## Related questions with answers

A current of $4.82 \mathrm{~A}$ exists in a $12.4 \mathrm{\Omega}$ resistor for $4.60 \mathrm{~min}$. (a) How much charge and (b) how many electrons pass through any cross section of the resistor in this time?

Solution

Verifieda)

From equation (29-1) we can find the total charge travelled trough the resistor in the given time:

$\begin{align*} i &= \frac{d q}{d t} \\ d q &= i \cdot d t \\ \Delta q &= \int i \cdot d t \\ \Delta q &= i \cdot \Delta t = 4.82 \ \text{A} \cdot (4.6 \cdot 60) \ \text{s} \end{align*}$

$\boxed{\Delta q = 1324.8 \ \text{C}}$

b)

To find the number of passed electrons we have to divide the total passed charge with the charge of a single electron $q = e$.

$\Delta N = \frac{\Delta q}{e} = \frac{1324.8 \ \text{C}}{1.602 \cdot 10^{-19} \ \text{C}}$

$\boxed{\Delta N \approx 8.269 \cdot 10^{21}}$

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