A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 220 V to a primary coil of 230 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages

Answers

Answer 1

Answer:

Answer:

Explanation:

A multipurpose transformer can act as step up as well as step down transformer according to the desired setting by a user.

When the voltage at the output is greater than the voltage at the input of the transformer then it acts as step-up transformer and vice-versa acting is a step down transformer.

Given that:

input (primary) voltage of the transformer, $V_i=220~V$

no. of turns in the primary coil, $N_i=230$

When the output voltage is 5.60 V:

$V_o=5.60~V$

$(N_i)/(N_o) =(V_i)/(V_o)$

$(N_o)/(230)=(5.60)/(220)$

$N_o=5.85\approx 6$ turns compensating the losses

When the output voltage is 12.0 V:

$V_o=12.0~V$

$(N_i)/(N_o) =(V_i)/(V_o)$

$(N_o)/(230)=(12.0)/(220)$

$N_o=12.45\approx 13$ turns compensating the losses

When the output voltage is 480 V:

$V_o=480~V$

$(N_i)/(N_o) =(V_i)/(V_o)$

$(N_o)/(230)=(480)/(220)$

$N_o=501.8\approx 502$ turns compensating the losses

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Answers

Answer:

The code is attached.

Explanation:

I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.

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FILLIN THE BLANK Buffering is an important op amp application because it solves _____ that can't easily be solved with purely resistive circuits.Group of answer choices

Answers

Buffering is an important op amp application because it solves impedance-matching problems that can't easily be solved with purely resistive circuits. This allows for the transfer of maximum power between two circuits without any loss of signal strength.

Buffering with operational amplifiers (op amps) is a crucial application in electronics because it helps to overcome impedance-matching problems that cannot be easily resolved with only resistive circuits. Impedance-matching issues can cause signal distortion, and buffering solves this problem by creating a high-input impedance and low-output impedance circuit that separates the input and output signals, preventing the signal from being affected by the load impedance.

By using op amp buffering, the signal can be efficiently transmitted and received without signal loss or distortion, making it a useful technique in many applications, such as audio amplification and signal conditioning.

Complete question

Buffering is an important op amp application because it solves _____ that can't easily be solved with purely resistive circuits.

Group of answer choices

delay issues

tolerance issues

temperature issues

loading effects

impedance-matching problems

You can learn more about Buffering at

brainly.com/question/29761016

#SPJ11

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°.

The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-kg block B. If the unstrectched spring has a stiffness k = 1000 N/m, determine the maximum compression of the spring due to the collision. Assume the collision is perfectly plastic. Take e=0.6 .

Answers

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

Shear modulus is analogous to what material property that is determined in tensile testing? (a)- Percent reduction of area (b) Yield strength (c)- Elastic modulus (d)- Poisson's ratio

Answers

Answer:

(c)- Elastic modulus

Explanation:

We know that in tensile test we measure the properties of the material like yield strength,ultimate tensile strength ,Poisson ratio.

In tensile test

σ = ε E

Where σ is the stress

ε is the strain.

E is the elastic modulus.

Now for shear tress

τ = Φ G

Where τ the shear stress

Φ is the shear strain.

G is the shear modulus.

So we can say that Shear modulus is analogous to Elastic modulus.

Use Newton's method to determine the angle θ, between 0 and π/2 accurate to six decimal places. for which sin(θ) = 0.1. Show your work until you start computing x1, etc. Then just write down what your calculator gives you.