JEE Main 2025 Apr 3 Shift 1, Physics Q25: Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

JEE Main2025Apr 3, Shift 1Physics

In the figure shown below, a resistance of $150.4 Ω$ is connected in series to an ammeter A of resistance $240 Ω$ . A shunt resistance of $10 Ω$ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.

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Solution

The ammeter (240 Ω) is in parallel with the 10 Ω shunt. Their combined resistance is

$R_{p} = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6 Ω.$

Total circuit resistance: $R_{eq} = 150.4 + 9.6 = 160 Ω.$

Main current from the battery: $I = \frac{V}{R _{eq}} = \frac{20}{160} = 0.125 A .$

This main current $I$ then divides at the ammeter-shunt node. The current through the ammeter branch is

$I_{A} = I \cdot \frac{R _{shunt}}{R _{A} + R _{shunt}} = 0.125 \cdot \frac{10}{250} = 5 \times 1 0^{- 3} A = 5 mA .$

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JEE Main 2025 Apr 3 Shift 1, Physics Q25: Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

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