While troubleshooting a 24-volt DC auxiliary signaling circuit, a technician attempts to measure the current drawn by an active warning lamp. When the technician places the multimeter test leads across the two terminals of the illuminated lamp, the lamp immediately goes out and the multimeter's internal fuse blows. Which of the following explains why the meter fuse blew?
- AThe meter was configured for direct current while the circuit was operating on alternating current, creating an overcurrent condition.
- The ammeter was connected in parallel across the load, creating a low-resistance short circuit path that drew excessive current.Answer
- CThe meter test leads were connected in series with the lamp, causing the full circuit voltage to overload the meter's internal circuit board.
- DThe selector switch was set to measure DC voltage, which caused the meter to act as a direct electrical short across the power source.
Answer
The ammeter was connected in parallel across the load, creating a low-resistance short circuit path that drew excessive current.
An ammeter is designed with extremely low internal resistance to avoid affecting the circuit when placed in series. When connected in parallel across an energized component, it creates a short circuit across the voltage source, causing a massive current surge that instantly blows the meter's protective fuse.
Step-by-Step Solution
Key Concept
Multimeter Current vs. Voltage Hookup Principles and Internal Meter Protection
Estimated Time:1m 0s