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
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
Answers
GPT-4o mini
To find the amperage (I) of the circuit, we can use Ohm's Law, which is given by the formula:
\[ V = I \times R \]
Where:
- \( V \) is the voltage,
- \( I \) is the current in amperes,
- \( R \) is the resistance in ohms.
First, we need to calculate the total voltage of the flashlight. Since there are four 1.5-volt batteries connected in series, the total voltage (\( V \)) is:
\[ V = 4 \times 1.5 \, \text{V} = 6 \, \text{V} \]
Now we can rearrange the formula to solve for current \( I \):
\[ I = \frac{V}{R} \]
Substituting the values we have:
\[ I = \frac{6 \, \text{V}}{2.4 \, \Omega} \]
Calculating the amperage:
\[ I = \frac{6}{2.4} = 2.5 \, \text{A} \]
Thus, the amperage of the circuit is **2.5 A**.
\[ V = I \times R \]
Where:
- \( V \) is the voltage,
- \( I \) is the current in amperes,
- \( R \) is the resistance in ohms.
First, we need to calculate the total voltage of the flashlight. Since there are four 1.5-volt batteries connected in series, the total voltage (\( V \)) is:
\[ V = 4 \times 1.5 \, \text{V} = 6 \, \text{V} \]
Now we can rearrange the formula to solve for current \( I \):
\[ I = \frac{V}{R} \]
Substituting the values we have:
\[ I = \frac{6 \, \text{V}}{2.4 \, \Omega} \]
Calculating the amperage:
\[ I = \frac{6}{2.4} = 2.5 \, \text{A} \]
Thus, the amperage of the circuit is **2.5 A**.