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

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​ R eq ​ = R 1 ​ + R 2 ​ R eq ​ = 150.4 + 250 240 × 10 ​ = 150.4 + 9.6 = 160Ω I 1 ​ = 240 I R 2 ​ ​ I 1 ​ = 240 I × 9.6 ​ = 160 20 ​ × 2400 9.6 ​ = 200 1 ​ = 5 × 1 0 − 3 A = 5 mA ​

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