Home
Class 12
PHYSICS
A milli voltmeter of 25 milli volt range...

A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will be

A

0.05 Ω

B

0.001 Ω

C

0.01 Ω

D

1 Ω

Text Solution

AI Generated Solution

The correct Answer is:
To convert a millivolt meter with a range of 25 millivolts into an ammeter with a range of 25 amperes, we need to calculate the value of the necessary shunt resistor (S) that should be connected in parallel with the voltmeter. ### Step-by-Step Solution: 1. **Identify the Given Values:** - Maximum voltage of the voltmeter (Vg) = 25 mV = 25 × 10^(-3) V - Maximum current of the ammeter (Im) = 25 A - The resistance of the voltmeter (R) is unknown. 2. **Calculate the Current through the Voltmeter (Ig):** - The current through the voltmeter when it reads its maximum voltage can be calculated using Ohm's Law: \[ I_g = \frac{V_g}{R} \] - Here, \(V_g = 25 \times 10^{-3} \, \text{V}\). 3. **Determine the Total Current (Im) and the Current through the Shunt (Is):** - The total current flowing through the circuit when the ammeter is at maximum is \(I_m = 25 \, \text{A}\). - The current through the shunt resistor (Is) can be expressed as: \[ I_s = I_m - I_g \] 4. **Apply Kirchhoff's Voltage Law:** - The voltage across the voltmeter (Vg) is equal to the voltage across the shunt resistor (Vs): \[ V_g = I_g \cdot R = I_s \cdot S \] - This leads to: \[ I_g \cdot R = (I_m - I_g) \cdot S \] 5. **Rearranging the Equation to Solve for S:** - From the above equation, we can rearrange to find S: \[ S = \frac{I_g \cdot R}{I_m - I_g} \] 6. **Substituting the Values:** - We know \(I_g = \frac{25 \times 10^{-3}}{R}\). - Substitute \(I_g\) into the equation for S: \[ S = \frac{\left(\frac{25 \times 10^{-3}}{R}\right) \cdot R}{25 - \frac{25 \times 10^{-3}}{R}} \] - This simplifies to: \[ S = \frac{25 \times 10^{-3}}{25 - 25 \times 10^{-3}/R} \] 7. **Assuming R is Large:** - Since R is much larger than the shunt resistance S, we can approximate: \[ S \approx \frac{25 \times 10^{-3}}{25} = 10^{-3} \, \text{ohm} \] 8. **Final Calculation:** - Thus, the value of the necessary shunt resistor (S) is: \[ S = 0.001 \, \text{ohm} \] ### Conclusion: The value of the necessary shunt resistor is **0.001 ohm**.

To convert a millivolt meter with a range of 25 millivolts into an ammeter with a range of 25 amperes, we need to calculate the value of the necessary shunt resistor (S) that should be connected in parallel with the voltmeter. ### Step-by-Step Solution: 1. **Identify the Given Values:** - Maximum voltage of the voltmeter (Vg) = 25 mV = 25 × 10^(-3) V - Maximum current of the ammeter (Im) = 25 A - The resistance of the voltmeter (R) is unknown. ...
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    A2Z|Exercise Section - C|1 Videos
  • CURRENT ELECTRICITY

    A2Z|Exercise AIIMS Questions|36 Videos
  • CURRENT ELECTRICITY

    A2Z|Exercise Section B - Assertion Reasoning|28 Videos
  • ATOMIC PHYSICS

    A2Z|Exercise Section D - Chapter End Test|30 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    A2Z|Exercise Section D - Chapter End Test|30 Videos

Similar Questions

Explore conceptually related problems

A millivoltmeter of 25 m V range is to be converted into an ammeter of 25 A range . The value (in ohm ) of necessary shunt will be

A voltmeter of 250 mV range having a resistance of 10 Omega is converted into an ammeter of 250 mA range. The value of necessary shunt is (nearly)

An ammeter of low range can be converted into an ammeter to higher range by connecting

A voltmeter of range 3V and resistance 200Omega cannot be converted to an ammeter of range

A voltmeter of range 3 V and resistance 200Omega connot be converted to an ammeter of range

A voltmeter of range 2 V and resistance 300 Omega cannot be converted into ammeter of range

A2Z-CURRENT ELECTRICITY-AIPMTNEET Questions
  1. A ring is made of a wire having a resistance R(0) = 12 Omega. Find the...

    Text Solution

    |

  2. 2A cell having an emf epsilon and internal resistance r is connected a...

    Text Solution

    |

  3. A milli voltmeter of 25 milli volt range is to be converted into an am...

    Text Solution

    |

  4. In the circuit shown the cells A and B have negligible resistance. For...

    Text Solution

    |

  5. If voltage across a bulb rated 220 volt-100 watt drops by 2.5 % of its...

    Text Solution

    |

  6. The power dissipated in the circuit shown in the figure is 30 watt. Th...

    Text Solution

    |

  7. A wire of resistance 4 Omega is stretched to twice its original length...

    Text Solution

    |

  8. The internal resistance of a 2.1 V cell which gives a current 0.2 A th...

    Text Solution

    |

  9. The resistance of the four arms P, Q, R and S in a Wheatstone's bridg...

    Text Solution

    |

  10. Two cities are 150 km apart. Electric power is sent from one city to a...

    Text Solution

    |

  11. The resistance in the two arms of the meter bridge are 5 Omega and R O...

    Text Solution

    |

  12. A potentiometer circuit has been setup for finding. The internal resis...

    Text Solution

    |

  13. In a ammeter 0.2% of main current passes through the galvanometer. If ...

    Text Solution

    |

  14. Across a metallic conductor of non-uniform cross-section a constant po...

    Text Solution

    |

  15. Two metal wires of identical dimesnios are connected in series. If sig...

    Text Solution

    |

  16. A potentiometer wire has length 4 m and resistance 8 Omega. The resist...

    Text Solution

    |

  17. A, B and C are voltmeters of resistances R, 1.5R and 3R respectively. ...

    Text Solution

    |

  18. A potentiometer wire of Length L and a resistance r are connected in s...

    Text Solution

    |

  19. A circuit contains an ammeter, a battery of 30 V and a resistance 40.8...

    Text Solution

    |

  20. The charge flowing through a resistance R varies with time t as Q = at...

    Text Solution

    |