Question

a circuit has a total voltage (potential difference) of 47.2 v due to a resistance of 159 \\(\omega\\). how many electrons pass through a resistor in this circuit in 0.954 s?

\\(1.77 \times 10^{18}\\) electrons
\\(3.79 \times 10^{18}\\) electrons
\\(8.99 \times 10^{19}\\) electrons
\\(2.44 \times 10^{20}\\) electrons

Explanation:

Calculate current using Ohm's Law

Using the Ohm's Law knowledge point

$$ I = \frac{V}{R} = \frac{47.2\text{ V}}{159\ \Omega} \approx 0.296855\text{ A} $$

Calculate total charge passing through

Using the Electric Current knowledge point

$$ q = I \cdot t = 0.296855\text{ A} \times 0.954\text{ s} \approx 0.28320\text{ C} $$

Calculate number of electrons

Using the Elementary Charge knowledge point

$$ N = \frac{q}{e} = \frac{0.28320\text{ C}}{1.602 \times 10^{-19}\text{ C/electron}} \approx 1.7678 \times 10^{18}\text{ electrons} $$

Match with options

Rounding to three significant figures gives \(1.77 \times 10^{18}\) electrons.

Answer:

• (A) \(1.77 \times 10^{18}\) electrons (Correct answer)
• (B) \(3.79 \times 10^{18}\) electrons
• (C) \(8.99 \times 10^{19}\) electrons
• (D) \(2.44 \times 10^{20}\) electrons