Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The resistance of an ammeter is 13 \Omega and its scale is graduated for a current up to 100 A. After an additional shunt has been connected to this ammeter, it becomes possible to measure currents up to 750 A by this ammeter. The value of shunt resistance is:

20 \Omega

2 \Omega

0.2 \Omega

2 k\Omega

Step-by-Step Solution

Concept: An ammeter (or galvanometer) with resistance $R_G$ and full-scale deflection current $I_g$ can have its range extended to measure a larger current $I$ by connecting a low resistance shunt ( $S$ ) in parallel with it .
Formula: The potential drop across the galvanometer and the shunt is the same in a parallel combination. Therefore: $I_g R_G = (I - I_g) S$ Rearranging for the shunt resistance $S$ : $S = \frac{I_g R_G}{I - I_g}$
Given Values:

Resistance of the initial meter ( $R_G$ ) = 13 \Omega
Initial full-scale current ( $I_g$ ) = 100 A
New full-scale current ( $I$ ) = 750 A

Calculation: $S = \frac{100 \times 13}{750 - 100}$ $S = \frac{1300}{650}$ $S = 2 \, \Omega$

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