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### Kirchhoff's laws With Examples

1. 1 Mohammad Waqas (19CH18) Presentation Topic: Kirchhoff's Laws Subject: Basic ELECTRICAL TECHNOLOGY Date: 24/7/2020
2. Kirchhoff's Laws  Kirchhoff's laws quantify how current flows through a circuit and how voltage varies around a loop in a circuit  There are two laws  Kirchhoff’s Current Law (KCL) or First Law  Kirchhoff’s Voltage Law (KVL) or Second Law Kirchhoff’s Current Law (KCL) or First Law  The total current entering a junction or a node is equal to the charge leaving the node as no charge is lost 2
3.  Put differently, the algebraic sum of every current entering and leaving the node has to be null. This property of Kirchhoff law is commonly called as Conservation of charge wherein, I(exit) + I(enter) = 0.  In the above figure, the currents I1, I2 and I3 entering the node is considered positive, likewise, the currents I4 and I5 exiting the nodes is considered negative in values. This can be expressed in the form of an equation:  I1 + I2 + I3 – I4 – I5 = 0 Figure 3
4.  Node  It is the point in a circuit at which at least two elements (active or passive) are joined.  Junction  It is the point in a circuit at which at least three elements (active or passive) are joined.  Note:-A junction must be a node but a node may or may not be a junction 4 Loop  A loop current is defined as a constant current that flows around a closed path or loop. (A closed path is a path through the network that ends where it starts.)
5. Kirchhoff’s Voltage Law (KVL) or Second Law  According to Kirchhoff’s Voltage Law,  The voltage around a loop equals to the sum of every voltage drop in the same loop for any closed network and also equals to zero.  Put differently, the algebraic sum of every voltage in the loop has to be equal to zero and this property of Kirchhoff’s law is called as conservation of energy. 5 Figure
6. 6  Kirchhoff's Law Rules
7.  Example Of KCL  Calculate the value of current I in the section of networks shown in figure At junction A : 𝐼2 + 8 = 15 ; 𝐼2 =7A At junction D : 𝐼3 + 5 = 8 ; 𝐼3 = 3A At junction B : 𝐼2 + 3 =𝐼4; 𝐼4 =10A At junction C : 𝐼1= 𝐼3 + 𝐼4 ; 𝐼1 =13 A  Solution  Figure 7
8.  EXAMPLE OF KVL  Find out V1 and V2 using KVL  Solution  -20 + 𝑉1+ 𝑉2 = 0 ….. Equation 1  𝑉1= 2 і  𝑉2 = 3 i  -20 + 2 i + 3 i = 0  5 i = 20  I = 4A  𝑉1= 2 i = 2(4) = 8V  𝑉2 = 3 i = 3(4) = 12 V Put In Equation 1  FIGURE 8
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