Electrical And Electronics Engineering: May 2016

Saturday, 21 May 2016

D.C Supply (9v & 12v)

In this kit we are using a.c. supply 230v single phase to convert it into d.c. supply by the circuit which is given below.

In this circuit we are using transformer to step down the a.c. supply which is centre tapped and a full bridge rectifier is connected which is conveting it into d.c. supply.
The positive cycle will be fed to the positive half cycle i.e. ic 7812 and the negative half cycle will be fed to the negative half cycle i.e. 7912. Thus, output is obtained.

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Voltage Regulator

Voltage is regulated by this circuit . A.c single phase 230v supply is used. A step down transformer 230/18v is used which is converting 230v supply into 18v. A full wave bridge rectifier is using to convert that step down a.c. supply into d.c supply.

The positive half cycle is going to the positive terminal and and negative half cycle is goint to the negative terminal.

In positive terminal there is a ic 7812 which gives the positive output voltage and in negative terminal there is ic 7912 which gives negative output voltage.

On the positive terminal there is another ic LM317 which two out of three terminals i.e. output and adjust are shprt circuited which gives the regulated output voltage and its input terminal is connected to the output terminal of ic7812.

Hence regulated output voltage is obtained at the positive terminal of the circuit.

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SuperPosition Theorem

Superposition Theorem

PROCEDURE:-

1- Superposition Theorem:- In a linear lumped bi-lateral network with more than one source of emf, the current flowing through any branch, when both the sources are active is equal to algebraic sum of current that would flow, if one source of emf is kept active and all other sources are dead.
2- The circuit connection are made as shown in diagram 1.
3- The current flowing through load resistor (R2) is noted (I) that is 10.14 mA.

4- The circuit connection are made as shown in circuit diagram 2, and the current through load resistor (R2) is noted (I1) that is 4.30 mA.
5- The circuit connection are made as shown in circuit diagram 3, and the current flowing through load resistor (R2) is noted (I2) that is 5.76 mA.
6- It is observed I = I1 + I2

Calculation
The value of current (I) = 10.01 mA
The value of current (I1) = 4.30 mA
The value of current (I2) = 5.76 mA

I = I1 + I2 = 4.30 + 5.76
= 10.06 mA ≈ I

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SCR Characteristics

Circuit used to plot characteristics of SCR is shown in diagram.

Procedure:-

Make circuit connections as shown in the diagram 1 using patch cords.

1- To plot the V-I characteristics proceed as follows.

2- Rotate both the potentiometer P1 and P2 in fully counter clockwise direction, 3- connect voltmeter to point ‘6’ & ground to read VG and at point ‘3’ & ground to read VAK.

Connect ammeter at point ‘1’ & ‘2’ to indicate the current IA and at point ‘4’ & ‘5’ to indicate the gate current IG.

4- Switch on the power supply.

5- Vary potentiometer P2 to set the gate current IG to a lower value (5.6 mA, 5.7 mA, 5.8 mA).

6- Increase anode voltage VA gradually by varying potentiometer P1.

7- Observe the current IA in the anode circuit. It shows almost zero current at the initial stage.

8- At certain point of positive anode voltage current IA shows sudden rise in reading & voltmeter reading falls down to almost zero. This action indicates the firing of SCR.

9- If this not happens, repeat the procedure from step 5 for slightly higher value of gate current IG.

10- Try the various value of gate current to get the firing of SCR.

11- Keeping gate current constant observe precisely the firing voltage of SCR and record it in the observation table.

12- Also record the anode voltage VA & anode current after firing of the SCR.

13- Plot the graph of VA versus IA.

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Ohm's Law

Ohm’s Law states that at constant temperature the current through a conductor is directly proportional to the potential difference across its ends.
If V is the potential difference and I is the current, then
I Proportional V
V/I = Constant
The constant is the resistance R of the conductor.

Procedure:-

1- Make connections as shown in circuit diagram.
2- Take least count of ammeter and voltmeter.
3- Plug the key. Note the readings of voltmeter and ammeter.
4- Repeat the experiment by adjusting the rheostat.
5- Draw a graph between V and I.

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Verification of Kirchhoff Law

Kirchoff’s Current law:-

(i)This law states that algebraic sum of all current entering a node and leaving a node is equal to zero.

(ii) The circuit connections are made as shown in circuit diagram.

(iii) At node “B” the incoming current is measured.

(iv) The outgoing current I2 and I3 are also measured.

(v) It is found that the sum of all the incoming and outgoing current is zero;

I1=I2+I3

Kirchoff’s voltage law:-

It states that algebraic sum of voltage rise is equal to algebraic sum of voltage rise is equal to algebraic sum of voltage drop in a closed path (circuit).

(ii) The circuit connection are made as shown in circuit diagram

(iii) The voltage across R1 and R2 and R 3 are measured.

