DC Circuit PDF

1. D.C.Circuits 1.1. Define following terms (a) Current Free electron Copper wire Conventional current V Electron current Figure 1.1Concept of electric...

1. D.C.Circuits 1.1. Define following terms (a) Current Free electron Copper wire Conventional current V Electron current Figure 1.1Concept of electric current  Flow of electron in closed circuit is called current.  Amount of charge passing through the conductor in unit time also called current.  Unit of current is charge/second or Ampere (A). Q I t Where , I  Current Q  Charge t  Time (b) Potential or Voltage  The capacity of a charged body to do work is called potential.  Unit of potential is joule/coulomb or Volt (V). W V Q Where, V = Potential or Voltage W = Workdone (c) Potential difference A B + 12 V +7V Conventional current Figure 1. 1Potential differences  The difference of electrical potential between two charged bodies is called potential difference.  Unit of Potential Difference is Volt (V).  If potential of body A is +12V and potential of body B is +7V then potential difference is +5V. i.e. (+12V) - (+7V) = +5V PiyushRupala, EE Department Basic Electrical Engineering (3110005) 1 1. D.C.Circuits (d) Electro Motive Force (emf)  The force is required to move electron from negative terminal to positive terminal of electrical source in electrical circuit is called emf.  Unit of emf is volt (V).  Emf is denoted as ε. (e) Energy  Ability to do work is called energy.  Unit of energy is Joule or Watt-sec or Kilowatt-hour (KWh).  1KWh is equal to 1 Unit. V 2t W  P  t  VIt  I 2Rt  R Where , W =Energy P =Power t =Time (f) Power  Energy per unit in time is called power.  Unit of Power is Joule/Second or Watt (W). W P t (g) Resistance  Property of a material that opposes the flow of electron is called resistance.  Unit of resistance is Ohm (Ω). V R= I Where , R  Resistance (h) Conductance  Property of a material that allows flow of electron.  It is reciprocal of resistance.  Unit of conductance is (Ω-1) or mho or Siemens(S). 1 G R Where , G  Conductance (i) Resistivity or Specific Resistance  Amount of resistance offered by 1m length of wire of 1m2 cross-sectional area.  Resistivity is denoted as a ρ.  Unit of Resistivity is Ohm-meter (Ω­m). l R a PiyushRupala, EE Department Basic Electrical Engineering (3110005) 2 1. D.C.Circuits l Rρ a Ra ρ l Where , R  Resistance ρ Resistivity l Length of wire a Cross section area of wire (j) Conductivity  Ability of a material to allow flow of electron of a given material for 1 m length & 1 m2cross-sectional area is called conductivity. Unit of conductivity isΩ-1m-1 or Siemens m- 1. 1 σ ρ Where , σ  Conductivity 1.2. Explain types of electrical energysource  Electrical source is an element which supplies energy to networks. There are two types of electrical sources. (a) Independent sources Independent voltage source Independent current source + v(t) V I i(t) - Figure 1. 2Independent voltage source Figure 1. 3Independent current source  It is a two terminal element that provide a  It is two-terminal elements that provide specific voltage across its terminal. a specific current across its terminal.  