A specialized weather station telemetry circuit is powered by a 96\text{ V} direct current… — Armed Services Vocational Aptitude Battery (ASVAB)

A specialized weather station telemetry circuit is powered by a $96\text{ V}$ direct current (DC) power source connected in series with four resistive components: a surge-limiting resistor $R_1 = 18\ \Omega$ , a status indicator $R_2 = 27\ \Omega$ , a variable humidity sensor $R_3$ , and a reference resistor $R_4 = 15\ \Omega$ . A technician measures the voltage drop across the status indicator ( $R_2$ ) to be $21.6\text{ V}$ . What is the resistance of the humidity sensor $R_3$ in ohms ( $\Omega$ )?

Answer: 60 $\Omega$

Answer

The resistance of the humidity sensor

R_3

60\ \Omega

In a series circuit, current is uniform throughout the entire loop. Using the measured voltage drop across

R_2

(

21.6\text{ V}

) and its known resistance (

27\ \Omega

), the loop current is

I = \frac{V_2}{R_2} = \frac{21.6\text{ V}}{27\ \Omega} = 0.8\text{ A}

. The total equivalent resistance connected across the

96\text{ V}

source is

R_T = \frac{V_{\text{source}}}{I} = \frac{96\text{ V}}{0.8\text{ A}} = 120\ \Omega

. Because series resistances are directly additive (

R_T = R_1 + R_2 + R_3 + R_4

), the unknown resistance is determined by subtracting the known values from the total:

R_3 = 120\ \Omega - (18\ \Omega + 27\ \Omega + 15\ \Omega) = 120\ \Omega - 60\ \Omega = 60\ \Omega

Step-by-Step Solution

Calculate loop current from the known component voltage drop and resistance.

I = \frac{21.6\text{ V}}{27\ \Omega} = 0.8\text{ A}

In a series circuit, current is identical through all components.

Calculate total circuit resistance using Ohm's Law and the source voltage.

R_T = \frac{96\text{ V}}{0.8\text{ A}} = 120\ \Omega

Total resistance equals source voltage divided by total circuit current.

Calculate the sum of all known resistances.

R_{\text{known}} = 18\ \Omega + 27\ \Omega + 15\ \Omega = 60\ \Omega

Resistances in a series circuit are directly additive.

Subtract the sum of known resistors from total circuit resistance to isolate

R_3

R_3 = 120\ \Omega - 60\ \Omega = 60\ \Omega

R_T = R_1 + R_2 + R_3 + R_4

, so

R_3 = R_T - (R_1 + R_2 + R_4)

Key Concept

Series Circuit Loop Current and Total Resistance Analysis

Alternative Method

Calculate individual voltage drops using loop current

I = 0.8\text{ A}

V_1 = 0.8\text{ A} \times 18\ \Omega = 14.4\text{ V}

and

V_4 = 0.8\text{ A} \times 15\ \Omega = 12.0\text{ V}

. By Kirchhoff's Voltage Law, the voltage across

R_3

V_3 = 96\text{ V} - (14.4\text{ V} + 21.6\text{ V} + 12.0\text{ V}) = 48.0\text{ V}

. Applying Ohm's Law directly to

R_3

yields

R_3 = \frac{V_3}{I} = \frac{48.0\text{ V}}{0.8\text{ A}} = 60\ \Omega

Estimated Time:1m 30s

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