To find the amperage (current) of the circuit, we can use Ohm's Law, which states:
\[ V = I \cdot R \]
Where:
- \( V \) is the voltage,
- \( I \) is the current (amperage),
- \( R \) is the resistance.
First, let's calculate the total voltage from the batteries. Since the flashlight has four 1.5-volt batteries in series, the total voltage \( V \) is:
\[ V = 4 \times 1.5 , \text{volts} = 6 , \text{volts} \]
Now, we can rearrange Ohm's Law to solve for current \( I \):
\[ I = \frac{V}{R} \]
Substituting the values we have:
\[ I = \frac{6 , \text{volts}}{2.4 , \text{ohms}} = 2.5 , \text{A} \]
So, the amperage of the circuit is 2.5 A.
The correct response is:
2.5 A