220 V, 100 W will get fused if it is made to c...

    Text Solution

    |

  7. An immersion heater rated 1000 W, 220 V is used to heat 0.01 m^3 of wa...

    Text Solution

    |

  8. An electric kettle used to prepare tea, takes 2 minutes to boil 4 cups...

    Text Solution

    |

  9. The coil of an electric bulb takes 40 watts to start glowing. If more ...

    Text Solution

    |

  10. The 2.0 Omega resistor show in figure is dipped into a calorimeter con...

    Text Solution

    |

  11. The temperatures of the junctions of a bismuth-silver thermocouple are...

    Text Solution

    |

  12. Find the thermo-emf developed in a copper-silver thermocopule when the...

    Text Solution

    |

  13. Find the neutral temperatuer and inversion temperature of copper-iron ...

    Text Solution

    |

  14. Find the charge required to flow through an electrolyte to liberate on...

    Text Solution

    |

  15. Find the amount of silver liberated at cathode if 0.500 A of current i...

    Text Solution

    |

  16. An electroplating unit plates 3.0 g of silver on a brass plate in 3.0 ...

    Text Solution

    |

  17. Find the time requried to liberate 1.0 litre of hydrogen at STP in an...

    Text Solution

    |

  18. Two voltameters, one having a solution of silver salt and the other of...

    Text Solution

    |

  19. A brass plate having surface area 200 cm^2 on one side is electroplate...

    Text Solution

    |

  20. Figure, shows an electrolyte of AgCI through which a current is passed...

    Text Solution

    |