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Zorluk: ZorCircuit Faults and Troubleshooting

A 24 V24\text{ V} DC power supply energizes a circuit consisting of a 100 Ω100\,\Omega series resistor (R1R_1) connected in series with a parallel network composed of two 200 Ω200\,\Omega branch loads. Under normal operating conditions, the circuit draws 120 mA120\text{ mA}, dropping 12 V12\text{ V} across R1R_1 and 12 V12\text{ V} across the parallel network. During a troubleshooting check, a technician records a total circuit current of 80 mA80\text{ mA}, a voltage drop of 8 V8\text{ V} across R1R_1, and a voltage drop of 16 V16\text{ V} across the parallel network. Which fault condition in the circuit explains these diagnostic readings?

  1. A
    A direct short circuit occurred across one of the two parallel branch loads.
  2. B
    A partial short circuit developed across the series resistor R1R_1.
  3. An open circuit occurred in one of the two parallel branch loads.Cevap
  4. D
    The series resistor R1R_1 has opened completely.

Cevap

An open circuit occurred in one of the two parallel branch loads.
When one parallel branch develops an open circuit, current can only flow through the remaining 200 Ω200\,\Omega branch. This increases the equivalent parallel resistance from 100 Ω100\,\Omega to 200 Ω200\,\Omega, bringing total circuit resistance to 300 Ω300\,\Omega. By Ohm's law, total current drops to I=24 V300 Ω=80 mAI = \frac{24\text{ V}}{300\,\Omega} = 80\text{ mA}. Consequently, the voltage across R1R_1 drops to 8 V8\text{ V} (80 mA×100 Ω80\text{ mA} \times 100\,\Omega) while the parallel network drops the remaining 16 V16\text{ V} (80 mA×200 Ω80\text{ mA} \times 200\,\Omega).

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1
Analyze normal baseline circuit parameters.
Normal equivalent parallel resistance is Rparallel=200 Ω×200 Ω200 Ω+200 Ω=100 ΩR_{\text{parallel}} = \frac{200\,\Omega \times 200\,\Omega}{200\,\Omega + 200\,\Omega} = 100\,\Omega. Total normal resistance is Rtotal=100 Ω+100 Ω=200 ΩR_{\text{total}} = 100\,\Omega + 100\,\Omega = 200\,\Omega. Normal current is I=24 V200 Ω=120 mAI = \frac{24\text{ V}}{200\,\Omega} = 120\text{ mA}, producing equal 12 V12\text{ V} drops across R1R_1 and the parallel bank.
Establishing normal values is required to determine whether total circuit resistance increased or decreased under fault conditions.
2
Calculate the faulted circuit's total resistance and component values using the measured readings.
With Imeasured=80 mAI_{\text{measured}} = 80\text{ mA} and Vsource=24 VV_{\text{source}} = 24\text{ V}, total circuit resistance is Rtotal, fault=24 V0.08 A=300 ΩR_{\text{total, fault}} = \frac{24\text{ V}}{0.08\text{ A}} = 300\,\Omega. Since R1R_1 drops 8 V8\text{ V} at 80 mA80\text{ mA} (R1=8 V0.08 A=100 ΩR_1 = \frac{8\text{ V}}{0.08\text{ A}} = 100\,\Omega, unchanged), the parallel bank now exhibits Rparallel, fault=16 V0.08 A=200 ΩR_{\text{parallel, fault}} = \frac{16\text{ V}}{0.08\text{ A}} = 200\,\Omega.
Ohm's law confirms where the additional resistance is located.
3
Determine the physical fault that produces a 200 Ω200\,\Omega parallel bank resistance.
Because each individual branch load is rated at 200 Ω200\,\Omega, an equivalent resistance of 200 Ω200\,\Omega across the bank means only one 200 Ω200\,\Omega branch is conducting, indicating the other branch has an open circuit.
Removing one branch of a two-branch identical parallel network doubles the network's equivalent resistance to the value of the remaining single branch.

Anahtar Kavram

In a series-parallel circuit, an open fault in a parallel branch increases total circuit resistance, reduces total line current, and causes a higher proportion of the source voltage to drop across the parallel section.
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