When a 12.0 V car battery runs a single 30.0 W headlight, how many electrons pass through it each second? Strategy To find the number of electrons, we must first find the charge that moved in 1.00 s. The charge moved is related to voltage and energy through the equation ΔPE = qΔV . A 30.0 W lamp

Question

When a 12.0 V car battery runs a single 30.0 W headlight, how many electrons pass through it each second?
Strategy
To find the number of electrons, we must first find the charge that moved in 1.00 s. The charge moved is related to voltage and energy through the equation ΔPE = qΔV . A 30.0 W lamp uses 30.0 joules per second. Since the battery loses energy, we have ΔPE = –30.0 J and, since the electrons are going from the negative terminal to the positive, we see that ΔV = +12.0 V .

Accepted Answer

To find the charge q moved, we solve the equation ΔPE = qΔV :

[latex]q=\frac{\Delta PE }{\Delta V}[/latex]. (19.10)

Entering the values for ΔPE and ΔV , we get

[latex]q=\frac{-30.0 J }{+12.0 V }=\frac{-30.0 J }{+12.0 J / C }=-2.50 C[/latex]. (19.11)

The number of electrons [latex]n _{ e }[/latex] is the total charge divided by the charge per electron. That is,

[latex]n _{ e }=\frac{-2.50 C }{-1.60 imes 10^{-19} C / e ^{-}}=1.56 imes 10^{19}[/latex] electrons. (19.12)

Discussion
This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so many being present in ordinary systems. In fact, electricity had been in use for many decades before it was determined that the moving charges in many circumstances were negative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determine which is moving or whether both are moving.

Question 19.2: When a 12.0 V car battery runs a single 30.0 W headlight, ho...

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