Examples In Electrical Calculations By Admiralty Pdf

How to Use Admiralty PDFs for Electrical Calculations: Examples and Tips

If you are looking for a reliable and comprehensive resource for electrical calculations, you might want to check out the Admiralty PDFs. These are a series of books published by the British Admiralty that cover various topics in electrical engineering, such as circuits, machines, measurements, power systems, and more. The Admiralty PDFs are especially useful for students and professionals who need to solve practical problems in electrical calculations. In this article, we will show you how to use the Admiralty PDFs for electrical calculations, with examples and tips to help you along the way.

examples in electrical calculations by admiralty pdf

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What are the Admiralty PDFs?

The Admiralty PDFs are a collection of books that were originally published by the British Admiralty's Naval Electrical Department in the 1960s. They were intended to provide a standard reference for naval electricians and engineers, as well as a self-study guide for those who wanted to learn more about electrical engineering. The Admiralty PDFs cover a wide range of topics in electrical calculations, such as:

• Basic electrical theory and principles

• Electrical circuits and components

• Electrical machines and transformers

• Electrical measurements and instruments

• Electrical power systems and distribution

• Electrical safety and protection

The Admiralty PDFs are written in a clear and concise style, with plenty of diagrams, tables, formulas, and examples to illustrate the concepts and methods. They also include exercises and solutions at the end of each chapter, so you can test your knowledge and skills. The Admiralty PDFs are available in both print and digital formats, so you can access them easily from anywhere.

How to use the Admiralty PDFs for electrical calculations?

The Admiralty PDFs are a great resource for anyone who needs to perform electrical calculations, whether for academic, professional, or personal purposes. Here are some tips on how to use them effectively:

• Choose the right book for your topic. The Admiralty PDFs are divided into several volumes, each covering a specific area of electrical engineering. For example, if you need to learn about electrical circuits, you can use the book Admiralty Solved Examples in Electrical Calculations: Circuits. If you need to learn about electrical machines, you can use the book Admiralty Solved Examples in Electrical Calculations: Machines. And so on.

• Read the theory and principles carefully. Before you start solving any problems, make sure you understand the basic theory and principles behind them. The Admiralty PDFs explain the concepts and formulas in a simple and logical way, with examples to help you grasp them. Pay attention to the definitions, symbols, units, and conventions used in the books.

• Follow the examples step by step. The Admiralty PDFs provide many solved examples for each topic, showing you how to apply the theory and formulas to real-world situations. You can follow these examples step by step, or try to solve them on your own before checking the answers. The examples also show you how to use different methods and techniques to solve the same problem.

• Practice with the exercises and solutions. The Admiralty PDFs include exercises at the end of each chapter, with varying levels of difficulty and complexity. You can use these exercises to practice your skills and check your understanding of the topic. The solutions are also provided at the end of each book, so you can compare your answers and learn from your mistakes.

• Use other resources if needed. The Admiralty PDFs are comprehensive and self-contained, but they are not exhaustive. You may encounter some topics or problems that are not covered by the books, or that require more advanced knowledge or tools. In that case, you can use other resources, such as textbooks, online courses, websites, calculators, software, etc., to supplement your learning.

By following these tips, you can use the Admiralty PDFs for electrical calculations effectively and efficiently.

Examples of electrical calculations using Admiralty PDFs

To give you a better idea of how to use the Admiralty PDFs for electrical calculations, we will present some examples from different topics and books. You can follow these examples or try to solve them yourself using the Admiralty PDFs as a reference.

Example 1: Electrical circuits

This example is taken from the book Admiralty Solved Examples in Electrical Calculations: Circuits. It shows how to calculate the equivalent resistance, current, voltage, and power of a series-parallel circuit.

The circuit diagram is shown below:

The given values are:

• R1 = 10 ohms

• R2 = 20 ohms

• R3 = 30 ohms

• R4 = 40 ohms

• V = 100 volts

The steps to solve the problem are:

• Find the equivalent resistance of the parallel branch R2 and R3. This can be done using the formula: $$\frac1R_eq = \frac1R_2 + \frac1R_3$$ Substituting the values, we get: $$\frac1R_eq = \frac120 + \frac130$$ Simplifying, we get: $$\frac1R_eq = \frac560$$ Taking the reciprocal, we get: $$R_eq = \frac605$$ Therefore, the equivalent resistance of the parallel branch is 12 ohms.

