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Zorluk: OrtaElectric Current and Resistance

A steady electric current of 3.2 A3.2\,\text{A} flows through a conductor for 5.0 minutes5.0\,\text{minutes}. Given that the elementary charge is 1.6×10−19 C1.6 \times 10^{-19}\,\text{C}, how many electrons pass through a cross-section of the conductor during this period?

  1. A
    1.0×10201.0 \times 10^{20}
  2. B
    3.6×10203.6 \times 10^{20}
  3. 6.0×10216.0 \times 10^{21}Cevap
  4. D
    1.536×10−161.536 \times 10^{-16}

Cevap

The number of electrons passing through the cross-section of the conductor is 6.0×10216.0 \times 10^{21}.
Electric current II is related to total electric charge QQ and time tt by Q=I×tQ = I \times t. Converting time into seconds gives t=5.0×60=300 st = 5.0 \times 60 = 300\,\text{s}. The total charge passed is Q=3.2 A×300 s=960 CQ = 3.2\,\text{A} \times 300\,\text{s} = 960\,\text{C}. Using the charge quantization formula Q=n⋅eQ = n \cdot e, the number of electrons nn is given by n=960 C1.6×10−19 C=6.0×1021n = \frac{960\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 6.0 \times 10^{21}.

Adım Adım Çözüm

1
Convert the time from minutes into seconds.
t=5.0 minutes=5.0×60 s=300 st = 5.0\,\text{minutes} = 5.0 \times 60\,\text{s} = 300\,\text{s}
SI units require time to be in seconds when calculating electric charge.
2
Calculate the total electric charge passing through the conductor.
Q=I×t=3.2 A×300 s=960 CQ = I \times t = 3.2\,\text{A} \times 300\,\text{s} = 960\,\text{C}
Electric current is defined as the rate of flow of charge (I=Q/tI = Q/t).
3
Determine the number of electrons using charge quantization.
n=Qe=960 C1.6×10−19 C=6.0×1021n = \frac{Q}{e} = \frac{960\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 6.0 \times 10^{21}
Total charge is equal to the number of carrier electrons multiplied by the elementary charge (Q=n⋅eQ = n \cdot e).

Anahtar Kavram

Quantization of Electric Charge and Current
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Konular

Electric Current and Resistance

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