Question 12 CEL02 - Chief Engineer - Limited (Alt)

The Correct Answer is A. ### Explanation for Option A (Correct) **A) the coil will probably burn out immediately** The operating coil of an AC contactor is an inductor (choke) designed to present a specific high impedance (reactance) to the AC supply. This high impedance limits the current flow to the required operating level. If many turns of the coil become short-circuited: 1. **Reduced Impedance:** The overall number of active turns in the coil decreases significantly. Since inductive reactance ($X_L$) is proportional to the number of turns and the inductance ($L$), reducing the turns drastically reduces the total impedance ($Z$). 2. **Increased Current (Overcurrent):** According to Ohm's Law ($I = V/Z$), if the impedance ($Z$) drops sharply while the voltage ($V$) remains constant, the current ($I$) drawn by the coil will increase dramatically (a massive overcurrent). 3. **Burning Out:** The current will far exceed the coil's design limits. Due to excessive $I^2R$ power dissipation (heat), the remaining insulation will fail rapidly, and the coil will quickly overheat and burn out, often described as "immediately" in industrial settings. ### Explanation for Other Options (Incorrect) **B) the coil will operate on reduced current** This is incorrect. Short-circuiting turns reduces impedance, which causes the current to **increase**, not decrease. **C) the coil will continue to operate** This is highly unlikely. While the contactor might pull in briefly (or even not at all if the current is high enough to trip protection immediately), the drastically increased current due to the shorted turns will cause rapid thermal damage, leading to failure and burnout shortly after power is applied. It cannot "continue to operate" reliably or safely. **D) the coil temperature will drop** This is incorrect. The temperature rise is directly related to the power dissipated ($P = I^2 R$). Since the current ($I$) increases dramatically due to the short circuit, the power dissipation increases massively, leading to a sharp **increase** in temperature and subsequent burnout.