a flashlight has four 1.5-volt batteries the bulb has a resistance of 2.4 ohms. what is the amerage of the curcuit
2.5 a need this quickly
The current in the circuit is calculated using Ohm's Law: I = V/R, where I is the current, V is the voltage, and R is the resistance. In this case, V = 6 volts (4 batteries x 1.5 volts each) and R = 2.4 ohms. Therefore, the current in the circuit is I = 6 volts / 2.4 ohms = 2.5 amps.
To calculate the current (amperage) in the circuit, we can use Ohm's Law, which states:
Current (I) = Voltage (V) / Resistance (R)
In this case, the voltage is the total voltage provided by the batteries (4 * 1.5V = 6V), and the resistance is given as 2.4 ohms. Therefore, the formula becomes:
I = V / R
I = 6V / 2.4 ohms
Now, we can calculate the current:
I = 6V / 2.4 ohms
I = 2.5 Amperes
Therefore, the amperage of the circuit is 2.5 Amperes.
To find the amperage of the circuit, we can use Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the resistance (R).
First, calculate the total voltage of the flashlight by multiplying the voltage of one battery by the number of batteries:
Total voltage = 1.5 V/battery * 4 batteries = 6 V
Next, use Ohm's Law:
I = V/R
I = 6 V / 2.4 Ω
I ≈ 2.5 A
Therefore, the amperage of the circuit is approximately 2.5 A.