1. An ideal transformer is shown in Figure 6.12. Use this model for a distribution transformer from VH = 24 kV to Vx = 240 V (RMS). This is, of course, a single phase transformer and is of the type commonly used to provide the 240 V service commonly used in the United States. It is center-tapped to provide 120 V for branch circuits. Assume this transformer is rated at 24 kVA. (a) What are the current ratings Ix and IH? (b) If the secondary ('X') side has 26 turns, how many turns does the primary ('H') side have? (c) If this transformer is loaded by a resistor of value R, what value of R results in the transformer being loaded to its rating? (d) If the secondary is loaded by a resistor of value R, what is the impedance looking into the primary side? 'H ¹x Vx Figure 6.12 Ideal transformer

Could you please apply transfer function to analyse the attached diagram and explain the performance of that system please.

Homework 4 - Microsoft Word Главная ictaska Разметка страницы Рассылки Рецектирование A upesan A Hailt - Calibri (OCHORHOl TE - 11 E- E E A T A A A A AaBbCcDd AaBbCcDd AaBbCc AaBbCc AaBI AaBbCct AaBbCcDd AaBbCcDd AoBbCcDe |т Обычный тБез ингер. Заголовок 1 Заголовок 2 Название Подзаголо... Слабое вы- выделение Сильное в Konponat Заменить BETARurt Вставить J eopuar no obpauy he x Aa- Выделить - fyoep obmena AGsau Crane Pegacposae 1- In the following sequential circuit. There are 2 D FF (A and B), 2 inputs (x and y), and 1 output z The output and next-states equations are given A(t+1) = x'y + xA %3D B(t+1) = x'B + XA z = B a- Draw the logic diagram of the circuit. b- List the state table for the sequential circuit. c- Draw the corresponding state diagram. Cipaua: 1 a t Hncno Cnoe: 126 O AHTAnicKti (CLIA) DOa 220% 15:10 P Type here to search My Questions. Telegram (22 n File Explorer Assignment- E Homewark 4. ENG 15/05/2021

How do I get the eight BIT Adder Subtracter Module (the square one) on Quartus 2?

In R/2nR DAC, the maximum input resistor to be used is Rmin*2^(n-1) where n is the number of bits. Rmin*(2^n)-1 where n is the number of bits. Rmin*2^n wheren is the number of bits. Rmin*[(2^n)-1] where n is the number of bits.