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

0.4 A
0.4 A

0.625 A
0.625 A

2.5 A
2.5 A

14.4 A

Answer 1

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**.