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Difficulty: MediumCircuit Faults and Troubleshooting

A direct-current circuit powers three identical 60 Ω60\,\Omega indicator lamps connected in parallel across a 12 V12\text{ V} DC bus. During a routine operational check, an ammeter placed in series with the main supply line reads 0.4 A0.4\text{ A} instead of the expected normal current of 0.6 A0.6\text{ A}. Which condition best explains this current reading?

  1. A
    One indicator lamp has suffered an internal short circuit.
  2. One indicator lamp has burned out and created an open circuit.Answer
  3. C
    A fourth identical indicator lamp was accidentally wired in parallel.
  4. D
    The DC bus supply voltage has dropped to zero volts.

Answer

One indicator lamp has burned out and created an open circuit.
In a parallel circuit, each branch operates independently across the full 12 V12\text{ V} supply. Under normal conditions, each 60 Ω60\,\Omega lamp draws I=12 V60 Ω=0.2 AI = \frac{12\text{ V}}{60\,\Omega} = 0.2\text{ A}, yielding a total current of 3×0.2 A=0.6 A3 \times 0.2\text{ A} = 0.6\text{ A}. When one lamp fails open (such as a broken filament), its branch current falls to 0 A0\text{ A} while the other two branches continue drawing 0.2 A0.2\text{ A} each. The sum of the remaining branch currents is 0.2 A+0.2 A=0.4 A0.2\text{ A} + 0.2\text{ A} = 0.4\text{ A}, which matches the measured ammeter reading.

Step-by-Step Solution

1
Calculate the normal operating current for a single parallel branch using Ohm's Law.
Ibranch=VR=12 V60 Ω=0.2 AI_{\text{branch}} = \frac{V}{R} = \frac{12\text{ V}}{60\,\Omega} = 0.2\text{ A}
In a parallel circuit, each branch receives the full source voltage of 12 V12\text{ V} independently.
2
Determine the expected total current under normal operating conditions.
Itotal=3×0.2 A=0.6 AI_{\text{total}} = 3 \times 0.2\text{ A} = 0.6\text{ A}
Total current in a parallel circuit is the sum of all individual branch currents (Itotal=I1+I2+I3I_{\text{total}} = I_1 + I_2 + I_3).
3
Analyze the measured current to identify the active number of branches and diagnose the fault.
Active branches=0.4 A0.2 A/branch=2 branches\text{Active branches} = \frac{0.4\text{ A}}{0.2\text{ A/branch}} = 2\text{ branches}
A drop from 0.6 A0.6\text{ A} to 0.4 A0.4\text{ A} indicates that exactly one 0.2 A0.2\text{ A} branch is no longer drawing current, which occurs when a lamp burns out into an open circuit.

Key Concept

Parallel Circuit Fault Diagnosis and Open-Circuit Current Reduction

Alternative Method

Calculate equivalent resistance before and after: Normal equivalent resistance is Req=603=20 ΩR_{\text{eq}} = \frac{60}{3} = 20\,\Omega, drawing I=12 V20 Ω=0.6 AI = \frac{12\text{ V}}{20\,\Omega} = 0.6\text{ A}. With the measured current of 0.4 A0.4\text{ A}, the circuit's total resistance is Req=12 V0.4 A=30 ΩR_{\text{eq}} = \frac{12\text{ V}}{0.4\text{ A}} = 30\,\Omega. Since two 60 Ω60\,\Omega resistors in parallel equal 30 Ω30\,\Omega (602=30 Ω\frac{60}{2} = 30\,\Omega), exactly one branch is open.
Estimated Time:1m 0s
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Topics

Circuit Faults and Troubleshooting

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