A 24\text{ V} DC power supply energizes a circuit consisting of a 100\,\Omega series resis… — Armed Services Vocational Aptitude Battery (ASVAB)

A $24\text{ V}$ DC power supply energizes a circuit consisting of a $100\,\Omega$ series resistor ( $R_1$ ) connected in series with a parallel network composed of two $200\,\Omega$ branch loads. Under normal operating conditions, the circuit draws $120\text{ mA}$ , dropping $12\text{ V}$ across $R_1$ and $12\text{ V}$ across the parallel network. During a troubleshooting check, a technician records a total circuit current of $80\text{ mA}$ , a voltage drop of $8\text{ V}$ across $R_1$ , and a voltage drop of $16\text{ V}$ across the parallel network. Which fault condition in the circuit explains these diagnostic readings?

A
A direct short circuit occurred across one of the two parallel branch loads.
B
A partial short circuit developed across the series resistor $R_1$ .
An open circuit occurred in one of the two parallel branch loads.Answer
D
The series resistor $R_1$ has opened completely.

Answer

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\,\Omega

branch. This increases the equivalent parallel resistance from

100\,\Omega

200\,\Omega

, bringing total circuit resistance to

300\,\Omega

. By Ohm's law, total current drops to

I = \frac{24\text{ V}}{300\,\Omega} = 80\text{ mA}

. Consequently, the voltage across

R_1

drops to

8\text{ V}

(

80\text{ mA} \times 100\,\Omega

) while the parallel network drops the remaining

16\text{ V}

(

80\text{ mA} \times 200\,\Omega

Step-by-Step Solution

Analyze normal baseline circuit parameters.

Normal equivalent parallel resistance is

R_{\text{parallel}} = \frac{200\,\Omega \times 200\,\Omega}{200\,\Omega + 200\,\Omega} = 100\,\Omega

. Total normal resistance is

R_{\text{total}} = 100\,\Omega + 100\,\Omega = 200\,\Omega

. Normal current is

I = \frac{24\text{ V}}{200\,\Omega} = 120\text{ mA}

, producing equal

12\text{ V}

drops across

R_1

and the parallel bank.

Establishing normal values is required to determine whether total circuit resistance increased or decreased under fault conditions.

Calculate the faulted circuit's total resistance and component values using the measured readings.

With

I_{\text{measured}} = 80\text{ mA}

and

V_{\text{source}} = 24\text{ V}

, total circuit resistance is

R_{\text{total, fault}} = \frac{24\text{ V}}{0.08\text{ A}} = 300\,\Omega

. Since

R_1

drops

8\text{ V}

80\text{ mA}

(

R_1 = \frac{8\text{ V}}{0.08\text{ A}} = 100\,\Omega

, unchanged), the parallel bank now exhibits

R_{\text{parallel, fault}} = \frac{16\text{ V}}{0.08\text{ A}} = 200\,\Omega

Ohm's law confirms where the additional resistance is located.

Determine the physical fault that produces a

200\,\Omega

parallel bank resistance.

Because each individual branch load is rated at

200\,\Omega

, an equivalent resistance of

200\,\Omega

across the bank means only one

200\,\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.

Key Concept

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.

Estimated Time:1m 30s

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