Question

In the given figure, an ammeter $A$ consists of a $240 \, \Omega$ coil connected in parallel to a $10 \, \Omega$ shunt. The reading of the ammeter is ____ $\text{mA}$.

Answer 1

The circuit has a total resistance composed of a $140.4 \, \Omega$ resistor in series with the parallel combination of $240 \, \Omega$ and $10 \, \Omega$ resistors.
The equivalent resistance $R_{\text{eq}}$ of the parallel combination is:
$R_{\text{eq}} = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6 \, \Omega$
Thus, the total resistance in the circuit becomes:
$R_{\text{total}} = 140.4 + 9.6 = 150 \, \Omega$
Now, the current $I$ in the circuit is:
$I = \frac{V}{R_{\text{total}}} = \frac{24}{150} = 0.16 \, \text{A} = 160 \, \text{mA}$
Therefore, the current in the ammeter is $160 \, \text{mA}$.