A direct-current power distribution box supplies power to two parallel branches. The first… — Armed Services Vocational Aptitude Battery (ASVAB)

A direct-current power distribution box supplies power to two parallel branches. The first branch has a resistance of $20\ \Omega$ , and the second branch has a resistance of $5\ \Omega$ . What is the total equivalent resistance of the network?

A
$25\ \Omega$
B
$12.5\ \Omega$
$4\ \Omega$ Answer
D
$100\ \Omega$

Answer

4\ \Omega

In a parallel circuit with two branches, the equivalent resistance is calculated using the product-over-sum formula

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

. Substituting

20\ \Omega

and

5\ \Omega

gives

\frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4\ \Omega

. Total resistance in parallel is always less than the lowest individual resistor value.

Step-by-Step Solution

Identify the formula for the 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}

Parallel branches provide multiple paths for current, resulting in a total resistance that is less than any individual branch resistance.

Substitute the given values (

R_1 = 20\ \Omega

and

R_2 = 5\ \Omega

) into the product-over-sum formula.

R_{\text{eq}} = \frac{20 \times 5}{20 + 5} = \frac{100}{25}

Simplifying the numerator and denominator allows direct calculation of equivalent resistance.

Divide the product by the sum.

R_{\text{eq}} = 4\ \Omega

Computing

\frac{100}{25}

yields the final equivalent resistance.

Key Concept

Total Equivalent Resistance in Parallel Circuits

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