Home
Class 12
PHYSICS
How many lamps each of 50Wand 100Wcan be...

How many lamps each of `50W`and `100W`can be connected in parallel across a `120V` battery of internal resistance `10 Omega` so the each glows to full power ?

A

2

B

4

C

6

D

8

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of how many lamps of 50W and 100W can be connected in parallel across a 120V battery with an internal resistance of 10Ω so that each glows at full power, we can follow these steps: ### Step 1: Calculate the resistance of each lamp The resistance \( R \) of each lamp can be calculated using the formula: \[ R = \frac{V^2}{P} \] where \( V \) is the voltage across the lamp and \( P \) is the power rating of the lamp. For the 50W lamp: \[ R_{50W} = \frac{120^2}{50} = \frac{14400}{50} = 288 \, \Omega \] For the 100W lamp: \[ R_{100W} = \frac{120^2}{100} = \frac{14400}{100} = 144 \, \Omega \] ### Step 2: Calculate the current required by each lamp The current \( I \) through each lamp can be calculated using the formula: \[ I = \frac{P}{V} \] For the 50W lamp: \[ I_{50W} = \frac{50}{120} = \frac{5}{12} \, \text{A} \] For the 100W lamp: \[ I_{100W} = \frac{100}{120} = \frac{5}{6} \, \text{A} \] ### Step 3: Determine the total current required for \( N \) lamps If \( N \) lamps are connected in parallel, the total current required for the 50W lamps is: \[ I_{total, 50W} = N \times I_{50W} = N \times \frac{5}{12} \, \text{A} \] For the 100W lamps: \[ I_{total, 100W} = N \times I_{100W} = N \times \frac{5}{6} \, \text{A} \] ### Step 4: Calculate the total resistance of the circuit The total resistance \( R_{total} \) of the circuit when \( N \) lamps are connected in parallel is given by: \[ R_{total} = R_{internal} + R_{lamp} \] Where \( R_{internal} \) is the internal resistance of the battery (10Ω) and \( R_{lamp} \) is the equivalent resistance of \( N \) lamps in parallel. For \( N \) lamps of 50W: \[ R_{lamp, 50W} = \frac{R_{50W}}{N} = \frac{288}{N} \] Thus, \[ R_{total, 50W} = 10 + \frac{288}{N} \] For \( N \) lamps of 100W: \[ R_{lamp, 100W} = \frac{R_{100W}}{N} = \frac{144}{N} \] Thus, \[ R_{total, 100W} = 10 + \frac{144}{N} \] ### Step 5: Apply Ohm's law to find the maximum \( N \) Using Ohm's law, we can set up the equation for the total current from the battery: \[ I = \frac{V}{R_{total}} \] For the 50W lamps: \[ \frac{N \times \frac{5}{12}}{1} = \frac{120}{10 + \frac{288}{N}} \] Cross-multiplying gives: \[ N \times \frac{5}{12} \left(10 + \frac{288}{N}\right) = 120 \] Solving this equation will yield the maximum number of 50W lamps. For the 100W lamps: \[ \frac{N \times \frac{5}{6}}{1} = \frac{120}{10 + \frac{144}{N}} \] Cross-multiplying gives: \[ N \times \frac{5}{6} \left(10 + \frac{144}{N}\right) = 120 \] Solving this equation will yield the maximum number of 100W lamps. ### Final Calculation Solving the equations for both types of lamps will give us the maximum number of lamps that can be connected in parallel while glowing at full power.

To solve the problem of how many lamps of 50W and 100W can be connected in parallel across a 120V battery with an internal resistance of 10Ω so that each glows at full power, we can follow these steps: ### Step 1: Calculate the resistance of each lamp The resistance \( R \) of each lamp can be calculated using the formula: \[ R = \frac{V^2}{P} \] where \( V \) is the voltage across the lamp and \( P \) is the power rating of the lamp. ...
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    PRADEEP|Exercise Jee (main and advance)|4 Videos
  • CURRENT ELECTRICITY

    PRADEEP|Exercise Interger Type|2 Videos
  • CURRENT ELECTRICITY

    PRADEEP|Exercise Value based|2 Videos
  • COMMUNICATION SYSTEMS

    PRADEEP|Exercise MODEL TEST PAPER-2|9 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    PRADEEP|Exercise Exercise|191 Videos

Similar Questions

Explore conceptually related problems

Two heating coils of resistance 10 Omega and 20 Omega are connected in parallel and connected to a battery of emf 12V and internal resistance 1 Omega . The power consumed by them is in the ratio

two heading coils of resistances 10Omega and 20 Omega are connected in parallel and connected to a battery of emf 12 V and internal resistance 1Omega Thele power consumed by the n are in the ratio

Two bulbs 100 W, 250 V and 200 W, 250 V are connected in parallel across a 500 V line. Then-

A 10 V cell of neglible internal resitsance is connected in parallel across a battery of emf 200 V and internal resistance 38 Omega as shown in the figure. Find the value of current in the circuit.

An inductor coil of resistance 10 Omega and inductance 120 mH is connected across a battery of emf 6 V and internal resistance 2 Omega . Find the charge which flows through the inductor in (a) 10 ms, (b) 20ms and (c ) 100ms after the connections are made.

A series combination of two resistors 1 Omega each is connected to a 12 V battery of internal resistance 0.4 Omega . The current flowing through it will be

A 25 W , 220 V bulb and a 100 W , 220 V bulb are connected in parallel across a 440 V line

How will you connect ( series and parallel ) 24 cells each of internal resistance 1 Omega to get maximum power output across a load of 10 Omega ?

PRADEEP-CURRENT ELECTRICITY-Exercise
  1. If two bulbs of 25 W and 100 W rated at 220 V are connected in series ...

    Text Solution

    |

  2. In the circuit shown in fig the heat produced in the 5 ohm resistor du...

    Text Solution

    |

  3. How many lamps each of 50Wand 100Wcan be connected in parallel across...

    Text Solution

    |

  4. A 100 W bulb B(1) and two 60 W bulbs B(2) and B(3), are connected to a...

    Text Solution

    |

  5. Incandescent bulbs are designed by keeping in mind that the resistance...

    Text Solution

    |

  6. If power dissipated in the 9 Omega resistor in the resistor shown is 3...

    Text Solution

    |

  7. The power dissipated in the circuit shown in the is 30 watts. The valu...

    Text Solution

    |

  8. The supply voltage to room is 120 V. The resistance of the lead wires ...

    Text Solution

    |

  9. A d.c. voltage with appreciable ripple expressed as V = V(1) + V(2) co...

    Text Solution

    |

  10. A heater boils 1kg water in time t(1) and another heater bolis the sam...

    Text Solution

    |

  11. In the above question , if the heaters are connected in series , the c...

    Text Solution

    |

  12. A time wire with a circuit-sectional radius of 0.02mm blows with a cur...

    Text Solution

    |

  13. A resistance of 4 Omega is connected across a cell Then it is replace...

    Text Solution

    |

  14. The charge flowing in a conductor varies with times as Q = at - bt^2. ...

    Text Solution

    |

  15. Two heaters designed for the same voltage V have different power ratin...

    Text Solution

    |

  16. In the circuit shown in fig. some potential difference is applied betw...

    Text Solution

    |

  17. Two bulbs consume same energy when operated at 200 V and 300 V , respe...

    Text Solution

    |

  18. For the circuit shown in figure

    Text Solution

    |

  19. For the circuit

    Text Solution

    |

  20. Heater of an electric kettle is made of a wire of length L and diamete...

    Text Solution

    |