The value of this voltage at any instant is  The value and direction of this current at independent of value or direction of the any instant is independent of value or current that flow through it. direction of the voltage that appears across the terminal of source PiyushRupala, EE Department Basic Electrical Engineering (3110005) 3 1. D.C.Circuits (b) Dependent sources Voltage controlled voltage source (VCVS) Voltage controlled current source (VCCS) + + + Icd + a c a c + Vab μ Vab - Vcd Vab g m Vab Vcd b d b d - - - - Figure 1.5VCVS Figure 1.6VCCS  Voltage controlled voltage source is four  Voltage controlled current source is four terminal network components that terminal network components that established a voltage Vcd between two- established a current icd in the branch of point c and d. circuit. Vcd  μVab icd  gmVab  The voltage Vcd depends upon the control  icd depends only on the control voltage Vab voltage Vab and μ is constant so it is and constant g m ,is called trans dimensionless. conductance or mutual conductance.  μ is known as a voltage gain.  Unit of transconductance is Ampere/Volt or Siemens(S). Current controlled voltage source (CCVS) Current controlled current source (CCCS) + i ab + + i ab icd + a c a c + r iab - Vcd β iab b d b d - - - - Figure 1.7CCVS Figure 1.8CCCS  Current controlled voltage source is four  Current controlled current source is four terminal network components that terminal network components that established a voltage Vcd between two- established a current Icd in the branch of point c and d. circuit. Vcd  riab icd  βiab  Vcd depends on only on the control  icd depends on only on the control current current iab and constant r and r is called iab and constant β and β is called current trans resistance or mutual resistance. gain. Current gain is constant.  Unit of transresistance is Volt/Ampere  Current gain is dimensionless. or Ohm (Ω). PiyushRupala, EE Department Basic Electrical Engineering (3110005) 4 1. D.C.Circuits 1.3. Explain source conversion  A voltage source with a series resistor can be converted into an equivalent current source with a parallel resistor. Conversely, a current source with a parallel resistor can be converted into a voltage source with a series resistor.  Open circuit voltages in both the circuits are equal and short circuit currents in both the circuit are equal.Source transformation can be applied to dependent source as well. R I V I=V/R R Figure 1. 9Source conversion Network simplification techniques + + V1 + - + V1 + V2 - V2 + - - - (a) + + V1 + - + V1 - V2 - - V2 + - - (V1 > V2) (b) PiyushRupala, EE Department Basic Electrical Engineering (3110005) 5 1. D.C.Circuits + + i1 i2 i1 + i2 - - (c) + + i1 i2 i1 - i2 - - (i1 > i2) (d) + + + + + V1 = V2 V1 - - V2 - - - (e) + + i1 i1 = i2 i2 - - (f) PiyushRupala, EE Department Basic Electrical Engineering (3110005) 6 1. D.C.Circuits + R + Vs Vs - - (g) R + + is is - - (h) Figure 1.10Rules under which source may be combined and separated 1.4. Explain ideal electrical circuit element.  There are major three electrical circuit elements which are discussed below. (a) Resistor  Resistor is element which opposes the flow of current. a l Figure 1.11Resistor Figure 1.12Conductor  Resistance is property of material which opposes the flow current. It is measured in Ohms (Ω).  Value of resistance of conductor is  Proportional to its length.  Inversely proportional to the area of cross section.  Depends on nature of material.  Depends on temperature of conductor. l R a ρl R a PiyushRupala, EE Department Basic Electrical Engineering (3110005) 7 1. D.C.Circuits (b) Inductor  An inductor is element which store energy in form of magnetic field.  The property of the coil of inducing emf due to the changing flux linked with it is known as inductance of the coil.  Inductance is denoted by L and it is measured in Henry (H). 