An automotive technician tests a lighting circuit consisting of two auxiliary lamps wired … — Armed Services Vocational Aptitude Battery (ASVAB)

An automotive technician tests a lighting circuit consisting of two auxiliary lamps wired in parallel. The resistance of the first lamp is $10\ \Omega$ , and the resistance of the second lamp is $15\ \Omega$ . What is the total equivalent resistance of this parallel circuit?

A
$25\ \Omega$
$6\ \Omega$ Answer
C
$12.5\ \Omega$
D
$5\ \Omega$

Answer

The total equivalent resistance of the parallel circuit is

6\ \Omega

The equivalent resistance of two resistors connected in parallel is calculated using

R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}

. Substituting

10\ \Omega

and

15\ \Omega

gives

\frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\ \Omega

Step-by-Step Solution

Identify the formula for equivalent resistance of two parallel resistors

R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}

\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}

Resistors in parallel provide multiple paths for current, resulting in an equivalent resistance smaller than any individual branch.

Substitute the given resistances (

R_1 = 10\ \Omega

and

R_2 = 15\ \Omega

) into the product-over-sum formula

R_{\text{eq}} = \frac{10 \times 15}{10 + 15} = \frac{150}{25}

Using the product-over-sum shortcut provides a direct calculation for two parallel branches.

Calculate the final equivalent resistance

R_{\text{eq}} = 6\ \Omega

Dividing

150

25

yields the total resistance of

6\ \Omega

Key Concept

Equivalent resistance in a parallel circuit is always less than the smallest individual branch resistance and is calculated using the reciprocal or product-over-sum rule.

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