Question

A 12-V battery powers three lightbulbs in a series circuit. The voltage reading is 6 V across the first lightbulb and 3 V across the second lightbulb. What portion of the battery's energy transforms in the third lightbulb?

Please show how you got the answer!

Answer 1

Answer:

$\displaystyle \frac{1}{4}$.

Step-by-step explanation:

The voltage across the three bulbs would add up to $12\; {\rm V}$ since the three bulbs are connected in serial.

The voltage across the third bulb would thus be: $12\; {\rm V} - 6\; {\rm V} - 3\; {\rm V} = 3\; {\rm V}$.

Also because the three bulbs are connected in serial, the current going through each of them be the same as the size of the current of the whole circuit. Let $I\; {\rm A}$ be the size of that current.

• The power supplied by the battery would be $P = V \, I = 12\; {\rm V} \, (I\; {\rm A}) = (12\, I)\; {\rm W}$.
• The power of the third bulb would be $P = V\, I = 3\; {\rm V} \, (I\; {\rm A}) = (3\, I)\; {\rm W}$.

Thus, the portion of the power consumed by the third bulb would be $(3\, I) / (12\, I) = 1/4$.