Determine the voltage across a 2-μF capacitor if the current through it is i(t) = 3e−6000t mA. Assume that the initial capacitor voltage is zero g

Answers

Answer 1
Answer:

Answer:

v = 250[1 - {e^(-6000t)}] mV

Explanation:

The voltage across a capacitor at a time t, is given by:

v(t) = (1)/(C) \int\limits^(t)_(t_0) {i(t)} \, dt + v(t_0)                 ----------------(i)

Where;

v(t) = voltage at time t

t_(0) = initial time

C = capacitance of the capacitor

i(t) = current through the capacitor at time t

v(t₀) = voltage at initial time.

From the question:

C = 2μF = 2 x 10⁻⁶F

i(t) = 3e^(-6000t) mA

t₀ = 0

v(t₀ = 0) = 0

Substitute these values into equation (i) as follows;

v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + v(0)    

v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + 0

v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt            

v = (3)/(2*10^(-6)) \int\limits^(t)_(0) {e^(-6000t)} \, dt             [Solve the integral]

v = (3)/(2*10^(-6)*(-6000))  {e^(-6000t)}|_0^t

v = (-3000)/(12)  {e^(-6000t)}|_0^t

v = -250 {e^(-6000t)}|_0^t

v = -250 {e^(-6000t)} - [-250 {e^(-6000(0))]

v = -250 {e^(-6000t)} - [-250]

v = -250 {e^(-6000t)} + 250

v = 250 -250 {e^(-6000t)}

v = 250[1 - {e^(-6000t)}]

Therefore, the voltage across the capacitor is v = 250[1 - {e^(-6000t)}] mV


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Answers

The four relevant pressures in a Rankine cycle with one stage of reheat are P1, P2, P3, and P4.

For a Rankine cycle with one stage of reheat between turbines, there are typically four relevant pressures:

  1. Boiler pressure (P1): This is the pressure at which the water is heated in the boiler before entering the first turbine.
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What is the deflection equation for a simply supported beam with a uniformly distributed load?

Answers

Answer:

\Delta _(max)=(5wL^4)/(384EI)

Explanation:

Given that

Load is uniformly distributed load and beam is simply supported.

Ra + Rb= wL

Ra = Rb =wL / 2

Lets x is measured from left side,then the deflection of beam at any distance x is given as

\Delta _x=(wx)/(24EI)(L^3-2Lx^2+x^3)

The maximum deflection of beam will at   x = L/2 (mid point )

\Delta _(max)=(w* (L)/(2))/(24EI)(L^3-2L* \left ((L)/(2)\right)^2+\left((L)/(2)\right)^3)

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Answers

Answer:

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Explanation:

Electric generating power plant generates a lot of steam and heat. On the efficiency aspect of the cooling tower with respect to the operation of the power plant, a cooling tower with its fan at the end of the airstream leaving the tower is best applied considering the pressure of steam coming out of the turbine. On the maintenance angle, it is easier to maintain.

can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and angle. thanks

Answers

Answer:

  • branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
  • branch 2: i = 21.466∠63.435°; pf = 0.447 leading
  • total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

  X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915*10^(-3)\approx j4.99984\,\Omega

The capacitive reactance is ...

  X_C=(1)/(j\omega C)=(-j)/(100\pi\cdot 318.31*10^(-6)F)\approx -j10.00000\,\Omega

Branch 1

The impedance of branch 1 is ...

  Z1 = 8 +j4.99984 Ω

so the current is ...

  I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

Branch 2

The impedance of branch 2 is ...

  Z2 = 5 -j10 Ω

so the current is ...

  I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

Total current

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

  It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

  It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

__

The phasor diagram of the currents is attached.

_____

Additional comment

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

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Answers

Answer:

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Explanation:

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Answers

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