An automated assembly fixture uses an industrial direct current (DC) solenoid valve to act… — Armed Services Vocational Aptitude Battery (ASVAB)

An automated assembly fixture uses an industrial direct current (DC) solenoid valve to actuate a pneumatic clamp. The solenoid coil has an internal resistance of $600\text{ }\Omega$ and requires a steady current of $35\text{ mA}$ to operate reliably. What DC voltage must the power supply deliver to the solenoid valve?

A
$17.1\text{ V}$
B
$2.1\text{ V}$
$21\text{ V}$ Answer
D
$21,000\text{ V}$

Answer

21\text{ V}

According to Ohm's Law, voltage is calculated by multiplying current by resistance (

V = I \times R

). Converting the current from milliamperes to amperes gives

35\text{ mA} = 0.035\text{ A}

. Multiplying

0.035\text{ A} \times 600\text{ }\Omega

results in

21\text{ V}

Step-by-Step Solution

Convert the operating current from milliamperes (

\text{mA}

) to base amperes (

\text{A}

35\text{ mA} = \frac{35}{1,000}\text{ A} = 0.035\text{ A}

Ohm's Law calculations require electrical quantities in standard SI base units.

Apply Ohm's Law (

V = I \times R

) to find the required supply voltage.

V = 0.035\text{ A} \times 600\text{ }\Omega = 21\text{ V}

Voltage equals the product of current and resistance.

Key Concept

Ohm's Law (

V = I \times R

) with metric prefix conversion

Estimated Time:1m 0s

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