Measuring Devices ( Galvanometer, Voltmeter and Ammeter & Meter Bridge ): Practice Problem & Solution
A circuit contains an ammeter, a battery of $30 \text{ V}$ and a resistance $40.8 \Omega$ all connected in series. If the ammeter has a coil of resistance $480 \Omega$ and a shunt of $20 \Omega$, the reading in the ammeter will be: (2015 Re)
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:
Total resistance of ammeter $R_A = \frac{480 \times 20}{480 + 20} = 19.2 \Omega$. Total resistance of circuit $R_{eq} = 40.8 + 19.2 = 60 \Omega$. Reading of ammeter $I = \frac{V}{R_{eq}} = \frac{30}{60} = 0.5 \text{ A}$.
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