An electronics technician evaluates four separate direct current (DC) test circuits on a w… — Armed Services Vocational Aptitude Battery (ASVAB)

An electronics technician evaluates four separate direct current (DC) test circuits on a workbench. Match each circuit description on the left with its corresponding calculated electrical quantity on the right using Ohm's law ( $V = I \times R$ ).

A circuit powered by a $48\text{ V}$ DC supply that draws a current of $12\text{ mA}$ A total resistance of $4\text{ k}\Omega$
A DC indicator lamp with a resistance of $250\text{ }\Omega$ operating at a current of $40\text{ mA}$ A voltage drop of $10\text{ V}$
A $1.5\text{ k}\Omega$ resistor connected across a $30\text{ V}$ DC power sourceA current flow of $20\text{ mA}$
A DC relay coil with a resistance of $80\text{ }\Omega$ connected across a $24\text{ V}$ lineA current flow of $300\text{ mA}$

Answer

The circuit drawing

12\text{ mA}

48\text{ V}

matches a total resistance of

4\text{ k}\Omega

; the lamp with

250\text{ }\Omega

40\text{ mA}

matches a voltage drop of

10\text{ V}

; the

1.5\text{ k}\Omega

resistor at

30\text{ V}

matches a current of

20\text{ mA}

; and the

80\text{ }\Omega

relay coil at

24\text{ V}

matches a current of

300\text{ mA}

Each circuit scenario directly applies one variation of Ohm's law after properly adjusting metric units:

R = \frac{V}{I}

gives

4\text{ k}\Omega

V = I \times R

gives

10\text{ V}

, and

I = \frac{V}{R}

gives

20\text{ mA}

and

300\text{ mA}

respectively.

Step-by-Step Solution

Calculate resistance for the first circuit (

V = 48\text{ V}

I = 12\text{ mA}

)

R = \frac{48\text{ V}}{0.012\text{ A}} = 4{}000\text{ }\Omega = 4\text{ k}\Omega

Convert milliamperes to amperes (

12\text{ mA} = 0.012\text{ A}

) and apply Ohm's law formula

R = \frac{V}{I}

Calculate voltage across the lamp (

R = 250\text{ }\Omega

I = 40\text{ mA}

)

V = 0.040\text{ A} \times 250\text{ }\Omega = 10\text{ V}

Convert milliamperes to amperes (

40\text{ mA} = 0.040\text{ A}

) and apply

V = I \times R

Calculate current through the resistor (

V = 30\text{ V}

R = 1.5\text{ k}\Omega

)

I = \frac{30\text{ V}}{1{}500\text{ }\Omega} = 0.020\text{ A} = 20\text{ mA}

Convert kilohms to ohms (

1.5\text{ k}\Omega = 1{}500\text{ }\Omega

) and apply

I = \frac{V}{R}

, expressing the answer in milliamperes.

Calculate current through the relay coil (

V = 24\text{ V}

R = 80\text{ }\Omega

)

I = \frac{24\text{ V}}{80\text{ }\Omega} = 0.30\text{ A} = 300\text{ mA}

Apply

I = \frac{V}{R}

and convert amperes to milliamperes (

0.30\text{ A} = 300\text{ mA}

Key Concept

Ohm's law governs the fundamental mathematical relationship between electromotive force (voltage in volts), current (amperes), and opposition to current (resistance in ohms):

V = I \times R

I = \frac{V}{R}

, and

R = \frac{V}{I}

, with standard metric prefix conversions (

1\text{ k}\Omega = 1{}000\text{ }\Omega

1\text{ A} = 1{}000\text{ mA}

Estimated Time:1m 30s

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