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Zorluk: Çok zorCircuit Faults and Troubleshooting

An industrial control module features a 36 V36\text{ V} DC source driving a resistor network. The configuration consists of a 150 Ω150\ \Omega input resistor (R1R_1) wired in series with a parallel pair of 300 Ω300\ \Omega load resistors (R2R_2 and R3R_3). A test point (TP1) is located at the junction separating R1R_1 from the parallel pair.

Under nominal conditions, the circuit draws 120 mA120\text{ mA} from the source, and a multimeter reads 18 V18\text{ V} at TP1 with respect to ground.

A system failure occurs, and a technician records new diagnostic readings:
- The voltage at TP1 with respect to ground is now 36 V36\text{ V}.
- The total current drawn from the 36 V36\text{ V} source is now 240 mA240\text{ mA}.

Based on these diagnostic readings, which of the following component failures has occurred?

  1. R1R_1 has suffered an internal short circuit.Cevap
  2. B
    TP1 has short-circuited directly to the chassis ground.
  3. C
    The common ground connection for R2R_2 and R3R_3 has become an open circuit.
  4. D
    R1R_1 has burned out, creating an open circuit.

Cevap

The component failure is that R1R_1 has suffered an internal short circuit.
For the voltage at TP1 to equal the source voltage (36 V36\text{ V}), there must be no voltage drop across the series resistor R1R_1. This occurs either when there is no current flowing, or when R1R_1 has zero resistance. Because the circuit is drawing 240 mA240\text{ mA} (a higher-than-normal current), the circuit is not open. Therefore, R1R_1 must have zero resistance due to an internal short circuit. When R1R_1 is shorted, the total circuit resistance drops to just the 150 Ω150\ \Omega equivalent resistance of the parallel pair, which mathematically accounts for the exactly 240 mA240\text{ mA} current draw (36 V/150 Ω36\text{ V} / 150\ \Omega).

Adım Adım Çözüm

1
Analyze the voltage reading at TP1 during the fault.
TP1 measures 36 V36\text{ V}, which equals the source voltage.
This indicates there is zero voltage drop across R1R_1. This can only happen if no current is flowing through R1R_1 (an open circuit downstream) or if R1R_1 has zero resistance (a short circuit).
2
Analyze the total current reading during the fault.
Total current is 240 mA240\text{ mA}, which is twice the normal operating current.
Since current is flowing (and is higher than normal), the circuit cannot be open. Therefore, the zero voltage drop across R1R_1 must be caused by a short circuit across R1R_1 itself.
3
Verify the shorted R1R_1 hypothesis mathematically using Ohm's Law.
If R1=0 ΩR_1 = 0\ \Omega, total resistance is 300 Ω/2=150 Ω300\ \Omega / 2 = 150\ \Omega. Total current = 36 V/150 Ω=0.24 A36\text{ V} / 150\ \Omega = 0.24\text{ A} (240 mA240\text{ mA}). Voltage at TP1 = 36 V−(0.24 A×0 Ω)=36 V36\text{ V} - (0.24\text{ A} \times 0\ \Omega) = 36\text{ V}.
Both mathematical results precisely match the technician's fault measurements, confirming R1R_1 is shorted.

Anahtar Kavram

Series-Parallel Circuit Troubleshooting and Ohm's Law
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Konular

Circuit Faults and Troubleshooting

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