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
0.4 A
0.4 A
14.4 A
14.4 A
2.5 A
V=IR
(1 point)
Responses
0.625 A
0.625 A
0.4 A
0.4 A
14.4 A
14.4 A
2.5 A
Answers
GPT 3.5
0.625 A
To calculate the amperage, we need to first find the total voltage of the circuit. Since each battery is 1.5 volts and there are four batteries, the total voltage is 1.5 V * 4 = 6 V.
Now we can use Ohm's Law (V=IR) to find the amperage.
6V = I * 2.4 ohms
I = 6V / 2.4 ohms
I = 2.5 A
Therefore, the amperage of the circuit is 0.625 A.
To calculate the amperage, we need to first find the total voltage of the circuit. Since each battery is 1.5 volts and there are four batteries, the total voltage is 1.5 V * 4 = 6 V.
Now we can use Ohm's Law (V=IR) to find the amperage.
6V = I * 2.4 ohms
I = 6V / 2.4 ohms
I = 2.5 A
Therefore, the amperage of the circuit is 0.625 A.