A diesel generator preheat circuit is powered by a 24\text{ V} DC source connected in seri… — Armed Services Vocational Aptitude Battery (ASVAB)

A diesel generator preheat circuit is powered by a $24\text{ V}$ DC source connected in series with a ballast resistor $R_1 = 1.8\ \Omega$ and two identical glow plugs ( $R_2$ and $R_3$ ). If an ammeter in the circuit measures a steady current of $4.0\text{ A}$ , what is the resistance of a single glow plug in ohms ( $\Omega$ )?

Answer: 2.1 Ω

Answer

The resistance of a single glow plug is

2.1\ \Omega

Applying Ohm's Law gives a total series loop resistance of

R_{\text{total}} = 24\text{ V} / 4.0\text{ A} = 6.0\ \Omega

. In a series circuit, total resistance is the sum of all individual resistances (

R_{\text{total}} = R_1 + R_2 + R_3

). Subtracting the ballast resistor (

1.8\ \Omega

) leaves

4.2\ \Omega

across the two identical glow plugs. Dividing

4.2\ \Omega

equally between them gives

2.1\ \Omega

per glow plug.

Step-by-Step Solution

Calculate the total circuit resistance using Ohm's Law.

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{24\text{ V}}{4.0\text{ A}} = 6.0\ \Omega

In a closed DC loop, total resistance equals total source voltage divided by total current.

Subtract the known ballast resistance from the total resistance to determine the combined resistance of both glow plugs.

R_2 + R_3 = R_{\text{total}} - R_1 = 6.0\ \Omega - 1.8\ \Omega = 4.2\ \Omega

Resistances in a series circuit are directly additive (

R_{\text{total}} = R_1 + R_2 + R_3

Divide the combined glow plug resistance equally between the two identical units.

R_{\text{plug}} = \frac{4.2\ \Omega}{2} = 2.1\ \Omega

Because the two glow plugs are identical, each provides exactly half of the remaining series resistance.

Key Concept

Series Total Resistance and Ohm's Law

Estimated Time:1m 0s

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