1.13Inductor  Value of inductance of coil is  Directly proportional to the square of number of turns.  Directly proportional to the area of cross section.  Inversely proportional to the length.  Depends on absolute permeability of magnetic material. F NI NI NIμ0 μr A Φ    S S l l μ0 μr A  NIμ0 μr A  N  N2μ μ A NΦ  l  Now , L   0 r I I l Where , L =Inductance of coil N= Number of turns of coil Φ = Flux link in coil F = Magneto motive force(MMF) I = Current in the coil l = Mean length of coil μ0 = Permiability of free space μr = Relative permiability of magnetic material A = Cross sectional area of magnetic material (c) Capacitor  Capacitor is an element which stored energy in form of charge.  Capacitance is the capacity of capacitor to store electric charge.  It is denoted by C and measured in Farad (F). Figure 1.14Capacitor  Value of capacitance is  Directly proportional to the area of plate.  Inversely proportional to distance between two plates.  Depends on absolute permittivity of medium between the plates. PiyushRupala, EE Department Basic Electrical Engineering (3110005) 8 1. D.C.Circuits A C d εA C d εεA C 0 r d Where , C =Capacitance of capacitor A =Cross sectional area of plates d =Distance between two plates ε = Abolute Permittivity ε0 = Permittivity of free space εr = Relative permittivity of dielectric material 1.5. Explain Ohm’s law and its limitations.  Current flowing through the conductor is directly proportional to the potential difference applied to the conductor, provided that no change in temperature. Voltage (V) Current (A) Figure 1.15Change in current w.r.t change in voltage for conducting material V I V  IR  Where R is constant which is called resistance of the conductor. V R  I  Limitations of Ohm’s Law:  It cannot be applied to non-linear device e.g. Diode, Zener diode etc.  It cannot be applied to non-metallic conductor e.g. Graphite, Conducting polymers  It can only be applied in the constant temperature condition. 1.6. State and explain the Kirchhoff’s current and voltage laws (a) Kirchhoff’s current law (KCL)  Statement: “Algebraic sum of all current meeting at a junction is zero”  Let, Suppose PiyushRupala, EE Department Basic Electrical Engineering (3110005) 9 1. D.C.Circuits  Branches are meeting at a junction ‘J’  Incoming current are denoted with (+ve) sign  Outgoing currents are denoted with (-ve) sign A R2 I1 J I3 R4 B I2 + E1 - R3 R5 R1 + E2 - E D C Figure 1.16Kirchhoff’s law diagram  Then, I  0 (  I1 )  ( I2 )  ( I3 )  0 I1  I2  I3  0 I1  I2  I3  Incoming current  Outgoing current (b) Kirchhoff’s voltage law (KVL)  Statement: “Algebraic sum of all voltage drops and all emf sources in any closed path is zero”  Let, Suppose  Loop current in clockwise or anticlockwise direction  Circuit current and loop current are in same direction than voltage drop is denoted by (-ve) sign.  Circuit current and loop current are in opposite direction than voltage drop is denoted by (+ve) sign.  Loop current move through (+ve) to (-ve) terminal of source than direction of emf is (-ve).  If Loop current move through (-ve) to (+ve) terminal of source than direction of emf is (+ve). R R + + I I V= -IR V= +IR E= -E1 E= +E1 Figure 1.17Sign convention for Kirchhoff’s voltage law PiyushRupala, EE Department Basic Electrical Engineering (3110005) 10 1. D.C.Circuits  IR   E  0 KVL to loop AJDEA I1R2  I2R3  E2  I1R1  E1  0 KVL to loop JBCDJ I3R4  I3R5  E2  I2R3  0 1.7. Explain series and parallel combination of resistor Series combination of resistor Parallel combination of resistor R1 R2 I1 R1 I1 I2 I V1 I V1 V2 R2 I2 + - + V2 V - V Figure 1.18Series combination of resistors Figure 1.19Parallel combinations of resistors Here , I1  I2  I Here ,V1  V2  V As per KVL, As per KCL, V  V1  V2 I  I1  I2 V  IR1  IR2 V V I   V  I(R1  R2 ) R1 R2 V 1 1   (R1  R2 ) I  V   I  R1 R2  Req  R1  R2 I 1 1  For n resistor are connected in series     V  R1 R2  Req  R1  R2  R3 .........  