Final answer:
Placing a wire across the terminals of bulb A creates a short circuit, causing the current to bypass the bulb, which would then not light up. Shorting a battery with a wire produces a high current flow, leading to heat generation due to the battery's internal resistance.
Step-by-step explanation:
If you put a wire across the two terminals of bulb A, you would create a path of lower resistance for the current to flow through, bypassing the bulb's filament. This is known as a short circuit. Since the resistance through the wire alone is negligible compared to the filament, nearly all the current will go through the wire, causing bulb A to effectively receive no current and therefore not light up. This is why, when you short out the terminals, the circuit bypasses the resistive load (bulb filament), preventing it from lighting up.
When a battery is shorted by connecting its terminals with a low-resistance path like a wire, it causes a high current to flow. This high current can lead to heat generation due to the internal resistance of the battery. The heat is generated because power (P), which is dissipated as heat, is given by P = I2R, where I is the current, and R is the resistance. A high current with any non-zero resistance, even a small one, will lead to a significant amount of power being dissipated as heat, which is why the battery gets hot.