An electronics maintenance panel is powered by a 120\text{ V} DC bus. The circuit features… — Armed Services Vocational Aptitude Battery (ASVAB)

An electronics maintenance panel is powered by a $120\text{ V}$ DC bus. The circuit features a primary $40\ \Omega$ load resistor ( $R_1$ ) connected in series with a parallel junction containing two secondary resistors: $R_2$ ( $60\ \Omega$ ) and $R_3$ ( $30\ \Omega$ ). During a routine inspection, a digital multimeter indicates a $48\text{ V}$ drop across $R_1$ . Assuming the source voltage is nominal, which specific component failure is causing this reading?

The $30\ \Omega$ secondary resistor ( $R_3$ ) has failed open.Cevap
B
The $60\ \Omega$ secondary resistor ( $R_2$ ) has failed open.
C
The primary resistor ( $R_1$ ) has degraded and reduced in value to $24\ \Omega$ .
D
The parallel junction has short-circuited to ground.

Cevap

The

30\ \Omega

secondary resistor (

R_3

) has failed open.

Based on the

48\text{ V}

drop across the

40\ \Omega

series resistor, the actual circuit current is

1.2\text{ A}

. To draw exactly

1.2\text{ A}

from a

120\text{ V}

source, the total circuit resistance must be

100\ \Omega

. Since the series resistor accounts for

40\ \Omega

, the parallel junction must be contributing exactly

60\ \Omega

. The only way the parallel junction (normally containing a

60\ \Omega

and a

30\ \Omega

resistor) equals

60\ \Omega

is if the

30\ \Omega

resistor has failed open, leaving only the

60\ \Omega

resistor in the path.

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Calculate the actual current flowing through the primary resistor (

R_1

) under fault conditions.

I_{total} = \frac{48\text{ V}}{40\ \Omega} = 1.2\text{ A}

By Ohm's Law, the measured voltage drop across a known resistance reveals the actual current flowing through that component.

Determine the new total circuit resistance (

R_{total}

) required to restrict the current to

1.2\text{ A}

R_{total} = \frac{120\text{ V}}{1.2\text{ A}} = 100\ \Omega

The total resistance determines how much current is drawn from the main voltage bus.

Calculate the equivalent resistance of the parallel network (

R_p

) under the current faulted condition.

R_p = R_{total} - R_1 = 100\ \Omega - 40\ \Omega = 60\ \Omega

Subtracting the series resistance from the total circuit resistance isolates the effective resistance of the parallel branch.

Identify the component failure that changes the parallel network's resistance from its normal

20\ \Omega

to exactly

60\ \Omega

The

30\ \Omega

resistor (

R_3

) must be an open circuit.

R_3

is open (infinite resistance), no current flows through it. The only path through the junction is via

R_2

(

60\ \Omega

), perfectly matching the required

60\ \Omega

equivalent resistance.

Anahtar Kavram

Troubleshooting Series-Parallel Circuits using Voltage Drops and Ohm's Law

Tahmini Süre:3m 0s

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