To verify Kirchhoff’slaws forD.C. Circuits

Experiments: First Year Electrical Engineering Lab [EEP151] 2019-20

EXPERIMENT NO: 1

Aim:To verify Kirchhoff'slaws forD.C. Circuits

Sr No. Apparatus

Range /raMtinagke

1 D.C Voltage source

Make

2 Rheostat

3 Ammeters(DC)

4 Voltmeter(DC)

RHS

1. Theory:Kirchhoff's laws

The two laws given by Gustav Robert Kirchhoff (1824-1887) form the fundamental principals used in writing circuit equations. These laws relate to the topology (i.e. the way the circuit elements are connected) of the circuit. The laws do not depend on the nature of the elements of the circuit.

A) Kirchhoff's Current Law (KCL): First law of Kirchoff's

It is also known as Kirchhoff's first law. It states that the algebraic sum of currents meeting at a junction in a circuit is zero. If there are k no of branches meeting at a junction (also called a node), then

=0

=1

Note: this law is just a restatement of principal of conservation of charge. Since charges can not accumulate at a junction , the amount of charge entering at a instant must be same as the amount of charge leaving it Explanation of KCL: Suppose some conductors are meeting at a point/node "A" as shown in fig 1.a. In some conductors, currents are incoming to the point "A" while in other conductors, currents are leaving or outgoing from point "A".Current entering the node shall be considered to be positive (+) whereas current leaving the node shall be considered as negative(-)". then with respect to fig-1a

I1 + (-I2) + (-I3) + (-I4) + I5 = 0 OR I1 + I5 -I2 -I3 -I4 = 0 OR I1 + I5 = I2 + I3 + I4

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Experiments: First Year Electrical Engineering Lab [EEP151] 2019-20

i.e. Incoming or Entering Currents = leaving or Outgoing Currents OrI Entering = I Leaving

For instance, 8A is coming towards a point and 5A plus 3A are leaving that point in fig 1.b, therefore, 8A = 5A + 3A 8A = 8A.

B)Kirchhoff's Voltage Law(KVL): Second law of Kirchhoff's

It states that at any instant the algebraic sum of voltages around a closed loop or circuit is zero.

For a closed loop having k elements = Vj = 0

=1

In other words it can be defined as in a closed loop/closed path (or circuit) in a network, the algebraic sum of the IR product is equal to the EMF in that path.

Explanation of KVL:Sign Conventions a) Battery e.m.f.:

A rise in voltage should be given a + ve sign and a fall in voltage a ?ve sign. Keeping this in mind, it is clear that as we move from negative terminal of source to positive terminal, there is a rise in potential, hence voltage E1 should be given a +ve sign. If, on the other hand, we move from +ve terminal to ?ve terminal of voltage source, then there is a fall in potential, hence it is to be considered as ?ve therefore E2 is given a negative sign.

b) Sign of IR Drop: -

Whenever we move in the direction of current there is a drop in voltage. Since the current always flows from point at higher potential to the point at lower potential. Hence Voltage drop in the current direction is taken as ?ve. However, if we go in a direction opposite to that of the current, then there is a rise in voltage.Hence Voltage drop in the opposite current direction is taken as +ve.

Consider a closed circuit is shown in fig below which contains two emf sources E1and E2. The overall sum of E.M.F's of the batteries is indicated by E1-E2. The assumed direction of current is also shown in the fig.E1 drive the current in such a direction which is supposed to be positive while E2 interfere in the direction of current (i.e. it is in the opposite direction of the assumed direction of

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Experiments: First Year Electrical Engineering Lab [EEP151] 2019-20

current) hence, it is taken as negative. The voltage drop in this closed circuit depends on the product of resistance and Current.

In the above fig, I1R1 and I2R2 is positive voltage drop and I3R3 and I4R4 are negative V.D. If we go around the closed circuit (or each mesh), and multiply the resistance of the conductor and the flowing current in it, then the sum of the IR is equal to the sum of the applied EMF sources connected to the circuit.

The overall equation for the above circuit is:

E1-E2 = i1R1 + i2R2 ? i3R3 ? i4R4

OR

E1-E2 - i1R1 - i2R2 + i3R3 + i4R4 = 0

2. Procedure:

1. Connect the circuit as per circuit diagram 2. Keep all the rheostats at maximum position. 3. Switch on the DC voltage supply, adjust the voltage to any suitable value(e.g.15V,20Vand 30V) 4. Take the reading of all the ammeters. 5. Measure the voltages across all the rheostats. 6. Change the voltage of power supply and repeat step (5) and (6). 7. Keeping the voltage constant, Change the position of rheostats and repeat step 4 and 5. 8. Verify Kirchhoff's laws.

3. Result & Conclusion:

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Experiments: First Year Electrical Engineering Lab [EEP151] 2019-20

4. Discussion Questions:

1. What is the internal resistance of the ideal voltage source and ideal current source? 2. Define a node, a branch and a loop. 3. What are the limitations of Kirchhoff's law? 4. For a given experimental set up, calculate the total power dissipated in the circuit and total

power supplied .Comment on the result. 5. Find the values of i2, i4and i5 if i1= 3A, i3= 1A and i6=1A

100 V

i1

i5

15 ohm

i6

5 ohm

i3

i2 10 ohm

30 ohm i4

20 ohm

6. Find the value of V if V1= 20 V and the value current source is 6 A.

R

2 ohm

+

+

V1

Y10ouorhrmoll no in ohms V2

V 5 ohm

-

-

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Experiments: First Year Electrical Engineering Lab [EEP151] 2019-20

LHS with Pencil

EXPERIMENT NO: 1

Aim:To verify the Kirchhoff's law for the given DC network

Sr No. 1 2 3 4

Apparatus D.C Voltage source Rheostat Ammeters Voltmeter

Range /raMtinagke

Make

1. Circuit diagram:

2. Observation table:

Sr. Source Voltage

Voltage

Voltage

No. Voltage Across R1 AcrossR2 Across R1 I1 I2 I3

V

VR1

VR2

VR3

1 2 3

3. Calculations:

i. Verification of KCL and KVL ii. Error calculation for current and voltage.

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