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

14.4 A
14.4 A

0.4 A
0.4 A

2.5 A
2.5 A

0.625 A
0.625 A

1 answer

To find the amperage (I) of the circuit, we first need to determine the total voltage supplied by the batteries. Since there are four 1.5-volt batteries, the total voltage (V) is calculated as follows:

\[ V = 4 \times 1.5 , \text{volts} = 6 , \text{volts} \]

Next, we can use Ohm's law, which is given by the formula:

\[ V = I \times R \]

We can rearrange this formula to solve for the current (I):

\[ I = \frac{V}{R} \]

Substituting the known values into the equation:

  • Voltage (V) = 6 volts
  • Resistance (R) = 2.4 ohms

Now we can calculate the current:

\[ I = \frac{6 , \text{volts}}{2.4 , \text{ohms}} = 2.5 , \text{A} \]

So, the amperage of the circuit is 2.5 A.