In the figure shown below, a resistance of 150.4 is connected in series to an ammeter A of resistance 240. A shunt resistance of 10 is connected in parallel with the ammeter. The reading of the ammeter is ______ mA. [2025]

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Solution

Q.
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. [2025]

Ans.
(5)

$R_{eq} = R_{1} + R_{2}$

$R_{eq} = 150.4 + \frac{240 \times 10}{250}$

$= 150.4 + 9.6 = 160 Ω$

$I_{1} = \frac{I R_{2}}{240} = \frac{I \times 9.6}{240}$

$= \frac{20}{160} \times \frac{96}{2400} = \frac{1}{200} A$

$= 5 \times 10^{- 3} A = 5 mA$

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