The current in two identical light bulbs connected in series is 0.25 A. The voltage across both bulbs is 110 V. The resistance of a single light bulb is
A)22 ohms.
B)44 ohms.
C)220 ohms.
D)440 ohms.
E)none of the above

Question

Quizplus · Accepted Answer

Since the two identical light bulbs are connected in series, the current through both of them is the same. Therefore, the current through each bulb is 0.25 A/2 = 0.125 A.
The voltage across each bulb can be found using Ohm's Law: V = IR, where V is the voltage, I is the current, and R is the resistance. Therefore, the voltage across each bulb is 110 V/2 = 55 V.
Using Ohm's law again, we can find the resistance of each bulb: R = V/I = 55 V/0.125 A = 440 ohms.
Since the two bulbs are identical, the resistance of a single light bulb is 440 ohms/2 = 220 ohms.

The Current in Two Identical Light Bulbs Connected in Series