Final answer:
If bulb B is replaced with a short circuit, the effect on the brightness of bulb A depends on the circuit configuration. In a series circuit, bulb A would shine brighter due to increased current. In a parallel circuit, bulb A's brightness would remain unchanged.
Step-by-step explanation:
When bulb B is replaced with a short circuit, it essentially removes that bulb from the circuit since the current takes the path of least resistance, which would be the new direct connection created by the short circuit. The brightness of bulb A would then depend on whether the bulbs were initially connected in series or parallel.
In a series circuit, replacing bulb B with a short would increase the current through the remaining parts of the circuit. Because brightness is related to power (P = I^2 * R, where P is power, I is current, and R is resistance), and the resistance of bulb A has not changed, the increase in current would cause bulb A to shine brighter, assuming it does not exceed its rated capacity.
However, in a parallel circuit, the current through each branch is independent of the other branches. If bulb B were shorted in a parallel circuit, bulb A would not be affected and its brightness would remain the same.