(iv) V= VR1+VR2+VR3

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AUTOMATIC LIGHTNING

It is a circuit which works as an automatic switch. It gets “On” & “OFF” automatically according to our desire. It works on the fact that its resistance decreases with increase in incident light i.e. daytime. And its resistance increases with the decrease in incident light i.e. night time.

CIRCUIT DIAGRAM OF LDR

At day time

I1= 230/ (30×103+100)

I1= 7.395 mA

Voltage across 30kΩ

V (30kΩ) = 30×103×7.395×10-3

V= 221.85V

VL=230-221.85V

VL= 8.15V

The Voltage across the capacitor is 8.15V. The capacitor cannot discharge.

(i) There won’t be any pulse to the triac.

(ii)The triac will be in open circuit state no current will flow through the load .The light is ‘OFF’.

At night

I2= 230/(30×103+1×103+100×103)

I2= 1.75mA

Voltage across 30KΩ

Substitute the value of eqn.(ii) in above eqn.

We get

V (30KΩ) = I2R

V (30KΩ) =30×103×1.75×10-5

V (30K Ω) = 52.5V

+VC=230-52.5

VC= 177.5V

The voltage across the capacitor is 177.5v .The capacitor will charge and discharge .

(i) There will be pulse through diac and hence gate pulse to the triac.

(ii) The triac will be in short circuit the lead. The light is “ON”.

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Time constant in charging

The time constant is the time taken by the capacitor(uncharged) to acquire 63% of the voltage being applied to it if it starts with zero volt. Equivalently, it is the time taken by a capacitor to acquire 63% of the difference between its initial voltage and voltage being supplied to it.
Calculating the time constant
Time Constant, t=RXC
R=330Ω and C=100µF
t=330ΩX100µF
=330ΩX100X10-6
=.0033second
Deriving the time constant from charging Curve
If we assume that the capacitor charges at a constant rate, it will add one volt in one second. In practice, this cannot happen. A capacitor adds up more volts in one second. To verify this, we will use experimental observation and plot the graph of voltage across the capacitor versus time .
Vcc=9V, R=1KΩ and C=1000µF
Let us first calculate the value of rc.
RC=1KX1000µF
=1 second
Putting the value of RC and t in equation (A) we get
V=9(1-e-1/1)
=9(1-e-1)
=9(1-1/e)
=9(1-1/2.71)
=9(1-0.36)
=9(0.63)
=63%(9)
Thus the voltage gained by the capacitor in 1 second is equal to 63% of Vcc.

Circuit Explaination

Charge the 100µF capacitor
Insert the wires attached to the meter probes in parallel with the series combination of LED and R. To do this, the red and black wires of the meter are connected to the left leg of R and the negative terminal of the LED.
Connect the charged capacitor in parallel with the series combination of the LED and the resistor in such a way that its positive terminal is connected to the end of the resistor to which the red probe is connected and its negative terminal is connected to the negative terminal of the LED.

Observation:-

1) The meter will show a continuously decreasing voltage reading. The reading becomes steady after sometime.

2) The LED glows and then slowly fades out.

3) The rate of decrease of voltage reading is initially fast but subsequently slows down with time.

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DIAC Characteristics

1- Make circuit as shown in the fig. using patch cords.
2- To plot V-I characteristics proceed as follows.
3- Rotate both the potentiometer P1 fully counter clockwise direction.
4- Connect voltmeter across point ‘3’ & ground to read voltage VA.
5- Connect ammeter between point ‘1’n& ‘2’ to indicate the current IA.
6- Switch ‘ON’ the power supply.
7- Put the +35 v switch ‘ON’.
8- Vary the potentiometer P1 so as to increase the value of DIAC voltage VA and measure the corresponding values of current IA in an observation table1.
9- Plot the curve between +VA and +IA
10- Rotate potentiometer P1 fully in counter clockwise direction.
11- Switch ‘OFF’ the power supply.
12- Put the switch towards -35V
13-Switch ‘ON’ the power supply.
14- Vary the potentiometer P1 so as to increase the value of DIAC voltage VA and measure the corresponding values of current IA in an observation table.
15- Plot the curve between -VA and –IA.

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Thevenin’s Theorem

Theorem:- In a linear lumped bi-lateral network, the current flowing through any branch or load resistor(R1) can be obtained by reading the circuit into a voltage source(VTM) and a series connected resistor(RTH).
IL= VTH/RTH+RL
2) The circuit connection are given as shown in the circuit diagram.

3) The Switches circuit SW1, SW2, and SW3 are closed and the current flowing through load resistor(RL) is noted .

4) The load resistor removed by opening switch SW3. The open circuit voltage is measured across terminal AB.

5) Short circuit the terminal AB and short circuit current ISC is noted. The value of RTHis given by RTH= VOC/ISC

6) The current flowing through the load resistor RL is calculated by using the thevenin’s formula.
IL=VTH/RTH+RL

7) By verifying the value of resistance by R1,R2 and R3 different set of reading obtained and are tabulated.

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