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

14.4 A
14.4 A

2.5 A
2.5 A

0.4 A

Answer 1

To find the amperage of the circuit, we can use Ohm's Law, which is given by the formula:

\[ V = I \cdot R \]

Where:
- \( V \) is the voltage,
- \( I \) is the current (amperage),
- \( R \) is the resistance.

Since the flashlight has four 1.5-volt batteries, we can find the total voltage:

\[ V = 4 \times 1.5 \, \text{volts} = 6 \, \text{volts} \]

Now, we can rearrange Ohm's Law to solve for \( I \) (current):

\[ I = \frac{V}{R} \]

Substituting in the values we have:

\[ I = \frac{6 \, \text{volts}}{2.4 \, \text{ohms}} \]

Calculating this gives:

\[ I = \frac{6}{2.4} = 2.5 \, \text{A} \]

So the amperage of the circuit is **2.5 A**.

The correct response is:

**2.5 A**