An electrical circuit is powered by a 1.5-volt battery and contains a light bulb producing 5 Ohms of resistance. There is also another 5 Ohm resistor in the circuit. How many amps of current are in this circuit? HINT: V = IR FAST PLEASE!!

if the resistor is fitted in series with the bulb , then the current flowing will be 0.15 A

if the resistor if fitted in parallel with the bulb ,

the current flowing will be 0.6 A

Explanation:

total potential difference = 1.5V

when resistors in series ,

total equivalent resistance is = 5 + 5 = 10 ohm

so current = 1.5 ÷ 10 = 0.15

when resistors in parallel ,

total equivalent resistance is = (5 × 5)÷(5 + 5) =2.5 ohm

so current = 1.5÷2.5 = 0.6 A

c. newton's third law of motion

it has to be b, about 97 percent.