In an electronic flash unit, a high-voltage trigger circuit transfers 0.72\text{ J} of ele… — Armed Services Vocational Aptitude Battery (ASVAB)

In an electronic flash unit, a high-voltage trigger circuit transfers $0.72\text{ J}$ of electrical energy by moving a charge of $1.6\text{ mC}$ across a gas-discharge tube. What is the electromotive force (voltage) developed across the tube during this discharge?

A
$0.45\text{ V}$
B
$1.15\text{ V}$
$450\text{ V}$ Answer
D
$2.22\text{ V}$

Answer

450\text{ V}

Electromotive force (EMF), or voltage, is defined as the amount of work or energy transferred per unit of electric charge, expressed by the formula

V = \frac{W}{Q}

, where

1\text{ volt} = 1\text{ joule per coulomb}

. Converting the charge from millicoulombs to coulombs gives

1.6\text{ mC} = 0.0016\text{ C}

. Dividing the total transferred energy (

0.72\text{ J}

) by the charge (

0.0016\text{ C}

) produces

\frac{0.72}{0.0016} = 450\text{ V}

Step-by-Step Solution

Identify the given physical quantities and their base units.

Energy

W = 0.72\text{ J}

and charge

Q = 1.6\text{ mC}

Standard electrical calculations require values expressed in base SI units.

Convert the electric charge from millicoulombs to coulombs.

Q = 1.6 \times 10^{-3}\text{ C} = 0.0016\text{ C}

The prefix milli- represents

10^{-3}

, so

1.6\text{ mC} = 0.0016\text{ C}

Apply the definition formula of electromotive force and voltage (

V = \frac{W}{Q}

V = \frac{0.72\text{ J}}{0.0016\text{ C}} = 450\text{ V}

One volt is defined as one joule of energy transferred per coulomb of charge.

Key Concept

Electromotive force (EMF) and voltage represent the potential energy transferred per unit electric charge (

1\text{ V} = 1\text{ J/C}

Estimated Time:1m 30s

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