An inductor with L = 9.55 mH is connected across an ac source that has voltage amplitude 49.0 V You may want to review (Pages 10
1 answer:
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The current flowing in silicon bar is 2.02 10^-12 A.
<u>Explanation:</u>
Length of silicon bar, l = 10 μm = 0.001 cm
Free electron density, Ne = 104 cm^3
Hole density, Nh = 1016 cm^3
μn = 1200 cm^2 / V s
μр = 500 cm^2 / V s
The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.
J = Je + Jh
J = n qE μn + p qE μp
where, n and p are electron and hole densities.
J = Eq (n μn + p μp)
we know that E = V / l
So, J = (V / l) q (n μn + p μp)
J = (1.6 10^-19) / 0.001 (104
1200 + 1016
500)
J = 1012480 10^-16 A / m^2.
or
J = 1.01 10^-9 A / m^2
Current, I = JA
A is the area of bar, A = 20 μm = 0.002 cm
I = 1.01 10^-9
0.002 = 2.02
10^-12
So, the current flowing in silicon bar is 2.02 10^-12 A.
Answer:
A.The spring constant for B is one quarter of the spring constant for A.
Explanation:
If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.
As , and the 2 springs have the same mass
So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.