TUTORIAL- 1 - ENGLISH GRAMMAR

Grammar  rules  are  created  in  order  to  give  structure  to  the  language.  Rules  reflect  the  usual

behaviour  of  a  grammatically  constructed  pattern.  A  rule  does  not  necessarily  have  to  behave

absolutely the same every time, but will behave according to the rule  most of the time.

SENTENCE STRUCTURE

                       SUBJECT - VERB - COMPLIMENT - MODIFIER
                       Eg: MANU - WROTE - A STORY - LAST WEEK

Subject

• The subject is the actor of a sentence in active voice. It is the person or thing that performs, or is responsible for, the action of the sentence. It usually begins the sentence and precedes the verb.
•  Every sentence in English must have a subject. Commands will not have a visible subject, however, the subject [you] is understood. Example: Run quickly! = You run quickly!

The subject can be a single noun.

1. Cats chase mice.

2. Children  like candy.

The subject can also be a noun phrase, which is a group of words ending with a noun. A noun

phrase CANNOT begin with a preposition.

1. The car  is in the garage.

2. That hot red dress  looks fabulous.

Examples of subjects:

1. Tom  likes to go fishing.
2. The English teacher  is a very nice person.
3. Susan and Alex went to the movie together.
4. Those boys are good basketball players.
5. We actors are a happy group.

Verb

• The verb follows the subject when the sentence is declarative. A verb normally reveals the action of the sentence. Every sentence in English must have a verb.
• The verb can be a single word.

1. Mary likes chocolate cake.
2. They play soccer.

• The verb can also be a verb phrase. A verb phrase contains one, or more, auxiliary verbs and one main verb.
• The main verb is always preceded by the auxiliary verbs.

1. Martha has been talking to her new friend.
2. Terry is visiting  his aunt today.

Examples of verbs and verb phrases:

1. Jerry has returned from lunch.
2. The storm made a lot of noise.
3. George is playing  in a tournament tomorrow.
4. I will go to bed soon.
5. Sally was jealous of Lisa’s new dress.

Complement (Object)

• A complement (object) provides more information about the verb. Often, it consists of a noun, or noun phrase, and will usually follow the verb in a sentence relaying active voice.
• A complement (object) CANNOT begin with a preposition
• A complement (object) answers the question what? or whom?

Examples of complements:

1. Jack threw the stone far.        (What did Jack throw?)
2. The hungry bird ate a worm.        (What did the bird eat?)
3. He called Janice after the party.      (Whom did he call?)
4. She was chewing gum  in class.      (What was she chewing?)
5. The ball hit Mike during the game last night.  (Whom did the ball hit?)

Modifier

• A modifier tells the time, place, or manner of action. The modifier usually follows the complement. Not every sentence requires a modifier.
• Prepositional phrases are commonly used as modifiers

Examples of prepositional phrases:

under the house, after breakfast, in the morning

 

 

Adverbs and adverbial phrases are also used as modifiers, or modifiers of time. A modifier of

time will usually come last when more than one modifier is used.

 

Examples of adverbs and adverbial phrases:

yesterday, quickly,  last semester, overhead, quite awful

 

A modifier answers the question of where? When? or how?

 

Examples of modifiers:

1. She is earning her degree at Cornell University.   (Where is she earning her degree?)
2. John fell down the stairs        (Where did John fall?)
3. yesterday.          (When did John fall?)
4. The cheetah was running quite fast.      (How was the cheetah running?)
5. We have an appointment at ten o’clock tomorrow.   (When do we have an appointment?)
6. The soldier fired the gun repeatedly .       (How did the soldier fire the gun?)

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TUTORIAL 4 - Kirchoff's Law Lab Practical

                                               Kirchoff's Law - Lab Practical
                   
                                    Note: You have to enable Java to use the Lab

TUTORIAL 3 - Thevenin Equivalent analysis with Practical Lab

This is a Lab Tutorial, So kindly use the Tool provided for better understanding

Note: Enable Java for using the Lab

Thevenin Equivalent

Here is a problem in which it's advantageous to use a Thevenin equivalent circuit. We want to determine the value of R_L that will cause i_b = 2 mA. (This is problem P5.5-12 from Introduction to Electric Circuits, 5e by R.C. Dorf and J.A. Svoboda.) We will solve this problem by finding a Thevenin or Norton equivalent circuit for the part of the circuit that is left of the terminals and then replacing that part of the circuit by its Thevenin or Norton equivalent circuit.

The row of buttons below the circuit correspond to various calculations that are used to find or to use a Thevenin or Norton equivalent circuit.

