Rankers Physics

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

In an ammeter $0.2%$ of main current passes through the galvanometer. If resistance of galvanometer is $G$, the resistance of ammeter will be (2014)
$\frac{1}{499} G$
$\frac{499}{500} G$
$\frac{1}{500} G$
$\frac{500}{499} G$

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 $S = \frac{G}{499}$. The equivalent resistance of the ammeter is $R_A = \frac{GS}{G+S} = \frac{G(G/499)}{G + G/499} = \frac{1}{500} G$.

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