A 24\text{ V} DC bridge circuit consists of two parallel branches connected between a cons… — Armed Services Vocational Aptitude Battery (ASVAB)

A $24\text{ V}$ DC bridge circuit consists of two parallel branches connected between a constant $+24\text{ V}$ DC supply rail and ground:
- Left branch: Resistor $R_1 = 1.0\,\text{k}\Omega$ connected in series with resistor $R_2 = 1.0\,\text{k}\Omega$ to ground, with Test Point A ( $V_A$ ) located between them.
- Right branch: Resistor $R_3 = 1.0\,\text{k}\Omega$ connected in series with resistor $R_4 = 1.0\,\text{k}\Omega$ to ground, with Test Point B ( $V_B$ ) located between them.

Under normal operation, the total current drawn from the supply is $24\text{ mA}$ , and $V_A = V_B = 12.0\text{ V}$ . During diagnostic testing of a faulty unit, a technician measures the following values:
- $V_A = 12.0\text{ V}$ (measured with respect to ground)
- $V_B = 24.0\text{ V}$ (measured with respect to ground)
- Total current drawn from the $24\text{ V}$ source = $36\text{ mA}$

Which of the following circuit faults accounts for these diagnostic readings?

A
An open circuit fault in resistor $R_4$
B
A short circuit fault across resistor $R_4$
A short circuit fault across resistor $R_3$ Answer
D
An open circuit fault in resistor $R_3$

Answer

A short circuit fault across resistor

R_3

A short circuit across resistor

R_3

bypasses the upper resistor, placing Test Point B directly at the

+24\text{ V}

supply rail. This leaves only resistor

R_4

(

1.0\,\text{k}\Omega

) in the right branch, increasing the right branch current to

24\text{ V} / 1.0\,\text{k}\Omega = 24\text{ mA}

. Combining this with the normal

12\text{ mA}

from the left branch produces the exact measured total current of

36\text{ mA}

Step-by-Step Solution

Analyze the operational status and current of the left branch.

Left branch current

I_{\text{left}} = 12\text{ mA}

Since

V_A = 12.0\text{ V}

across

R_2 = 1.0\,\text{k}\Omega

, the left branch is normal and draws

I_{\text{left}} = 24\text{ V} / (1.0\,\text{k}\Omega + 1.0\,\text{k}\Omega) = 12\text{ mA}

Determine the current flowing through the right branch under fault conditions.

Right branch current

I_{\text{right}} = 24\text{ mA}

Subtract the normal left branch current from the measured total current:

I_{\text{right}} = I_{\text{total}} - I_{\text{left}} = 36\text{ mA} - 12\text{ mA} = 24\text{ mA}

Calculate the effective resistance of the right branch and evaluate voltage drop across components.

R_{\text{right}} = 1.0\,\text{k}\Omega

, with

0\text{ V}

dropped across

R_3

R_{\text{right}} = V_{\text{source}} / I_{\text{right}} = 24\text{ V} / 24\text{ mA} = 1.0\,\text{k}\Omega

. Since

R_4

1.0\,\text{k}\Omega

and

V_B = 24.0\text{ V}

, the full

24\text{ V}

drops across

R_4

, meaning

R_3

has

0\text{ V}

drop and zero resistance (a direct short circuit).

Key Concept

Distinguishing between open and short circuit faults using branch voltage and total current analysis

Estimated Time:1m 30s

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