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Zorluk: ZorElectric Current and Resistance

Two cylindrical metallic conductors, XX and YY, are connected in series across a direct-current source. Wire XX has a diameter of 1.0mm1.0\,\text{mm} and a free-electron density of 6.0×1028m36.0 \times 10^{28}\,\text{m}^{-3}. Wire YY has a diameter of 3.0mm3.0\,\text{mm} and a free-electron density of 2.0×1028m32.0 \times 10^{28}\,\text{m}^{-3}. What is the ratio of the drift velocity of free electrons in wire XX to that in wire YY?

  1. 3.03.0Cevap
  2. B
    0.330.33
  3. C
    1.01.0
  4. D
    9.09.0

Cevap

The ratio of the drift velocity in wire X to that in wire Y is 3.0.
Because the wires are connected in series, the same current passes through both (IX=IYI_X = I_Y). Using the formula for electric current in terms of drift velocity I=nAevdI = n A e v_d, where cross-sectional area A=πd2/4A = \pi d^2 / 4, we find vd1/(nd2)v_d \propto 1 / (n d^2). Taking the ratio yields vX/vY=(nYdY2)/(nXdX2)=(2.0×1028×3.02)/(6.0×1028×1.02)=18/6=3.0v_X / v_Y = (n_Y d_Y^2) / (n_X d_X^2) = (2.0 \times 10^{28} \times 3.0^2) / (6.0 \times 10^{28} \times 1.0^2) = 18 / 6 = 3.0.

Adım Adım Çözüm

1
Relate electric current to drift velocity and conductor geometry.
I=nAevd=n(πd24)evdI = n A e v_d = n \left( \frac{\pi d^2}{4} \right) e v_d
Electric current II depends on free-electron density nn, cross-sectional area AA, elementary charge ee, and electron drift velocity vdv_d.
2
Apply the series connection constraint.
IX=IY    nXdX2vX=nYdY2vYI_X = I_Y \implies n_X d_X^2 v_X = n_Y d_Y^2 v_Y
In a series circuit, the steady current flowing through every conductor is identical.
3
Rearrange to solve for the drift velocity ratio vX/vYv_X / v_Y.
vXvY=nYdY2nXdX2\frac{v_X}{v_Y} = \frac{n_Y d_Y^2}{n_X d_X^2}
Isolating vX/vYv_X / v_Y demonstrates inverse proportionality to electron density and the square of conductor diameter.
4
Substitute the given values into the ratio expression.
\frac{v_X}{v_Y} = \frac{(2.0 \times 10^{28}\,\text{m}^{-3}) \times (3.0\,\text{mm})^2}{(6.0 \times 10^{28}\,\text{m}^{-3}) \times (1.0\,\text{mm})^2} = \frac{2.0 \times 9.0}{6.0 \times 1.0} = \frac{18.0}{6.0} = 3.0
Numerical calculation yields the simplified dimensionless ratio.

Anahtar Kavram

Drift Velocity and Current Density in Series Conductors
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