Question

The resistance of each brake lightbulb on an automobile is 6.6 Ω. Use the fact that cars have 12-V electrical systems to compute the current that flows in each bulb if they are connected in series.

Answer 1

In a series circuit, the total resistance is the sum of individual resistances. The current flowing in each brake lightbulb connected in series in a car's 12-V electrical system would be approximately 0.909 A.

Step-by-step explanation:

In a series circuit, the total resistance is equal to the sum of the individual resistances. Since the brake lightbulbs are connected in series, their resistances simply add up. In this case, each brake lightbulb has a resistance of 6.6 Ω, so the total resistance would be 2 times 6.6 Ω, which is 13.2 Ω.

Now we can use Ohm's law, which states that I = V/R, to find the current that flows in each bulb. Given that the electrical system in cars is 12 V, we can substitute these values into the formula to find I. Calculating I = 12 V / 13.2 Ω gives us a current of approximately 0.909 A for each bulb.

Answer 2

current is 0.91 A

Step-by-step explanation:

given data

resistance = 6.6 Ω

voltage = 12 V

to find out

current

solution

we consider here two brake light bulbs in automobile

we know here total resistance for automobile is 6.6 + 6.6 = 13.2 Ω

so current will be voltage / resistance

current = voltage / resistance ...................1

so current = 12 / 13.2

current = 0.91

so current is 0.91 A