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

0.4 A
0.4 A

0.625 A
0.625 A

2.5 A
2.5 A

14.4 A

1 answer

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

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

Now, we can use Ohm's law, which states:

\[ V = I \times R \]

Where:

  • \( V \) is the voltage (in volts),
  • \( I \) is the current (in amperes),
  • \( R \) is the resistance (in ohms).

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

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

Substituting the values we have:

\[ I = \frac{6 , \text{V}}{2.4 , \Omega} \]

Calculating the current:

\[ I = 2.5 , \text{A} \]

So the amperage of the circuit is 2.5 A.