Rn 1 1 1      Req  R1 R2  For n resistor are connected in Parallel 1 1 1 1 1    .........  Req R1 R2 R3 Rn  Value of equivalent resistance of series  Value of equivalent resistance of parallel circuit is bigger than the biggest value of circuit is smaller than the smallest value of individual resistance of circuit. individual resistance of circuit. PiyushRupala, EE Department Basic Electrical Engineering (3110005) 11 1. D.C.Circuits 1.8. Explain Voltage divider law and current divider Law. Voltage Divider Law Current Divider Law R1 I1 R1 R2 I1 I2 I V1 I V1 V2 I2 R2 + V2 + - V - V Figure 1.20Voltage divider circuit Figure 1.21Current divider circuit Here , I1  I2  I Here ,V1  V2  V As per KVL, As per KCL, V  V1  V2 I  I1  I2 V  I1R1  I2R2 V1 V2 I   V  IR1  IR2 R1 R2 V  I(R1  R2 ) V V I   R1 R2 V I  I1  I2  (R1  R2 ) 1 1 I  V   Now ,V1  I1R1  R1 R2  V I 1 1 V1  R1     R1  R2 V  R1 R2   R1   RR  V1  V   V  V1  V2  I  1 2   R1  R2   R1  R2  Now ,V2  I2R2 V Now , I1  1 V R1 V2  R2 R1  R2  RR  I 1 2   R2  R  R2  V2  V   I1   1  R1  R2  R1  R2  I1  I    R1  R2  V Now , I2  2 R2  RR  I 1 2  R  R2  I2   1 R2  R1  I2  I    R1  R2  PiyushRupala, EE Department Basic Electrical Engineering (3110005) 12 1. D.C.Circuits 1.9. Derive the equation of delta to star and star to delta transformation 1 1 R12 R1 R12 R1 R31 R31 R2 R3 R2 R3 2 2 R23 R23 3 3 Figure 1.22Delta connected network Figure 1.23Star connected network Resistance between terminal 1 &  2 Resistance between terminal 1 & 2  R12 (R23  R31 )  R1  R2 R12 (R23  R31 ) Resistance between terminal 2 & 3  R12  R23  R31  R2  R3 Resistance between terminal  2 & 3 Resistance between terminal 3 & 1  R23 (R12  R31 )  R3  R1 R23 (R12  R31 )  R12  R23  R31 Resistance between terminal 3 & 1   R31 (R12  R23 ) R31 (R12  R23 )  R12  R23  R31 Resistance between terminals 1 & 2 in delta equal to resistance between terminals 1 & 2 in star R12 (R23  R31 ) R1  R2  (i ) R12  R23  R31 Similarly , R23 (R12  R31 ) R2  R3  (ii ) R12  R23  R31 R31 (R12  R23 ) R3  R1  (iii ) R12  R23  R31 (a) Delta to star conversion Simplify  i    ii    iii  on both the side of equations R12 (R23  R31 ) R23 (R12  R31 ) R31 (R12  R23 ) R1  R2  R2  R3 - R3 - R1  + - R12  R23  R31 R12  R23  R31 R12  R23  R31 PiyushRupala, EE Department Basic Electrical Engineering (3110005) 13 1. D.C.Circuits (R12R23  R12R31 ) (R23R12  R23R31 ) (R31R12  R31R23 )  + - R12  R23  R31 R12  R23  R31 R12  R23  R31 (R12R23  R12R31  R23R12  R23R31 - R31R12 - R31R23 )  (R12  R23  R31 ) 2R12R23 2R2  R12  R23  R31 R12R23 R2  R12  R23  R31 R12R31 Similarly , R1  R12  R23  R31 R23R31 R3  R12  R23  R31 (b) Star to delta conversion Simplify  i  ii    ii  iii    iii  i  on both the side of equation (R1  R2 )(R2  R3 )  (R2  R3 )(R3  R1 )  (R3  R1 )(R1  R2 )  R (R  R31 )  R23(R12  R31 )   R23(R12  R31 )   R31 (R12  R23 )   R31 (R12  R23 )  R12(R23  R31 )    12 23        R12  R23  R31  R12  R23  R31   R12  R23  R31   R12  R23  R31   R12  R23  R31  R12  R23  R31  R1R2  R1R3  R22  R2R3  R2R3  R2R1  R32  R3R1  R3R1  R3R2  R12  R1R2  R R  R12R31   R23R12  R23R31   R23R12  R23R31  R31R12  R31R23   