To find the amperage (current) of the circuit, we first need to calculate the total voltage supplied by the batteries. Since there are four 1.5-volt batteries, the total voltage (V) is:
\[ V = 4 \times 1.5 , \text{V} = 6 , \text{V} \]
Now, we can use Ohm's law, which states:
\[ V = I \times R \]
Where:
- \( V \) is the voltage (in volts),
- \( I \) is the current (in amperes),
- \( R \) is the resistance (in ohms).
We can rearrange this equation to solve for current (I):
\[ I = \frac{V}{R} \]
Substituting the values we have:
\[ I = \frac{6 , \text{V}}{2.4 , \Omega} \]
Calculating the current:
\[ I = 2.5 , \text{A} \]
So the amperage of the circuit is 2.5 A.