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Difficulty: EasyCircuit Faults and Troubleshooting

A technician is troubleshooting a simple series circuit powered by a 12-volt DC battery and containing a single incandescent light bulb. When the switch is closed, the bulb does not light up. Measuring with a digital voltmeter directly across the bulb's socket terminals shows a reading of 12 volts. What fault condition does this measurement indicate?

  1. A
    A direct short circuit across the light bulb socket
  2. An open circuit in the light bulb filamentAnswer
  3. C
    A blown fuse at the battery positive terminal
  4. D
    An incorrect multimeter setting configured for resistance measurement

Answer

An open circuit in the light bulb filament
When a filament burns out, it creates an open circuit. Because the circuit is open, current cannot flow (I=0 AI = 0\text{ A}), meaning no voltage is dropped across the closed switch or connecting wires. Consequently, connecting a voltmeter across the open filament places the meter across the full potential of the 12-volt battery.

Step-by-Step Solution

1
Analyze the circuit state and the voltmeter reading.
The voltmeter measures the full supply voltage of 12 V12\text{ V} directly across the bulb terminals, yet no light is produced.
In a series circuit with an open fault, current flow ceases entirely (I=0 AI = 0\text{ A}). With no current flowing, there are no voltage drops across any normal conductors or closed switches.
2
Apply voltage laws to identify the location of the open fault.
The full source potential difference (12 V12\text{ V}) appears across the open gap in the filament.
According to Kirchhoff's Voltage Law, the sum of voltage drops must equal the source voltage. Because the open filament has infinite resistance compared to the wiring, the entire 12 V12\text{ V} drops across the break.

Key Concept

An open component in a series circuit stops all current flow and drops the full source voltage across its terminals.
Estimated Time:45s
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Topics

Circuit Faults and Troubleshooting

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