An emergency backup lighting system connects three identical 15\,\Omega indicator lamps in… — Armed Services Vocational Aptitude Battery (ASVAB)

An emergency backup lighting system connects three identical $15\,\Omega$ indicator lamps in series across a $45\text{ V}$ direct-current power source. What is the electric current flowing through each lamp?

A
$0.33\text{ A}$
B
$3.0\text{ A}$
$1.0\text{ A}$ Answer
D
$9.0\text{ A}$

Answer

1.0\text{ A}

In a series circuit, the total electrical resistance is the sum of all individual resistances in the path:

R_{\text{total}} = 15\,\Omega + 15\,\Omega + 15\,\Omega = 45\,\Omega

. Applying Ohm's law (

I = \frac{V}{R}

), the current drawn from the

45\text{ V}

source is

I = \frac{45\text{ V}}{45\,\Omega} = 1.0\text{ A}

. In any series circuit, the electric current is identical at every point, meaning exactly

1.0\text{ A}

flows through each lamp.

Step-by-Step Solution

Calculate the total equivalent resistance of the series circuit.

R_{\text{total}} = 15\,\Omega + 15\,\Omega + 15\,\Omega = 45\,\Omega

In a series circuit, the total electrical resistance equals the direct sum of all individual resistances.

Apply Ohm's law to determine the total circuit current.

I = \frac{V}{R_{\text{total}}} = \frac{45\text{ V}}{45\,\Omega} = 1.0\text{ A}

Current is calculated by dividing total source voltage by the total resistance of the circuit.

Determine the current flowing through each individual lamp.

I_{\text{lamp}} = 1.0\text{ A}

Because there is only one continuous path for electrical charge in a series circuit, the current is uniform through all components.

Key Concept

Ohm's law and total resistance in series circuits

Estimated Time:1m 0s

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