Question

A)
A flashlight has four 1.5-volt batteries. The bulb has a resistance of 2.4 ohms. What is the
amperage of the circuit?
V = IR
(1 point)
0.4 A
14.4 A
0.625 A
2.5 A

Answer 1

To find the amperage of the circuit, we first need to determine the total voltage supplied by the batteries and then apply Ohm's Law, which states \( V = IR \).

1. **Total voltage**: The flashlight has four 1.5-volt batteries. Therefore, the total voltage \( V \) is:
\[
V = 4 \times 1.5 \text{ volts} = 6 \text{ volts}
\]

2. **Resistance**: The resistance of the bulb is given as \( R = 2.4 \) ohms.

3. **Applying Ohm's Law**: Rearranging Ohm's Law to find current (amperage) \( I \):
\[
I = \frac{V}{R}
\]
Substituting the values we have:
\[
I = \frac{6 \text{ volts}}{2.4 \text{ ohms}} = 2.5 \text{ A}
\]

Thus, the amperage of the circuit is **2.5 A**. The correct answer is **2.5 A**.