Question

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)
Responses

0.625 A
0.625 A

2.5 A
2.5 A

0.4 A
0.4 A

14.4 A

Answer 1

To find the amperage of the circuit, we first need to calculate the total voltage provided by the batteries.

Given that there are four 1.5-volt batteries, the total voltage (V) is:
\[
V = 4 \times 1.5 \, \text{volts} = 6 \, \text{volts}
\]

Now, we can use Ohm's Law, which states that:
\[
V = I \times R
\]
where:
- \(V\) is the voltage (in volts),
- \(I\) is the current (in amperes),
- \(R\) is the resistance (in ohms).

Rearranging the formula to solve for current \(I\):
\[
I = \frac{V}{R}
\]

Substituting the values we have:
\[
I = \frac{6 \, \text{volts}}{2.4 \, \text{ohms}} = 2.5 \, \text{A}
\]

Thus, the amperage of the circuit is:
\[
\boxed{2.5 \, A}
\]