Answer to Question #221147 in Electric Circuits for abcde

Question

when a 12 V car battery powers a single 30.0 W headlight how many electron pass through it each second.

Answer 1

Answer:- first we will find the charge that moved in 1.00 s , to find the number of electrons

charge moved is related to voltage and energy

$\Delta PE =q\Delta V$

30.0 W uses 30.0 joules per second. Since the battery loses energy, we have

$\Delta PE =-30 J$

and also the electrons are going from the negative terminal to the positive, so

$V=+12V$

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

$q=\boldsymbol{\frac{\Delta \textbf{PE}}{\Delta V}}$

Entering the values for ΔPE and ΔV, we get

$q=\boldsymbol{\frac{-30.0 \;\textbf{J}}{+12.0 \;\textbf{V}}} =-2.50 C$

The number of electrons ne is the total charge divided by the charge per electron. That is,

$n_e=\boldsymbol{\frac{-2.50 \;\textbf{C}}{-1.60 \times 10^{-19} \;\textbf{C} / \textbf{e}^{-}}}\\ \boldsymbol{= 1.56 \times 10^{19} \;\textbf{electrons.}}$