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
2.5 A
2.5 A
0.625 A
0.625 A
14.4 A
14.4 A
0.4 A
0.4 A
V=IR
(1 point)
Responses
2.5 A
2.5 A
0.625 A
0.625 A
14.4 A
14.4 A
0.4 A
0.4 A
Answers
GPT-4o mini
To find the amperage (I) of the circuit, we first need to calculate the total voltage of the batteries. Since the flashlight has four 1.5-volt batteries, the total voltage (V) is:
\[
V = 4 \times 1.5 \, \text{volts} = 6 \, \text{volts}
\]
Now, using Ohm's Law, which states that \( V = I \times R \), we can rearrange this to solve for current (I):
\[
I = \frac{V}{R}
\]
Given the resistance (R) of the bulb is 2.4 ohms, we can substitute the values into the equation:
\[
I = \frac{6 \, \text{volts}}{2.4 \, \text{ohms}}
\]
Calculating this gives:
\[
I = 2.5 \, \text{A}
\]
Therefore, the amperage of the circuit is **2.5 A**.
\[
V = 4 \times 1.5 \, \text{volts} = 6 \, \text{volts}
\]
Now, using Ohm's Law, which states that \( V = I \times R \), we can rearrange this to solve for current (I):
\[
I = \frac{V}{R}
\]
Given the resistance (R) of the bulb is 2.4 ohms, we can substitute the values into the equation:
\[
I = \frac{6 \, \text{volts}}{2.4 \, \text{ohms}}
\]
Calculating this gives:
\[
I = 2.5 \, \text{A}
\]
Therefore, the amperage of the circuit is **2.5 A**.