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. Numerical answer type. Work it out, then open the answer.

Answer: 5. Ammeter and shunt in parallel: (240 × 10)/(250) = 9.6 Ω Total resistance = 150.4 + 9.6 = 160 Ω, so I = 20/160 = 125 mA Current through the ammeter coil: IA = 125 × 10/250 = 5 mA Answer: 5 mA

In the figure shown below, a resistance of 150.4 Ω is connected in… | JEE Main 2025 (3 Apr, Shift 1)

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Kirchhoff's Laws & Measuring Instruments

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

Numerical answer type. Work it out, then open the answer.

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Answer: 5

Ammeter and shunt in parallel: $\frac{240 \times 10}{250} = 9.6\ \Omega$
Total resistance $= 150.4 + 9.6 = 160\ \Omega$ , so $I = \frac{20}{160} = 125\text{ mA}$
Current through the ammeter coil: $I_A = 125 \times \frac{10}{250} = 5\text{ mA}$
Answer: 5 mA

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