KVL & KCL

This is a ppt of a college project of the topic kvl and kcl ..do read this..i have such interest in science projects and do maake aa lot of money by doing freelancing in embedded system so make sure to check this ppt for more updtes

1. KVL & KCL Name- SAIKAT BISWAS University roll no- 14200321028 Meghnad Saha Institute of Technology 1

2. 2 Terminology: Nodes and Branches Node : A point where two or more circuit elements are connected Branch : A path that connects two nodes

3. 3 Kirchhoff’s Laws Kirchhoff’s Current Law (KCL) : The algebraic sum of all the currents entering any node in a circuit equals zero. (An expression of the conservation of charge.) Kirchhoff’s Voltage Law (KVL) : The algebraic sum of all the voltages around any loop in a circuit equals zero. (As a result of conservation of energy.)

4. 4 Alternative Formulations of Kirchhoff’s Current Law (Charge stored in node is zero.) Formulation 1 : Sum of currents entering node = sum of currents leaving node Formulation 2 : Algebraic sum of currents entering node = 0 Currents leaving are included with a minus sign. Formulation 3 : Algebraic sum of currents leaving node = 0 Currents entering are included with a minus sign.

5. Explanation of KCL: Suppose some conductors are meeting at a point “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”. Consider the incoming or entering currents as “Positive (+) towards point “A” while the leaving or outgoing currents from point “A” is “Negative (-)”. then: I 1  + (– I 2 ) + (– I 3 ) + (– I 4 ) +  I 5  = 0 OR I 1  +  I 5  –  I 2  –  I 3  –  I 4  = 0 OR I 1  +  I 5  =  I 2  +  I 3  +  I 4  = 0 i.e. Incoming or Entering Currents = Leaving or Outgoing Currents Or Σ I  Entering = Σ I  Leaving 5

6. Kirchhoff’s Current Law, KCL 6 KCL at node 2: Shade indicates a node At node 1, V 1 =3 v. due to source At node 0, V =0 v. due to ground KCL at node 3:

7. 7 Formulations of Kirchhoff’s Voltage Law (Conservation of energy) Formulation 1 : Sum of voltage drops around loop = sum of voltage rises around loop Formulation 2 : Algebraic sum of voltage drops around loop = 0 Voltage rises are included with a minus sign. Formulation 3 : Algebraic sum of voltage rises around loop = 0 Voltage drops are included with a minus sign.

8. Explanation of KVL A closed circuit is shown in fig which contains two  connections of batteries  E 1  and E 2 . The overall sum of E.M.F’s of the batteries is indicated by E 1 -E 2 . The imaginary direction of current is also shown in the fig. E 1  drives the current in such a direction which is supposed to be positive while E 2  interferes in the direction of current (i.e. it is in the opposite direction of the supposed direction of current) hence, it is taken as negative. The voltage drop in this closed circuit depends on the product of  Voltage and Current . The  voltage drop  occurs in the supposed direction of current is known as Positive voltage drop while the other one is negative voltage drop. In the above fig, I 1 R 1  and I 2 R 2  are positive voltage drops and I 3 R 3  and I 4 R 4  are negative V.D. 8

9. Example Resistors of R 1 = 10Ω, R 2  = 4Ω and  R 3  = 8Ω are connected up to two batteries (of negligible resistance) as shown. Find the current through each resistor. Assume currents to flow in directions indicated by arrows. Therefore, current in mesh ABC =  i 1 Current in Mesh CA =  i 2 Then current in Mesh CDA =  i 1  –  i 2 Now, Apply KVL on Mesh ABC, 20V are acting in clockwise direction. Equating the sum of IR products, we get; 10 i 1  + 4 i 2  = 20   …   (1) In mesh ACD, 12 volts are acting in clockwise direction, then: 8( i 1  –  i 2 ) – 4 i 2  = 12 8 i 1  – 8 i 2  – 4 i 2  = 12 8 i 1  – 12 i 2  = 12   …   (2) 9

10. Multiplying equation (1) by 3; 30 i 1  + 12 i 2  = 60 Solving for  i 1 30 i 1  + 12 i 2  = 60 8 i 1  – 12 i 2  = 12 ___________ 38 i 1  = 72 i 1  = 72 ÷ 38 =  1.895 Amperes  = Current in 10 Ohms resistor Substituting this value in (1), we get: 10 (1.895) + 4 i 2  = 20 4 i 2  = 20 – 18.95 i 2  =  0.263 Amperes  = Current in 4 Ohms Resistors. Now, i 1   –   i 2 = 1.895 – 0.263 =  1.632 Amperes 10

11. Applications of Kirchhoff’s Laws Kirchhoff’s laws can be used to determine the values of unknown values like  current  and  Voltage  as well as the direction of the flowing values of these quintets in the circuit. These laws can be applied on any circuit* (See the limitation of Kirchhoff’s Laws at the end of the article), but useful to find the unknown values in complex circuits and networks. Also used in Nodal and Mesh analysis to find the values of current and voltage. Current through each independent loop is carried by applying KVL (each loop) and current in any element of a circuit by counting all the current (Applicable in Loop Current Method). Current through each branch is carried by applying KCL (each junction) KVL in each loop of a circuit (Applicable in Loop Current Method). Kirchhoff’s Laws are useful in understanding the transfer of energy through an electric circuit. 11

12. THANK YOU 12