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Sav [38]
3 years ago
12

Recall from an earlier lecture that the relationship between the current through an inductor and the voltage across it is given

by What is the current through a 10 mH inductor if 1 V is applied across it for exactly 20 ms? Assume that initially there is no current through the inductor. Show your calculations in your engineering notebook.
Engineering
1 answer:
serious [3.7K]3 years ago
7 0

Answer: Current = 2A

Explanation:

Inductor L = 10mH

E.m.f = 1 V

Time t = 20 ms

Using

V = Ldi/dt

di/dt = V/ L

di/dt = 1/10×10^-3

di/dt = 100

di = 100dt

Integrate both sides

I = 100(t1 - to)

Assume that initially there is no current through the inductor.

to = 0

I = 100 × t

I = 100 × 20×10^-3

I = 2A

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What is the first step in the Design Process *
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Find an expectation value in the n th state of the harmonic oscillator.Find an expectation value in the n th state of the harmon
s344n2d4d5 [400]

The classical motion for an oscillator that starts from rest at location x₀ is

                                           x(t) = x₀ cos(ωt)

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Thus,

ρ =   \frac{1}{\pi / w} \int\limits^\pi_0 {d(x - x_o cos(wt))} \, dt

Limit is 0 to π/ω

We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt)  so that

ρ(x) = -\frac{w}{\pi } \int\limits^x_x {\frac{d ( x - y)}{x_ow sin(wt)} } \, dy

Limit is x₀ to -x₀

\frac{1}{\pi } \int\limits^x_x {\frac{d (x-y)}{x_o\sqrt{1 - cos^2(wt)} } } \, dy

Limit is -x₀ to x₀

= \frac{1}{\pi } \int\limits^x_x {\frac{d(x-y)}{\sqrt{x_o^2 - y^2} } } \, dy\\ \\= \frac{1}{\pi\sqrt{x_o^2 - x^2}  }

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6 0
4 years ago
In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery
Dima020 [189]

Answer:

  248.756 mV

  49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

  voltage: (100 V)(1000/(1000 +1000) = 50 V

  impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

  voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

  impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

__

The open-circuit voltage is the difference between these terminal voltages:

  (50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

__

The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

  (50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

8 0
3 years ago
A TV USE 75 WATTS WHILE IN USED ASSMING THAT ITIS USED 4 HOURS EVERY DAY HOW MUCH ENERGY IN 4 IN KWH WOULD THE TV CONSUME ANNUAL
prohojiy [21]

Answer:

i don't think i understand the question

Explanation:

7 0
2 years ago
Read 2 more answers
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