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

`\displaystyle (1)/(4)`.

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`.