A 24\text{ V} direct-current (DC) power source is connected to a circuit where resistor R_… — Armed Services Vocational Aptitude Battery (ASVAB)

A $24\text{ V}$ direct-current (DC) power source is connected to a circuit where resistor $R_1$ ( $4\ \Omega$ ) is in series with a parallel pair consisting of $R_2$ ( $12\ \Omega$ ) and $R_3$ ( $6\ \Omega$ ). Under normal operation, the total current is $3.0\text{ A}$ and the voltage drop across the parallel node is $12\text{ V}$ . An electronics technician records abnormal multimeter readings indicating different single-component faults. Match each set of diagnostic symptoms to the corresponding circuit fault.

Total current decreases to $2.4\text{ A}$ ; voltage across the parallel node rises to $14.4\text{ V}$ . $R_2$ has failed open (infinite resistance).
Total current decreases to $1.5\text{ A}$ ; voltage across the parallel node rises to $18.0\text{ V}$ . $R_3$ has failed open (infinite resistance).
Total current increases to $6.0\text{ A}$ ; voltage across the parallel node drops to $0\text{ V}$ . $R_2$ or $R_3$ has short-circuited (zero resistance).
Total current drops to $0\text{ A}$ ; voltmeter placed directly across $R_1$ reads $24\text{ V}$ . $R_1$ has failed open (infinite resistance).

Answer

The symptoms map to the faults as follows: 2.4 A and 14.4 V indicates R2 is open; 1.5 A and 18.0 V indicates R3 is open; 6.0 A and 0 V indicates a short in the parallel section (R2 or R3); 0 A and 24 V across R1 indicates R1 is open.

Each diagnostic reading mathematically corresponds to a specific component failure. When a component fails open, its resistance becomes infinite, altering the equivalent circuit resistance and redistributing voltage drops. When a component short-circuits, it bypasses that section of the circuit with near-zero resistance, causing a current spike and eliminating the voltage drop across that section. Placing a voltmeter across an open component in a series path will read the full source voltage.

Step-by-Step Solution

Analyze the normal circuit parameters to establish a baseline for comparison.

Normal equivalent parallel resistance (

R_p

) is

4\ \Omega

. Total resistance (

R_t

) is

8\ \Omega

. Normal current is

3.0\text{ A}

Baseline parameters are necessary to evaluate how mathematical deviations correlate to specific component failures.

Evaluate the symptom with

I = 2.4\text{ A}

and

V_p = 14.4\text{ V}

Total resistance increased to

10\ \Omega

(

24\text{ V} / 2.4\text{ A}

). Since

R_1

4\ \Omega

, the parallel section must now be

6\ \Omega

. This matches the value of

R_3

alone.

R_2

fails open, its branch has infinite resistance, leaving only

R_3

(

6\ \Omega

) to carry current in the parallel node.

Evaluate the symptom with

I = 1.5\text{ A}

and

V_p = 18.0\text{ V}

Total resistance increased to

16\ \Omega

(

24\text{ V} / 1.5\text{ A}

). The parallel section must now be

12\ \Omega

, which matches the value of

R_2

alone.

R_3

fails open, its branch has infinite resistance, leaving only

R_2

to carry current in the parallel section.

Evaluate the symptom with

I = 6.0\text{ A}

and

V_p = 0\text{ V}

Total resistance dropped to

4\ \Omega

, exactly equal to

R_1

. The parallel section resistance is

0\ \Omega

A short circuit in either

R_2

R_3

bypasses the resistance of the parallel node entirely, dropping its voltage to zero and causing a current spike.

Evaluate the symptom with

I = 0\text{ A}

and

24\text{ V}

measured directly across

R_1

An open circuit exists at

R_1

When a component fails open, the entire source voltage drops across the break because there is no voltage drop across the rest of the intact circuit components (since current is zero).

Key Concept

Circuit Fault Diagnostics using Ohm's Law and Series-Parallel Analysis

Estimated Time:3m 0s

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