Final answer:
In an electric circuit with three identical bulbs, the potential difference across each bulb is equal when connected in parallel. The brightness of a bulb is proportional to the power dissipated, which depends on both voltage and current. In parallel, a bulb with less resistance will be brighter, while in series, a bulb with greater resistance will be brighter.
The correct answer is c. A = B = C = D = E = F = G = H = I = J = K = L.
Step-by-step explanation:
When considering the potential differences across light bulbs in an electric circuit, we must recognize that they will depend on the bulbs' arrangement—whether in series or parallel—as well as their respective resistances.
Given three identical light bulbs (A, B, and C) in the circuit, if they are connected in parallel, the potential difference across each bulb would be the same because there are no components between any two points across a single bulb that would result in a potential drop.
Therefore, the correct ranking by potential difference across each bulb is option c, which states A = B = C.
The brightness of the bulbs is dependent on the power dissipated, which is related to both the voltage across the bulbs and the current flowing through them.
The brightness is thus proportional to the power dissipation (P=VI), assuming they all share the same voltage and consistent with the electric power definition in Section 19.4.
The bulb with the greater resistance will dissipate more power and be brighter if the bulbs are connected in series, as each bulb in a series circuit has the same current passing through it.
However, in a parallel circuit, the bulb with the less resistance will be brighter, as the voltage across each bulb is the same but the current through the bulb is inversely proportional to its resistance.
The correct answer is c. A = B = C = D = E = F = G = H = I = J = K = L.