An electric vehicle telemetry subsystem is powered by a 45\text{ V} direct current (DC) so… — Armed Services Vocational Aptitude Battery (ASVAB)

An electric vehicle telemetry subsystem is powered by a $45\text{ V}$ direct current (DC) source connected in series with three resistive sensors: $R_1 = 7\ \Omega$ , $R_2 = 11\ \Omega$ , and $R_3 = 12\ \Omega$ . What is the voltage drop across the $11\ \Omega$ sensor ( $R_2$ )?

A
$15.0\text{ V}$
$16.5\text{ V}$ Answer
C
$45.0\text{ V}$
D
$1.5\text{ V}$

Answer

16.5\text{ V}

(the voltage drop across the

11\ \Omega

resistor)

The correct answer is

16.5\text{ V}

. In a series circuit, total resistance is found by adding all individual resistors (

7\ \Omega + 11\ \Omega + 12\ \Omega = 30\ \Omega

). Dividing the

45\text{ V}

source by the

30\ \Omega

total resistance yields a circuit current of

1.5\text{ A}

. Multiplying this

1.5\text{ A}

current by the

11\ \Omega

resistance of the specified resistor gives a voltage drop of

16.5\text{ V}

Step-by-Step Solution

Calculate the total series resistance (

R_T

) by summing the three individual resistances.

R_T = R_1 + R_2 + R_3 = 7\ \Omega + 11\ \Omega + 12\ \Omega = 30\ \Omega

In a series circuit, equivalent resistance equals the sum of all individual resistances in the loop.

Calculate the total loop current (

I

) using Ohm's Law.

I = \frac{V_T}{R_T} = \frac{45\text{ V}}{30\ \Omega} = 1.5\text{ A}

Current remains constant and identical through all components in a series loop.

Calculate the voltage drop across

R_2

using Ohm's Law.

V_2 = I \times R_2 = 1.5\text{ A} \times 11\ \Omega = 16.5\text{ V}

The voltage drop across any individual resistor is the product of the common loop current and that resistor's resistance.

Key Concept

In a series circuit, current is identical through every component, and the source voltage divides across each resistor in direct proportion to its resistance value (

V_x = I \times R_x

Alternative Method

Use the voltage divider formula:

V_2 = V_T \times \left(\frac{R_2}{R_T}\right) = 45\text{ V} \times \left(\frac{11\ \Omega}{30\ \Omega}\right) = 16.5\text{ V}

Estimated Time:1m 0s

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Step-by-Step Solution

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