GANDHINAGAR INSTITUTE OF Technology.

ACTIVE LEARNING ASSIGnMENTCIRCUITS AND NETWORKS(2130901)Prepared byGroup - 1 Div- BSem-3rd Branch-ElectricalGuided by: Prof. Megha MamGANDHINAGAR INSTITUTE OF Technology.

Brijesh Patel -140120109030Kavan Patel -140120109032Darshil Shah -140120109050

Group : 1

PREPARED BY: ENROLLMENT NO.

Topic name:Kirchhoffs laws.

TYPES OF LAWS:Kirchhoffs laws.KIRCHHOFFS CURRENT LAW(KCL)KIRCHHOFFS VOLTAGE LAW (KVL) These laws are first derived by Guatov Robert Kirchhoff and hence these laws are also referred as Kirchhoff Laws.

Circuit DefinitionsNode any point where 2 or more circuit elements are connected togetherWires usually have negligible resistanceEach node has one voltage (w.r.t. ground)Branch a circuit element between two nodesLoop a collection of branches that form a closed path returning to the same node without going through any other nodes or branches twice

Kirchhoff's Current Law (KCL)The algebraic sum of currents entering a node is zeroAdd each branch current entering the node and subtract each branch current leaving the node currents in - currents out = 0Or currents in = currents out

(IB + IC + ID) - IA = 0Then, the sum of all the currents is zero. This can be generalized as follows,

Kirchhoff's Voltage Law (KVL)The algebraic sum of all voltages around a closed loop must be equal to zero. voltage drops - voltage rises = 0Or voltage drops = voltage rises

That means, V1 + V2 + V3 + ................... + Vm Vm + 1 Vm + 2 Vm + 3 + ..................... Vn = 0.So accordingly Kirchhoff Second Law, V = 0.

Ideal voltage and current source.

VOLTAGE SOURCECURRENT SOURCE

Voltage rise and voltage drop:

Voltage drop

Voltage rise

Example.Find the current through 8 resistor using KIRCHHOFFS LAWS.

EXAMPLE OF KCL

EXAMPLE OF KCL

Rearranging the terms:

Solving the linear equations with the Cramers Rule:

EXAMPLE OF KCLCurrents:I1k = = -3.252 mA ()I2k = = 2.626 mA ()I3k = = 1.928 mA ()I4k = = 1.928 mA ()I5k = = 0.699 mA ()I6k = = 0.625 mA ()

EXAMPLE OF KVL

EXAMPLE OF KVL Junction J1:I = I1 + I2 (equation 1)Junction J2:I1 + I2 = I (equation 2)

Loop A: (start from the upper left corner and move clockwise)-I1 x (100 ) + 1.5V = 0 (equation 3)Therefore: I1 = 0.015 A Loop B:-9V I2 x (200 ) + I1 x (100 ) = 0 (equation 4)Substituting the value of I1 into equation 3 yields: -9 I2 x (200 ) + (0.015)(100 ) = 0 -7.5 = (200) x I2 therefore: I2 = -0.0375 A And then I = -0.0225 AVR1 = I1 x R1 = (0.015 A) x (100 ) therefore VR1 = 1.5VVR2 = I2 x R2 = (-0.0375 A) x (200 ) therefore VR2 = -7.5V

REFERENCES.www.electricals4u.comwww.eefundamentals.comwww.electricaleasy.comwww.wikipedia.com/KCL,KVLCIRCUITS & NETWORKS book