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3 years ago

6

Consider a step-down transformer with 15 turns in the primary and 6 turns in the secondary windings. Calculate the input impedan

ce of this transformer loaded to an 8 Ω resistor, assuming that series resistances of the windings are negligible.

1 answer:

AURORKA [14]3 years ago

8 0

Answer:

Input impedance of this transformer is 50 ohms.

Explanation:

Given that,

Number of turns in the primary coil, $N_p=15$

Number of turns in the secondary coil, $N_s=6$

Output impedance of the transformer, $V_o=8\ \Omega$

The number of turns and the impedance ratio in the step down transformer is given by :

$\dfrac{15}{6}=\sqrt{\dfrac{Z_p}{Z_s}}\\\\\dfrac{15}{6}=\sqrt{\dfrac{Z_p}{8}}\\\\Z_p=50\ \Omega$

So, the input impedance of this transformer is 50 ohms. Hence, this is the required solution.

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An engine flywheel initially rotates counterclockwise at 6.55 rotations/s. Then, during 20.9 s, its rotation rate changes to 2.1

11Alexandr11 [23.1K]

Answer:

The average angular acceleration is -2.628 rad/s²

Explanation:

Counterclockwise = positive

Clockwise = -negative

Given;

initial rotation of the flywheel, θ₁ = 6.55 rotation/s

final rotation of the flywheel, θ₂ = - 2.19 rotation/s

The average angular acceleration is given by;

$\alpha = \frac{\delta \theta}{\delta t}\\\\ \alpha =\frac{\theta _2 - \theta_ 1}{t}\\\\ \alpha =\frac{-2.19 -6.55}{20.9} \\\\ \alpha =\frac{-8.74}{20.9}\\\\ \alpha = -0.4182 \ rotation / s^2\\\\ \alpha = \frac{-0.4182 \ rotation}{s^2}*\frac{2\pi \ radian}{rotation}\\\\ \alpha = -2.628 \ rad/s^2$

Therefore, the average angular acceleration is -2.628 rad/s²

7 0

3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

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