R31R12  R31R23  R12R23  R12R31    12 23        R12  R23  R31   R12  R23  R31   R12  R23  R31  R12  R23  R31   R12  R23  R31  R12  R23  R31  3R1R2  3R2R3  3R3R1  R12  R22  R32  R 2 R 2  R R 2 R  R 2 R R  R R R 2   R 2 R R  R R 2 R  R R R 2  R 2 R 2   R 2R R  R 2R 2  R R 2 R  R R R 2    23 12 12 23 31 12 23 31 12 23 31    12 23 31 12 23 31 12 23 31 23 31    12 23 31 12 31 12 23 31 12 23 31    R  R  R  2    R  R  R  2    R  R  R  2   12 23 31   12 23 31   12 23 31  R232R122  R12R232R31  R122R23R31  R12R23R312  R122R23R31  R12R232R31  R12R23R312  R232R312  R122R23R31  R122R312  R12R232R31  R12R23R312   R12  R23  R31  2 (R12R232R31  R122R23R31  R12R23R312  R122R23R31  R12R232R31  R12R23R312  R122R23R31  R12R232R31  R12R23R312 )  (R232R122  R232R312  R122R312 )   R12  R23  R31  2 R12R23R31 (R23  R12  R31  R12  R23  R31  R12  R23  R31 ) (R232R122  R232R312  R122R312 )    R12  R23  R31   R12  R23  R31  2 2 R12R23R31 (3R12  3R23  3R31 )  R232R122 R232R312 R122R312        R12  R23  R31  2  R  R  R  R  R  R  R  R  R   2 2 2  12 23 31 12 23 31 12 23 31  3R R R (R  R23  R31 )  2 R23 R122 2 R23 R31 2 2 R12 R31 2   12 23 31 12      R12  R23  R31  2   R  R  R 2  R  R  R 2  R  R  R 2   12 23 31 12 23 31 12 23 31   3R3R12  R22  R32  R12 Now equation become 3R1R2  3R2R3  3R3R1  R12  R22  R32  3R3R12  R22  R32  R12 3R1R2  3R2R3  3R3R1  3R3R12 R1R2 R12  R1  R2  R3 Similarly R2R3 R23  R2  R3  R1 R3R1 R31  R3  R1  R2 PiyushRupala, EE Department Basic Electrical Engineering (3110005) 14 1. D.C.Circuits 1.10. Explain Node analysis R1 R3 R5 A B + + V1 - R2 R4 - V2 C Figure 1.24Node analysis network  Node: Node refers to any point on circuit where two or more circuit elements meet.  Node analysis based on Kirchhoff’s current law states that algebraic summation of currents meeting at junction is zero.  Node C is taken as reference node in this network. If there are n nodes in any network, the number of equation to be solved will be (n-1).  Node A,B and C are shown in given network and their voltages areVA ,VB andVC. Value of node VC is zero because VC is reference node.  Steps to follow in node analysis:  Consider node in the network, assign current and voltage for each branch and node respectively.  Apply the KCL for each node and apply ohm’s law to branch current.  Solve the equation for find the unknown node voltage.  Using these voltages, find the required branch currents.  Node A R1 VA R3 VB R5 I1 I3 I2 + + V1 - R2 R4 - V2 VC Figure 1.25Node analysis network for node A Apply KCL at node A ,  I1    I2    I3   0 I1  I2  I3  0 VA - V1 VA  VC VA - VB   0 R1 R2 R3 PiyushRupala, EE Department Basic Electrical Engineering (3110005) 15 1. D.C.Circuits 1 1 1  1 V VA      VB     1 (i )  R1 R2 R3   R3  R1  Node B R1 VA R3 R5 I 3 VB I 5 I4 + + V1 - R2 R4 - V2 VC Figure 1.26Node analysis network for node B Apply the KCL at node B ,  I3    I4    I5   0 I3  I4  I5  0 VB - VA VB  VC VB - V2   0 R3 R4 R5  1 1 1 1 V VA     VB      2 ( ii )  R3   R3 R4 R5  R5 From, equation (i) & (ii) 1 1 1 1   V1  R R R   V R3    1 2 3   A    R1   1 1 1 1   VB   V2          R3 R3 R4 R5   R5   One can easily find branch current of this network by solving equation (i) and (ii),if V1 , V2 and all resistance value are given. 1.11. Explain Mesh analysis R1 R3 R5 + + I1 I2 I3 V1 - R2 R4 - V2 Figure 1.27Mesh analysis network  Mesh: It is defined as a loop which does not contain any other loops within it. PiyushRupala, EE Department Basic Electrical Engineering (3110005) 16 1. D.C.Circuits  The current in different meshes are assigned continues path that they do not split at a junction into a branch currents.  