Rankers Physics

Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ): Practice Problem & Solution

The resistance of an ammeter is $13 \Omega$ and its scale is graduated for a current upto $100 \text{ amps}$. After an additional shunt has been connected to this ammeter, it becomes possible to measure currents upto $750 \text{ amperes}$ by this meter. The value of shunt resistance is : (2007)
$2 \Omega$
$0.2 \Omega$
$2k \Omega$
$20 \Omega$

Solution Explained:

To solve this problem, we apply the core principles of Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ). Understanding the underlying formula is key to arriving at the correct answer below:

The shunt resistance is calculated as $S = \frac{I_g G}{I - I_g} = \frac{100 \times 13}{750 - 100} = \frac{1300}{650} = 2 \Omega$.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): Electric appliances with a metallic body have three pin connections whereas an electric bulb has two pin connections. Reason (R): Three pin connection reduces heating of connecting cable.  
2. A galvanometer of $50\text{ }\Omega$ resistance has $25$ divisions. A current of $4 \times 10^{-4}\text{ A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25\text{ V}$, it should be connected with a resistance of: (2004)
3. If voltage across a bulb rated $220\text{ Volt} - 100\text{ Watt}$ drops by $2.5\%$ of its rated value, the percentage of the rated value by which the power would decrease is: (2012 Pre)
4. Consider the following statements and choose the correct option. Statement (A): Kirchhoff's junction law follows the conservation of charge. Statement (B): Kirchhoff's loop law follows the conservation of energy.  
5. Resistance of a galvanometer coil is $8\text{ }\Omega$ and $2\text{ }\Omega$ shunt resistance is connected with it. If main current is $1\text{ A}$ then the current flowing through $2\text{ }\Omega$ resistance will be: (1998)