Thevenin and Norton equivalent circuits involve three parameters:

• V_oc, the open circuit voltage
• I_sc, the short circuit current
• R_th, the Thevenin Resistance

The buttons labeled "Voc", "Isc" and "Rth" show how to find a Thevenin or Norton equivalent circuit for the part of the circuit that is left of the terminals:

1. Left-click on the "Voc" button to see the circuit used to determine V_oc, the open circuit voltage. The voltmeter measures the value of V_oc in volts. Analyze this circuit, e.g. by writing and solving a mesh equation, to verify the value of V_oc.
2. Left-click on the "Isc" button to see the circuit used to determine I_sc, the short circuit current. The ammeter measures the value of I_sc in mA. Analyze this circuit, e.g. by writing and solving a mesh equation, to verify the value of I_sc.
3. Left-click on the "Rth" button to see the circuit used to determine R_th, the Thevenin Resistance. The current source causes the current in the equivalent resistance to be 1 mA. The voltmeter measures the value of the voltage across the equivalent resistance in volts. Thus Analyze this circuit, e.g. by writing and solving a mesh equation, to verify the value of R_th.

As expected (Section 5.5 of Introduction to Electric Circuits by RC Dorf and JA Svoboda), V_oc, I_sc and R_th are related by

V_oc = R_th * I_sc

The buttons labeled "Thevenin" and "Norton" show the circuits that result when the part of the circuit that is left of the terminals is replaced by its Thevenin or Norton equivalent circuit.

Left-click on the "Thevenin" button to see the circuit that we get when we use the Thevenin equivalent circuit. Analysis of this circuit shows

Next, left-click on the "Norton" button to see the circuit that we get when we use the Norton equivalent circuit. Analysis of this circuit shows

Of course, we expect both of these calculations to yield the same value of R_L.

Finally, left-click on the "circuit" button to return to the original circuit. Verify that setting R_L = 0 does indeed cause i_b=2 mA.

TUTORIAL 2 - Inductor

Inductor 

 

An inductor is an element which stores a magnetic field. An inductor is
a wire coiled around a material called a core. The core is typically made of 
a magnetic material however the core can be anything from a toilet paper roll to a piece of wood.
(The behavior of the inductor IS effected by the core material though.)

Recall from Physics that static charge sets up an electrostatic force. Also recall that
moving charge sets up a magnetic field called flux around the current-carrying wire.

If the wire is then coiled as shown below, then the flux increases linearly with every coil turn.




The flux is proportional to the current flowing through the inductor. 
We can use the following equation where L is the inductance of the inductor.(L is constant)

Flux = Li

Through experimentation it was found that if the flux changes then a voltage is induced.
If the flux is constant then there is no voltage across the inductor.

V = d(Flux)/dt
V = d(Li)/dt


V = L di/dt
When the current is changing a voltage is induced across the inductor.
This voltage opposes the change in the current.
When the current stops changing and stays constant, then the voltage collapses.
Therefore when conditions are constant (unchanging) the inductor appears like a
wire or short circuit.

Current is continuous in an inductor. In other words, current can not
change instantaneously. If current could change instantaneously then an infinite
voltage would be induced across the capacitor. Infinite voltage is a practical impossibility. 
Therefore current can not change instantaneously through an inductor.
Voltage across an inductor can change instantaneously.

The energy stored in an inductor is:

E = 1/2 Li²

Using the above concepts, let's analyze the following circuit:

This circuit has both a switch and an inductor: The switch closes at t=0 The switch is open for t<0 and is closed for t>0. This can be seen by inspecting the switch's arrow direction. We will not try to analyze the circuit at t=0 since the circuit's state at t=0 is unkonwn. Instead we will look at the circuit at t=0- (the time right before the switch moves) and t=0+ (the time right after the switch moves). In this problem it is given that i6 at t=0- is 2A With this information find the following things at t=0- and t=0+: The current through the inductor iL The current through the 6 ohm, i6 diL/dt for the inductor

Here are snapshots of the circuit at t=0- and t=0+: _______________________________________________________ t = 0-: At t=0- the switch is open. Since we know that i6 is given to be 2A, we can apply KCL to the marked node to find iL. Σ  ientering = Σ  ileaving 5A = i6 + iL 5A = 2A + iL iL = 3A at t=0-. Now to find diL/dt: diL/dt = VL/L We need to find VL first: Apply KVL to the right-most loop: VL + 4iL - (2A)(6 Ohms) = 0 VL + 4(3A) - 12 = 0 VL = 12 - 12 VL = 0 diL/dt = VL/L = 0/2 = 0 _______________________________________________________ t = 0+: iL(0-) = iL(0+)  This is always true! Therefore iL(0+) = 3A Current through an inductor is continuous  We also need to fine i6: Note that the 6 Ohm resistor has been shorted out by the closed switch. A resistor in parallel with a wire has zero voltage across is. If the 6 Ohm resistor has zero volts across it then the current through the 6 Ohm has to be zero due to Ohm's Law, V=IR. Thus i6(0+) = 0 Now to find diL/dt: diL(0+)/dt = VL(0+)/L We need to find VL(0+) first: VL + 4iL - 0 = 0 VL + 4(3A) = 0 VL = -12 diL/dt = VL/L = -12/2 = -6A/s

TUTORIAL 1 - Capacitor

Capacitor

A Capacitor is an element which stores charge. It is comprised of two
conducting plates sepparated by a non-conducting material called a dielectric.

