Resistance of a light bulb
A 100 W light bulb is placed in series with a 40 W light bulb. When they are connected to a 120-volt wall socket, which bulb will be brighter? It turns out to be the 40 W bulb.
Use the power equation to calculate the resistance of the two bulbs.
The bulbs only produce the power stamped on them when the potential difference is 120 V.
| For the 100-watt bulb: R |
= |
| DV2
|  |
| P
|
|
= |
| 1202
| |
| 100
|
|
= |
144 W
|
| For the 40-watt bulb: R |
= |
| DV2
| |
| P
|
|
= |
| 1202
| |
| 40
|
|
= |
360 W
|
We can use Ohm's Law to find the current each resistor usually gets:
| For the 100-watt bulb: I |
= |
| DV
| |
| R
|
|
= |
| 120
| |
| 144
|
|
= |
0.83 A
|
| For the 40-watt bulb: I |
= |
| DV
| |
| R
|
|
= |
| 120
| |
| 360
|
|
= |
0.33 A
|
When they are placed in series the total resistance is about 500 W, so each bulb sees a current of 120/500 = 0.24 A.
Use a power equation to find the power for each bulb when they're in series with one another. The more power, the brighter the bulb.
For the 100-watt bulb: P = I2 R = 0.242 * 144 = 8 W
For the 40-watt bulb: P = I2 R = 0.242 * 360 = 21 W
This is actually an underestimate, but it does give us the relative brightness. The bulbs aren't as hot as usual so they're resistances are each a little lower than we have assumed above.