In a test circuit, a 1.2\text{ k}\Omega resistor is connected across an adjustable DC powe… — Armed Services Vocational Aptitude Battery (ASVAB)

In a test circuit, a $1.2\text{ k}\Omega$ resistor is connected across an adjustable DC power supply. When the supply voltage is increased by $18\text{ V}$ , the current through the resistor reaches $35\text{ mA}$ . What was the initial current flowing through the resistor in milliamperes ( $\text{mA}$ )?

Answer: 20 mA

Answer

The initial current flowing through the resistor was

20\text{ mA}

Applying Ohm's Law (

I = \frac{V}{R}

), an

18\text{ V}

voltage increase across a

1.2\text{ k}\Omega

(

1{}200\ \Omega

) resistor causes a current increase of

\Delta I = \frac{18\text{ V}}{1{}200\ \Omega} = 0.015\text{ A} = 15\text{ mA}

. Subtracting this

15\text{ mA}

change from the final current of

35\text{ mA}

yields an initial current of

20\text{ mA}

Step-by-Step Solution

Convert resistance from kilohms (

k\Omega

) to ohms (

\Omega

R = 1.2\text{ k}\Omega = 1{}200\ \Omega

Ohm's Law calculations require standard base SI units (

V

in volts,

I

in amperes,

R

in ohms).

Calculate the increase in current (

\Delta I

) resulting from the

18\text{ V}

voltage increase.

\Delta I = \frac{\Delta V}{R} = \frac{18\text{ V}}{1{}200\ \Omega} = 0.015\text{ A} = 15\text{ mA}

According to Ohm's Law (

I = \frac{V}{R}

), an increase in voltage produces a directly proportional increase in current across a constant resistance.

Find the initial current by subtracting the current increase from the final current.

I_{\text{initial}} = 35\text{ mA} - 15\text{ mA} = 20\text{ mA}

The final current represents the sum of the initial current and the incremental current produced by the added voltage.

Key Concept

Ohm's Law defines the direct relationship between voltage and current across a constant resistance (

I = \frac{V}{R}

), meaning a change in voltage causes a proportional change in current.

Alternative Method

Calculate the final voltage first:

V_{\text{final}} = I_{\text{final}} \times R = 0.035\text{ A} \times 1{}200\ \Omega = 42\text{ V}

. Then subtract the

18\text{ V}

increase to find the initial voltage:

V_{\text{initial}} = 42\text{ V} - 18\text{ V} = 24\text{ V}

. Finally, compute the initial current:

I_{\text{initial}} = \frac{24\text{ V}}{1{}200\ \Omega} = 0.020\text{ A} = 20\text{ mA}

Estimated Time:50s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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