Final answer:
Adding a third bulb to a parallel circuit will decrease the overall resistance. In a parallel circuit, more paths for current are created, reducing total resistance according to Ohm's law.
Step-by-step explanation:
When a third bulb is added in parallel to a two-bulb parallel circuit, the total resistance of the circuit will decrease. Adding more paths for current to flow effectively reduces the overall resistance.
The principle that governs this is Ohm's law, which states that the total resistance in a parallel circuit is found by taking the reciprocal of the sum of the reciprocals of each individual resistance (1/Rtotal = 1/R1 + 1/R2 + ... + 1/Rn).
When you increase the number of resistors (or bulbs), you're adding another term to this sum, which will result in a smaller total resistance when you recalculate 1/Rtotal.
Looking at some related practice problems:
- If you double the voltage across an ohmic resistor, according to Ohm's law (V = IR), the current through the resistor will also double.
- The resistance of a bulb can be calculated using Ohm's law. If the current is 1.25 A when a 4 V voltage supply is connected, resistance (R) can be found by R = V/I, giving R = 4V / 1.25A = 3.2 ohms. If the voltage supply is increased to 7 V and the bulb's resistance remains constant, the new current will be I = V/R, so I = 7V / 3.2 ohms = 2.1875 A.
- Two identical resistors connected in parallel with one resistor's resistance increased results in the current through the unaffected resistor remaining the same, since it is still driven by the same voltage. The unaffected resistor will not change its resistance or the voltage across it, based on parallel circuit rules.