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  • 1. Kirchhoff’s LawsKirchhoff’s laws consist of two statements,Current LawAt any junction (node) in an electric circuit the total current flowing towardsthat junction is equal to the total current flowing away from the junction.Voltage LawIn any closed loop in a network, the algebraic sum of the voltage drops takenaround the loop is equal to the resultant emf acting in that loop.
  • 2. Kirchhoff’s LawsCurrent LawAt any junction (node) in an electric circuit the total current flowing towardsthat junction is equal to the total current flowing away from the junction.orThe algebraic sum of currents at a node is zeroI1 + I2 = I3 + I4 + I5orI1 + I2 – (I3 + I4 + I5 )=0I3I4I5I2I1
  • 3. Kirchhoff’s Current LawActivity1. Write the formula and deduce the current I1 for the circuit shown given thatI2 = 2A, I3 = 1A, I4 = 3A and I5 = 4A.I3I4I5I2I12. Determine the unknown currents.I1I4I5I3ECBFDA40A120A50A20A 15AI2
  • 4. Kirchhoff’s LawsVoltage LawIn any complete electric circuit, the algebraic sum of the source voltage mustequal the algebraic sum of the voltage drops.orIn any closed loop, the algebraic sum of all voltages is zero-I+E+ + +- - -R2R1 R3L1VR1 + VR2 + VR3 = EE - VR1 - VR2 - VR3 = 0 orVR1 = IR1 VR2 = IR2 and VR3 = IR3IR1 + IR2 + IR3 = E
  • 5. Kirchhoff’s LawsVoltage LawExample of a combinational circuit.L2+-E+-R2+ -R1+-R3I1 + I2L1I2I1VR1 + VR2 = EE - VR1 - VR2 = 0IR1 + IR2 = E . . . . . .(1)Loop 1VR1 + VR3 = EE - VR1 – VR3 = 0Loop 2IR1 + IR3 = E . . . . . .(2)Equations (1) and (2) are then solved simultaneously
  • 6. Kirchhoff’s Voltage LawActivityGenerate the voltage equations and show that all voltages sum to zero for thecircuits shown. Determine all network currents.Label direction of current flow and polarity at terminals of all resistorsTrace your loops in a clockwise direction(direction of conventional current flow)-+20VR1 = 2ΩR3 = 4ΩR2 = 6Ω-+10VR1 = 4ΩR3 = 2ΩR2 = 6Ωcircuit 1 circuit 2
  • 7. Kirchhoff’s LawsVoltage LawExample of a dual source network.Equations (1) and (2) are then solved simultaneously+-E1+-R3+ -R1I1 + I2L1I2I1+-E2L2+-R2VR1 + VR3 = E1E1 - VR1 – VR3 = 0I1R1 + R3(I1+I2 ) = E1 . . . . . .(1)Loop 1VR2 + VR3 = E2E2 – VR2 – VR3 = 0Loop 2I2R2 + R3(I1+I2 ) = E2 . . . . . .(2)
  • 8. Kirchhoff’s Voltage LawActivity1. Determine the network currents and the power dissipated in theload resistor RL.RL = 12ΩR1 = 3Ω R2 = 2Ω-+6VE1 E2-+4VLabel directions of current flow and polarity at terminals of all resistorsTrace your loops following the direction of conventional current2. After a period of time E2 becomes discharged to 2 volts. Determine thenetwork currents under these conditions and state reasons.
  • 9. Kirchhoff’s Voltage LawActivityA more practical use of Kirchhoff’s law is that of vehicle electrics. The batterywhen fully charged has an open circuit voltage of 12.6 volts and an internalresistance of 0.1Ω. The alternator provides a dc (rectified) output of 16 voltsand is used to supply current to both the load and battery thus maintaininga charge on the battery. Determine the current in the load and its source.Investigate this further to see when the battery begins to take a charge from thegenerator using computer simulation.RLOAD1.2ΩRgen0.2Ω-+EGEN 16V EBAT12.6VRbat0.1Ω
  • 10. Kirchhoff’s Voltage LawActivityA more practical use of Kirchhoff’s law is that of vehicle electrics. The batterywhen fully charged has an open circuit voltage of 12.6 volts and an internalresistance of 0.1Ω. The alternator provides a dc (rectified) output of 16 voltsand is used to supply current to both the load and battery thus maintaininga charge on the battery. Determine the current in the load and its source.Investigate this further to see when the battery begins to take a charge from thegenerator using computer simulation.RLOAD1.2ΩRgen0.2Ω-+EGEN 16V EBAT12.6VRbat0.1Ω