Home
Class 12
PHYSICS
A 50V battery is connected across a 10 o...

A `50V` battery is connected across a 10 ohm resistor. The current is `4.5` amperes. The internal resistance of the battery is

A

Zero

B

0.5 ohm

C

1.1 ohm

D

5.0 ohm

Text Solution

Verified by Experts

The correct Answer is:
C
`I = E/(R + r) rArr 4.5 = 50/(10 + r) rArr 45 + 4.5 r = 50`
`rArr r = 5/4.5 = 1.1 Omega`
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    ERRORLESS|Exercise NCERT BASED QUESTIONS (DIFFERENT MEASURING INSTRUMENTS) |74 Videos

  • CURRENT ELECTRICITY

    ERRORLESS|Exercise NCERT BASED QUESTIONS (HEATING EFFECT OF CURRENT)|67 Videos

  • CURRENT ELECTRICITY

    ERRORLESS|Exercise NCERT BASED QUESTIONS (Grouping of Resistances)|69 Videos

  • ATOMIC AND NUCLEAR PHYSICS

    ERRORLESS|Exercise Assertion and Reason|23 Videos

  • ELECTROMAGNETIC INDUCTION

    ERRORLESS|Exercise ASSERTION & REASON|20 Videos

Similar Questions

Explore conceptually related problems

When a battery connected across a resistor of 16 ohm , the voltage across the resistor is 12V. When the same battery is connected across a resistor of 10 ohm , voltage across it is 11V. The internal resistance of the battery in ohm is

When a battery connected across a resistor of 16Omega the voltage across the resistor is 12 V. When the same battery is connected across a resistor of 10Omega voltage. Across it is 11 V. The internal resistance of the battery in ohms is

When a battery connected across a resistor of 16 Omega , the voltage across the resistor is 12 V. When the same battery is connected across a resistor of 10 Omega , voltage across it is 11 V. The internal resistance of the battery (in ohm) is

A current of 2 A flows through a 2 Omega resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Omega resistor. The internal resistance of the battery is

A galvanometer, together with an unknown resistance in series, is connected across two identical batteries of each 1.5 V . When the batteries are connected in series,the galvanometer records a current of 1 A , and when the batteries are connected in parallel, the current is 0.6 A . In this case, the internal resistance of the battery is 1//'**' Omega . What is the value of '**'?

A current of 2A flows through a 2 Omega resistor when connected across a battery. The same battery supplies a current of 0.5A when connected across a 9Omega resistor The internal resistance of the battery is

ERRORLESS-CURRENT ELECTRICITY-NCERT BASED QUESTIONS (KIRCHHOFF.S LAW, CELLS)
  1. A group of N cells where e.m.f. varies directly with the internal resi...

    Text Solution

    |

  2. A primary cell has an e.m.f. of 1.5 volts, when short-circuited it giv...

    Text Solution

    |

  3. A 50V battery is connected across a 10 ohm resistor. The current is 4....

    Text Solution

    |

  4. n identical cells each of e.m.f. E and internal resistance r are conne...

    Text Solution

    |

  5. A cell of internal resistance "r" is connected to an external resistan...

    Text Solution

    |

  6. Two batteries of emf E1 and E2 (E2 gt E1) and internal resistances r1 ...

    Text Solution

    |

  7. Two identical calls send the same current in 2 Omega resistance, wheth...

    Text Solution

    |

  8. The internal resistances of two cells shown are 0.1Omega and 0.3 Omega...

    Text Solution

    |

  9. In the circuit shown, the current through the 5 Omega resistor is-

    Text Solution

    |

  10. Two cells having e.m.f. 4 V, 2V and internal resistances 1 Omega, 1 Om...

    Text Solution

    |

  11. A battery having e.m.f. 5 V and internal resistance 0.5 Omega is conne...

    Text Solution

    |

  12. Two batteries A and B each of e.m.f. 2 V are connected in series to a...

    Text Solution

    |

  13. When a resistor of 11 Omega is connected in series with an electric ce...

    Text Solution

    |

  14. A capacitor is conneted to a cell emf E having some internal resistanc...

    Text Solution

    |

  15. Two batteries, one of emf 18 volts and internal resistance 2 Omega and...

    Text Solution

    |

  16. 5 cells, each of emf 0.2V and internal resistance 1 Omega are connecte...

    Text Solution

    |

  17. In the circuit shown here, E(1) = E(2) = E(3) = 2 V and R(1) = R(2) = ...

    Text Solution

    |

  18. Two batteries V(1) and V(2) are connected to three resistors as shown ...

    Text Solution

    |

  19. Two cells having an internal resistance of 0.2 Omega and 0.4 Omega are...

    Text Solution

    |

  20. The n rows each containing m cells in series are joined parallel. Maxi...

    Text Solution

    |