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
0.625 A
14.4 A
0.4 A
0.625 A
To calculate the amperage of the circuit, we can use Ohm's Law, which states that V (voltage) is equal to I (current) multiplied by R (resistance).
Given that the voltage of each battery is 1.5 volts and there are four batteries in the flashlight, the total voltage can be calculated as follows:
V_total = 1.5 V/battery * 4 batteries = 6 V
Now, we can apply Ohm's Law to find the current (I):
V = IR
6 V = I * 2.4 ohms
Solving for I:
I = 6 V / 2.4 ohms = 2.5 A
Therefore, the amperage of the circuit is 2.5 A.
To find the amperage of the circuit in this scenario, we can use Ohm's Law:
Ohm's Law states that current (I) is equal to the voltage (V) divided by the resistance (R):
I = V / R
First, let's calculate the total voltage of the circuit. Since the flashlight has four 1.5-volt batteries connected in series, the total voltage is:
V = 1.5 V * 4 = 6 V
Next, we can substitute the known values into Ohm's Law:
I = 6 V / 2.4 Ω
Calculating this expression gives us:
I = 2.5 A
Therefore, the amperage of the circuit is 2.5 A.