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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.

To determine the reading of the ammeter, we first need to calculate the equivalent resistance of the circuit. The ammeter has an internal resistance of \(240 \, \Omega\) with a shunt resistance of \(10 \, \Omega\) in parallel. The equivalent resistance \(R_{\text{eq}}\) of these two resistances is given by:

\( \frac{1}{R_{\text{eq}}} = \frac{1}{240} + \frac{1}{10} \)

\( \frac{1}{R_{\text{eq}}} = \frac{1}{240} + \frac{24}{240} = \frac{25}{240} \)

\( R_{\text{eq}} = \frac{240}{25} = 9.6 \, \Omega \)

The total resistance in the circuit is the sum of the series resistance \(150.4 \, \Omega\) and the equivalent resistance of \(9.6 \, \Omega\):

\( R_{\text{total}} = 150.4 + 9.6 = 160 \, \Omega \)

Using Ohm’s law, the total current \(I\) through the circuit can be calculated using the voltage \(V = 20 \, V\):

\( I = \frac{V}{R_{\text{total}}} = \frac{20}{160} = 0.125 \, \text{A} \, (or \, 125 \, \text{mA}) \)

The current through the ammeter is the current through the shunt resistance, given by:

\( I_{\text{shunt}} = \frac{10}{10 + 240} \times 125 = \frac{10}{250} \times 125 = 5 \, \text{mA} \)

The reading of the ammeter is \(5 \, \text{mA}\), which falls within the expected range of 5 to 5 mA.

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