D.C Supply (9v & 12v)

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 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

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

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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