Home
Class 12
PHYSICS
Maximum current produced by a cell of em...

Maximum current produced by a cell of emf 10 V and internal resistance 2`Omega ` is

A

10A

B

2A

C

5A

D

0.16666666666667

Text Solution

Verified by Experts

The correct Answer is:
C
`I_(max)= ( E )/( r)`
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT(SECTION-A(OBJECTIVE TYPE QUESTIONS))|69 Videos

  • CURRENT ELECTRICITY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION-B(OBJECTIVE TYPE QUESTION ))|44 Videos

  • CURRENT ELECTRICITY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION-J|10 Videos

  • COMMUNICATION SYSTEMS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION D (Assertion-Reason)|10 Videos

  • DUAL NATURE OF RADIATION AND MATTER

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (SECTION-D)|10 Videos

Similar Questions

Explore conceptually related problems

A resistance of 4 Omega and a wire of length 5 meters and resistance 5 Omega are joined in series and connected to a cell of e.m.f. 10 V and internal resistance 1 Omega . A parallel combination of two identical cells is balanced across 300 cm of wire. The e.m.f. E of each cell is

Find the current drawn from a cell of emf 2 V and internal resistance 2 Omega connected to the network given below.

In a potentiometer circuit, the emf of driver cell is 2V and internal resistance is 0.5 Omega . The potentiometer wire is 1m long . It is found that a cell of emf 1V and internal resistance 0.5 Omega is balanced against 60 cm length of the wire. Calculate the resistance of potentiometer wire.

A voltmeter of resistance 998 Omega is connected across a cell of emf 2 V and internal resistance 2 Omega . Find the p.d. across the voltmeter, that across the terminals of the cell and percentage error in the reading of the voltmeter.

Two resistors of resistances 2Omega and 6Omega are connected in parallel. This combination is then connected to a battery of emf 2 V and internal resistance 0.5 Omega . What is the current flowing through the battery ?

There are ten identical cells each of emf 10 V and internal resistance 0.25 Omega are short circuited as shown in figure. The value of potential difference (V_x -V_y) is

An accumulator of emf 2 V and negligible internal resistance is connected across a uniform wire of length 10 m and resistance 30 Omega The appropriate terminals of a cell of emf 1.5 V and internal resistance 1 Omega is connected to one end of the wire and the other terminal of the cell is connected through a sensitive galvanometer to a slider on the wire. What is the length of the wire that will be required to produce zero deflection of the galvanometer? How will the balancing length change? (a) When a coil of resistance 5Omega is placed in series with the accumulator. (b) The cell of 1.5 V is shunted with 5 Omega resistor?

The wire of potentiometer has resistance 4 Omega and length 1m. It is connected to a cell of e.m.f. 2V and internal resistance 1 Omega . The current flowing through the potentiometer wire is

Two cells e_(1) and e_(2) connected in opposition to each other as shown in figure. The cell of emf 9 V and internal resistance 3Omega the cell is of emf 7V and internal resistance 7Omega . The potential difference between the points A and B is

The wire of potentiometer has resistance 4Omega and length 1m . It is connected to a cell of emf 2 volt and internal resistance 1Omega . If a cell of emf 1.2 volt is balanced by it, the balancing length will be

AAKASH INSTITUTE ENGLISH-CURRENT ELECTRICITY-Try Yourself
  1. Current l in the network shwon in figure is

    Text Solution

    |

  2. Potential difference across AB in the network shown in figure.

    Text Solution

    |

  3. Maximum current produced by a cell of emf 10 V and internal resistan...

    Text Solution

    |

  4. When the positive and negative electrodes of a cell are denoted as P ...

    Text Solution

    |

  5. About internal resistance of a cell , the correct statements that it i...

    Text Solution

    |

  6. Find the current through the 10 Omega resistance connected across AB

    Text Solution

    |

  7. Four identical cells each having emf E and internal resistance r are c...

    Text Solution

    |

  8. The current l through the cell in the network shown is

    Text Solution

    |

  9. In the network shown below , potential difference across CD is

    Text Solution

    |

  10. The current l in the circuit shown below is

    Text Solution

    |

  11. Twelve identical resistance R each form a cube as shown in figure. Re...

    Text Solution

    |

  12. In the network shown , find the resistance across AB.

    Text Solution

    |

  13. In the metre bridge, the balancing length AB =l=20 cm. The unknown res...

    Text Solution

    |

  14. An unknown resistance R(1) is connected is series with a resistance of...

    Text Solution

    |

  15. Two cells of emf E(1) and E(2) are to be compared in a potentiometer...

    Text Solution

    |

  16. A simple potentiometer circuit is shown in figure. The internal resi...

    Text Solution

    |

  17. Consider the circuit shown in Figure . Ther terminal voltage of the 2...

    Text Solution

    |

  18. N sources of current with different emf's are connected as shown in Fi...

    Text Solution

    |

  19. Voltmeter V and Ammeter A are used to measure the total potential dif...

    Text Solution

    |

  20. Could the equation for R(3) lead to a negative valuefor this resistanc...

    Text Solution

    |