All questions of Practice Tests for Electrical Engineering (EE) Exam

Heater resistance R =
RMS value of output voltage =
Therefore power absorbed by heater element = = 500W

Microfarad per phase, then determine the short circuit current available at the bus bar if a fault occurs at the CB.

To find the short circuit current, we need to first calculate the total impedance of the system. The reactance of the alternator per phase is given as 3 ohms, and the distributed capacitance per phase up to the CB is 0.01 microfarad.

The reactance of the distributed capacitance can be calculated as:

Xc = 1 / (2πfC) = 1 / (2π x 50 x 0.01 x 10^-6) = 318.3 ohms

The total impedance per phase is then given by:

Z = √(R^2 + Xl^2 + Xc^2) = √(3^2 + 318.3^2 + 3^2) = 318.9 ohms

Since we have a 3 phase system, the total impedance will be 3 times this value, which is 956.7 ohms.

The short circuit current can be calculated using Ohm's law:

I = V / Z

Where V is the voltage of the system, which is 11 kV or 11,000 volts.

I = 11,000 / 956.7 = 11.5 kA

Therefore, the short circuit current available at the bus bar is 11.5 kA.

Probability that A hits the target = 4/5
Probability that B hits the target = 3/4
Probability that C hits the target = 2/3
P (at least 2) = P(2) + P(3)

It will Jump only on zero. so for dcr b b=01 ,so nonzero it will not jump. so 0 loop

Given data:

DC voltage source = 300 V

Latching current for SCR = 150 mA

Load, L = 0.3 H

Load, R = 30 Ω

We know that SCR needs a gate pulse to turn on. The minimum width of the gate pulse is given by:

$t_{on} = \frac{\pi}{2 \omega_{r}}$

Where,

$\omega_{r} = \frac{1}{\sqrt{LC}}$

L and C are the inductance and capacitance of the load respectively. In this case, we have only given the value of L, so we need to calculate the value of C first.

$Z = R + j\omega L$

$Z = \sqrt{R^{2} + (\omega L)^{2}} \angle \tan^{-1}(\frac{\omega L}{R})$

At resonance, Z is minimum and the phase angle is zero. Hence,

$\omega L = R$

$\omega = \frac{R}{L}$

$\omega_{r} = \frac{R}{L} = \frac{30}{0.3} = 100 \text{ rad/s}$

$C = \frac{1}{\omega_{r}^{2} L} = \frac{1}{(100)^{2} \times 0.3} = 3.33 \times 10^{-6} \text{ F}$

Now, we can calculate the minimum width of the gate pulse as:

$t_{on} = \frac{\pi}{2 \omega_{r}} = \frac{\pi}{2 \times 100} = 0.0159 \text{ s}$

The gate pulse current required to turn on the SCR can be calculated using the formula:

$I_{g} = \frac{I_{L}}{\beta}$

Where,

$I_{L}$ is the load current

$\beta$ is the current gain of the SCR

We know that the latching current of the SCR is 150 mA. So, we can assume that the current gain is at least 50.

$I_{L} = \frac{V_{dc}}{R + j \omega L} = \frac{300}{30 + j 100 \times 0.3} = 7.5 \text{ A}$

$I_{g} = \frac{7.5}{50} = 0.15 \text{ A}$

Therefore, the minimum width of gate pulse current required to turn-on this SCR is 0.0159 s and its value is 0.15 A.

Let henry’s one day work be x
Let ford’s one day work be y

Solving 2 equations we get, y = 1/30
Hence, Ford can complete the job working alone in 30 days

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Rating of the machine = 90 MVA, Inertia constant (H) = 3 MJ/MVA

K.E. = 270 MJ

Understanding the Given Differential Equation
The equation provided is:
y log y dx + (x – log y) dy = 0
This is a first-order differential equation that can be analyzed for an integrating factor.
Identifying the Components
To find the integrating factor, we rewrite the equation in the standard form:
M(x, y) = y log y,
N(x, y) = x - log y.
Here, M and N are functions of x and y.
Calculating the Partial Derivatives
Next, we calculate the partial derivatives:
- ∂M/∂y = log y + 1
- ∂N/∂x = 1
We check for exactness:
- If ∂M/∂y ≠ ∂N/∂x, the equation is not exact.
Since log y + 1 ≠ 1, the equation is not exact.
Finding the Integrating Factor
For the equation to become exact, we seek an integrating factor that depends solely on y. The integrating factor is typically of the form µ(y).
The ratio of the differences of the partial derivatives gives us a clue:
µ(y) = (∂N/∂x) / (∂M/∂y - ∂N/∂x)
Substituting the values:
µ(y) = 1 / (log y + 1 - 1) = 1 / log y
Thus, the integrating factor simplifies to:
µ(y) = log y
Conclusion
The integrating factor for the given differential equation is indeed log y, confirming that option 'B' is the correct answer. This factor can be used to multiply the entire equation, making it exact and solvable.