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Basic Electronics

Laboratory

Submitted by: VADLAPUDI L S V RATNA CHOWDARY Roll No: 19CE

LAB REPORT (Experiment-1)

Basic Circuit Theorems

Kirchhoff’s Voltage Law (KVL)

Objective:

To set up a circuit to verify Kirchhoff’s Voltage Law (KVL).

Components Used:

SL.NO APPARATUS SPECIFICATIONS QUATITY

1. Breadboard - 1
2. Digital Multimeter - 1
3. Connecting wires - -
4. Voltage Source 10v 1
5. Resistor 5.7 k Ohms 1
6. Resistor 2.8 k Ohms 3
7. Resistor 3.7 k Ohms 2

Brief Theory:

KVL states that the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero. This is because a circuit loop is a closed conducting path, so no energy is lost. It is basically, conservation of energy as moving around a closed loop. For Example, in the above circuit diagram R 1 , R2, R 3 are three resistors connected in series across a voltage source V. V 1 , V2, V 3 are the voltage drops across resistors R 1 , R2, R 3 respectively, and assuming that current is flowing in clock wise direction. So now according to Kirchhoff’s Voltage Law V + (-V 1 ) + (-V 2 ) + (-V 3 ) = V - V 1 - V 2 - V 3 = V = V 1 + V 2 + V 3

Observation Table:

*All the units of below values are in volts

For the Loop A-B-C-D-E-A V(A) V(B) V(C) V(D) V(E) 10 4.08356 2.72238 1.36119 0 For the Loop B-F-E-D-C-B V(B) V(F) V(E) V(D) V(C) 4.08356 2.04178 0 1.36119 2. For the Loop A-B-F-E-A V(A) V(B) V(F) V(E) 10 4.08356 2.04178 0

Calculations:

*All the units of below values are in volts

For the Loop A-B-C-D-E-A V(AB) V(BC) V(CD) V(DE) V(EA) ALGEBRAIC SUM

• 5.91644 - 1.36118 - 1.36119 - 1.36119 10 0 For the Loop B-F-E-D-C-B V(BF) V(FE) V(ED) V(DC) V(CB) ALGEBRAIC SUM
• 2.04178 - 2.04178 1.36119 1.36119 1.36118 0 For the Loop A-B-F-E-A V(AB) V(BF) V(FE) V(EA) ALGEBRAIC SUM
• 5.91644 - 2.04178 - 2.04178 10 0

Conclusion:

From the above Experiment I got that Algebraic sum of Voltages for all the 3 Loops which are taken in the positive direction as “0”. This Indicates that the Kirchhoff’s Voltage Law (KVL) is Verified.

Kirchhoff’s Current Law (KCL)

Objective:

To set up a circuit to verify Kirchhoff’s Current Law (KCL).

Components Used:

SL.NO APPARATUS SPECIFICATIONS QUATITY

1. Breadboard - 1
2. Digital Multimeter - 1
3. Connecting wires - -
4. Voltage Source 5 v 1 5. Voltage Source 7v 1
5. Resistor 2 k Ohms 2
6. Resistor 4.3 k Ohms 3

Brief Theory:

Kirchhoff’s Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a junction must equal zero”. This is because it has no other place to go as no charge is lost. It is basically, conservation of charge entering and leaving a junction. From the above circuit, using KCL we can write, by assuming that currents away from the Junction as Positive. i 1 + (-i 2 ) + (-i 3 ) + i 4 = 0

Observation Table:

Here all the currents are taken that they flow away from the Junction A, i.e. currents flowing away from junction are taken as positive.

*All the units of below values are in ampere

I(R 1 ) I(R 2 ) I(R 3 ) I(R 4 ) I(R 5 )

Calculations:

According to KCL the algebraic sum of all currents entering and exiting a junction must equal zero. i.e. From the above circuit I(R 1 ) + I(R 2 ) + I(R 3 ) + I(R 4 ) + I(R 5 ) = 0 (-0.000732877) + 0.000821918 + 0.000821918 + 0.000821918 + (-0.00173288) ≈ 0 Note: Here even though I am using LT Spice stimulator (i.e. ideal situation) I didn’t get exact 0, because the decimals are exceeded the range of stimulator.

Conclusion:

From the above experiment I came to know that the algebraic sum of all currents entering and exiting a junction must equal zero, i.e. charges entering and leaving the junction are being conserved, so finally the Kirchhoff’s Current Law is verified.

Superposition Principle

Objective:

To find Current passing through a Resistor in the given circuit using Superposition Principle

Components Used:

SL.NO APPARATUS SPECIFICATIONS QUATITY

1. Breadboard - 1
2. Digital Multimeter - 1
3. Connecting wires - -
4. Voltage Source 5 v 1 5. Voltage Source 7v 1
5. Resistor 2 k Ohms 2
6. Resistor 4.3 k Ohms 3

Brief Theory:

For any linear and bilateral networks, consisting of two or more Independent Source (Voltage / Current Source), then current through (or voltage across) an element is the algebraic sum of the currents through (voltages across) that element caused by each independent source acting alone with all other sources are replaced by their internal resistances. i.e. replace the voltage sources with short circuit and replace the current source with the open circuit.

Circuit Diagram:

Step-2: Secondly, I had considered the effect of only V 1 source, i.e. I had short

circuited the V 2 Voltage source, and the current is measured as I 2.

Step-3: Finally, I had considered effect of all the sources, and the current is

measured as IR3.

Observation Table:

I 1 I 2 IR

Calculation:

I 1 + I 2 = 0.000479452 + 0.

= 0.000821918 = IR

Now according to Superposition Principle, total current passing through R 3 is I 1 + I 2 , which is nothing but equal to IR

Conclusion:

From this experiment I had calculated the current passing through R 3 Resistor using Superposition Theorem, and even I had calculated the current through normal method, and both methods gave same output. i.e. Superposition Theorem is verified.

Circuit Diagram:

Stimulation Results:

First, measure the current through the path between points A and B (i.e. through R2). Note this as Iref (reference measurement).

Now, disconnect R2 from the circuit, i.e. there should be an open circuit in place of R2. And measure the voltage between points A and B. This gives the Thevenin’s voltage Vth. Now, deactivate the two voltage sources 5V and 10 V. Keeping R2 disconnected, measure the resistance between points A and B. This gives the Thevenin’s resistance Rth.

So now Ical = 0. And now Imeas, 1 is calculating the current passing through the resistor R 2 using the information of Thevenin’s equivalent Imeas, 1 = Vth / (Rth + R 2 ) = 5.87678 / ((1.270142878+5.7) * ( 3 )) = 0. Iref Ical Imeas, 1 Imeas, 2 0.000843136 0.0008431547647 0.0008431362316 0.

Conclusion:

From the above Experiment I found that current passing through resistor R 2 is approximately 0.000843136, and by calculating the current value directly from the circuit gives the almost same value of current measured by Thevenin’s Theorem. So Thevenin’s Theorem is verified. And when I calculated the current through pen and paper, there is more deviation of current from the current measured through stimulator.