• Find the total equivalent resistance of the circuit. This can be done by adding the resistances of the series elements R1, R4, and R_eq. Using the formula: $$R_T = R_1 + R_4 + R_eq$$ Substituting the values, we get: $$R_T = 10 + 40 + 12$$ Simplifying, we get: $$R_T = 62$$ Therefore, the total equivalent resistance of the circuit is 62 ohms.

• Find the total current in the circuit. This can be done by applying Ohm's law, which states that the current is equal to the voltage divided by the resistance. Using the formula: $$I_T = \fracVR_T$$ Substituting the values, we get: $$I_T = \frac10062$$ Simplifying, we get: $$I_T = 1.61$$ Therefore, the total current in the circuit is 1.61 amperes.

• Find the voltage across each resistor. This can be done by multiplying the current by the resistance of each resistor. Using the formula: $$V_R = I_R \times R$$ For R1, we have: $$V_1 = I_T \times R_1$$ Substituting the values, we get: $$V_1 = 1.61 \times 10$$ Simplifying, we get: $$V_1 = 16.1$$ Therefore, the voltage across R1 is 16.1 volts.

For R2 and R3, we have to find the current in each resistor first. This can be done by applying Kirchhoff's current law, which states that the current entering a node is equal to the current leaving it. In this case, the node is where R2 and R3 are connected. Using the formula: $$I_T = I_2 + I_3$$ Substituting the values, we get: $$1.61 = I_2 + I_3$$ <

We can also use Ohm's law to find the current in each resistor. For R2, we have:<

$I_2=\fracV_2R_2$<

We know that V2 is equal to V3 because they are in parallel. We can use Ohm's law again to find V2 and V3. For R_eq, we have:<

$V_eq=I_eq\times R_eq$<

We know that I_eq is equal to IT because they are in series. Substituting the values, we get:<

$V_eq=1.61\times 12$<

Simplifying, we get:<

$V_eq=19.32$<

Therefore, V2 and V3 are both equal to 19.32 volts.<

We can now find I2 by substituting V2 and R2 into Ohm's law:<

$I_2=\frac19.3220$<

Simplifying, we get:<

$I_2=0.966$<

We can now find I3 by subtracting I2 from IT:<

$I_3=I_T-I_2$<

$I_3=1.61-0.966$<

$I_3=0.644$<

We can now find V4 by subtracting V1 and V_eq from V:<

$V_4=V-V_1-V_eq$<

$V_4=100-16.1-19.32$<

$V_4=64.58$<

We can now find I4 by dividing V4 by R4:<

$I_4=\fracV_4R_4$<

$I_4=\frac64.5840$<

$I_4=1.61$<

We can now find VR for each resistor by multiplying IR by R:<

$V_R=I_R\times R$<

$V_2=0.966\times 20$<

$V_2=19.32$<

$V_3=0.644\times 30$<

$V_3=19.32$<

$V_4=1.61\times 40$<

$V_4=64.58$<

We have already found V1 and V_eq earlier.<

Find the power dissipated by each resistor. This can be done by multiplying the voltage by the current of each resistor. Using the formula:<

$P_R=V_R\times I_R$<

For R1, we have:

$P_1=V_1\times I_T$

$P_1=16.1\times 1.61$

$P_1=25.92$

Therefore, the power dissipated by R1 is 25.92 watts.

For R2, we have:

$P_2=V_2\times I_2$

$P_2=19.32\times 0.966$

$P_2=18.66$

Therefore, the power dissipated by R2 is 18.66 watts.

For R3, we have:

$P_3=V_3\times I_3$

$P_3=19.32\times 0.644$

$P_3=12.44$

Therefore, the power dissipated by R3 is 12.44 watts.

For R4, we have:

$P_4=V_4\times I_T$

$P_4=64.58\times 1.61$

$P_4=104$

Therefore, the power dissipated by R4 is 104 watts.