For every + unit charge put on one plate, there is an equal - unit charge on the
the other plate. Thus the entire capacitor is charge neutral. Since the + and - charges are
separated by a dielectric there is an electric field (or voltage) across the capacitor.

If positive charge is put on one plate then work is done because like-charges repel each other.Recall that voltage is related to work done on charge by the equation V=J/C.

The more charge on the capacitor's plates, the more work had to be done to put the charge there,
and the higher the voltage across the capacitor.

In this class all capacitors will be considered linear and will obey the following
charge-voltage relationship:

q = CV
If this equation is differentiated we get:dq/dt = C(dV/dt)

or      i = C(dV/dt) 

The capacitor on the left is charging, therefore the voltage is increasing.
The capacitor on the right is actually discharging, therefore the voltage is decreasing.
Note the sign difference.


If voltage is not changing then dV/dt=0 and i=0. So in static (unchanging conditions) 
the capacitor behaves like an 'open circuit' since no current flows.
Charge on a capacitor can not instantly leave. It takes some finite amount of time
for the charge to leave the plate. Therefore charge on the capacitor is continuous.
Since V = q/C then voltage across a capacitor is also continuous.
(i.e. The voltage across a capacitor can never change instantaneously.)

However the current can change instantaneously since i = C dV/dt.
This is because current is the rate of charge moving over time.

Energy stored in a capacitor is:
E = 1/2 CV²

Using the above concepts, let's analyze the following circuit:

This circuit has both a switch and a capacitor: The switch opens at t=0 The switch is closed for t<0 and is open for t>0. This can be seen by inspecting the switch's arrow direction. We will not try to analyze the circuit at t=0 since the circuit's state at t=0 is unkonwn. Instead we will look at the circuit at t=0- (the time right before the switch moves) and t=0+ (the time right after the switch moves). In this problem it is given that V4 = 8V at t=0-.  With this information find the following things at t=0- and t=0+: The voltage across the capacitor. The current i1 The current in the capacitor (going down) dV/dt in the capacitor

Here are snapshots of the circuit at t=0- and t=0+: _______________________________________________________ t = 0-: At t=0- the switch is closed. Since V4 = 8V and  and the capacitor is in parallel with V4, then VC = 8V at t=0-. To determine i1 we need to find the voltage across the horizontal 4 ohm resistor. To find this, we will apply KVL: -20V + 4i1 + 8V = 0 4i1 = 20 - 8 = 12V i1 = 12/4 = 3A To find the current in the capacitor we have to apply KCL at the marked node: i1 = i + iC 3A = 2A + iC iC = 1A dV/dt = iC/C = 1A/(.25F) = 4V/s _______________________________________________________ t = 0+: At t=0+ the switch is open. Since the voltage across a capacitor can not change instantly, then VC(0-) = VC(0+)Therefore VC(0+) = 8V Since the switch is now open no current flows through the horizontal resistor, so i1 = 0 To find the current in the capacitor we have to apply KCL at the marked node: i1 = i + iC 0A = 2A + iC iC = -2A dV/dt = iC/C = -2A/(.25F) = -8V/s dV/dt is negative, therefore for t>0 the capacitor is discharging. The energy stored in the capacitor is being absorbed by the resistor. Eventually all the initial energy stored in the capacitor will be absorbed by the resitor.

PIC MICROCONTROLLER AND EMBEDDED SYSTEMS - M.A.MAZIDI

PIC Microcontroller and Embedded Systems offers a systematic approach to PIC programming and interfacing using the Assembly and C languages. Offering numerous examples and a step-by-step approach, it covers both the Assembly and C programming languages and devotes separate chapters to interfacing with peripherals such as timers, LCDs, serial ports, interrupts, motors and more. A unique chapter on the hardware design of the PIC system and the PIC trainer round out coverage, while text appendices and online support make it easy to use in the lab and classroom.

Features

• Systematic coverage of the PIC18 family of microcontrollers.
• Coverage of C language programming of the PIC18-starting from Chapter 7.
• Chapters (9-17) on programming and interfacing the PIC with peripherals.
• An entire chapter (Chapter 8) dedicated to the design of the PIC Trainer.

Table Of Contents

• The PIC Microcontrollers: History and Features
• PIC Architecture & Assembly Language Programming
• Branch, Call, and Time Delay Loop
• PIC I/O Port Programming
• Arithmetic, Logic Instructions, and Programs
• Bank Switching, Table Processing, Macros, and Modules
• PIC Programming in C
• PIC18F Hardware Connection and ROM Loaders
• PIC18 Timer Programming in Assembly and C
• PIC18 Serial Port Programming in Assembly and C
• Interrupt Programming in Assembly and C
• LCD and Keyboard Interfacing
• ADC, DAC, and Sensor Interfacing
• CCP and ECCP Programming
• Radio wave Propagation
• SPI Protocol and DS1306 RTC Interfacing
• Motor Control: Relay, PWM, DC, and Stepper Motors 

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