Basically, this analysis consists of writing mesh equation by Kirchhoff’s voltage law in terms of unknown mesh current.  Steps to be followed in mesh analysis:  Identify the mesh, assign a direction to it and assign an unknown current in it.  Assigned polarity for voltage across the branches.  Apply the KVL around the mesh and use ohm’s law to express the branch voltage in term of unknown mesh current and resistance.  Solve the equations for unknown mesh current.  Loop 1 R1 R3 R5 I1 + + V1 - R2 R4 - V2 I1 I2 Figure 1.28Mesh analysis network for loop-1 Now apply the KVL in loop  1, - I1R1 -  I1  I2  R2  V1  0 - I1R1 - I1R2  I2R2  V1  0 -  R1  R2  I1  R2 I2  V1 (i)  Loop 2 R1 R3 R5 I2 + + V1 R2 R4 - V2 - I1 I2 I2 I3 Figure 1.29Mesh analysis network for loop-2 Now Apply the KVL loop  2, - I2R3 -  I2  I3  R4 -  I2  I1  R2  0 - I2R3 - I2R4 +I3R4  I2R2  I1R2  0 I1R2 - I2  R3  R4  R2  +I3R4  0 R2 I1 -  R3  R4  R2  I2 +R4 I3  0 (ii) PiyushRupala, EE Department Basic Electrical Engineering (3110005) 17 1. D.C.Circuits  Loop 3 R1 R3 R5 I3 + + V1 R2 R4 - V2 - I1 I3 Figure 1.30Mesh analysis network for loop-3 Now Apply the KVL loop  3, - I3R5 - V2 -  I3  I2  R4  0 - I3R5 - V2 -I3R4  I2R4  0 I2R4 - I3  R5  R4   V2 R4 I2 -  R5  R4  I3  V2 (iii) From equation (i ),(ii ) & (iii )  -  R1  R2  R2 0   I1   V1         R2  R3  R4  R2  R4   I2    0    0 R4   R5  R4    I3  V   2   -  R1  R2  R2 0    Δ R2  R3  R4  R2  R4    0 R4   5 4   R  R  V1 R2 0    Δ1   0  R3  R4  R2  R4  V  2 R4   R5  R4    -  R1  R2  V1 0    Δ2   R2 0 R4    0 V 2   5 4   R  R  -  R1  R2  R2 V1    Δ3   R2  R3  R4  R2  0   0 R4 V2   Now , Δ Δ Δ I1  1 , I2  2 , I3  3 Δ Δ Δ 1.12. Explain Superposition theorem  The superposition theorem states that in any linear network containing two or more sources, the current in any element is equal to the algebraic sum of the current caused by individual sources acting alone, while the other sources are inoperative. PiyushRupala, EE Department Basic Electrical Engineering (3110005) 18 1. D.C.Circuits  According to the application of the superposition theorem. It may be noted that each independent source is considered at a time while all other sources are turned off or killed. To kill a voltage source means the voltage source is replaced by its internal resistance whereas to kill a current source means to replace the current source by its internal resistance.  To consider the effects of each source independently requires that sources be removed and replaced without affecting the final result. To remove a voltage source when applying this theorem, the difference in potential between the terminals of the voltage source must be set to zero (short circuit) removing a current source requires that its terminals be opened (open circuit).  Any internal resistance or conductance associated with the displaced sources is not eliminated but must still be considered.  The total current through any portion of the network is equal to the algebraic sum of the currents produced independently by each source.  That is, for a two-source network, if the current produced by one source is in one direction, while that produced by the other is in the opposite direction through the same resistor, the resulting current is the difference of the two and has the direction of the larger.  If the individual currents are in the same direction, the resulting current is the sum of two in the direction of either current. This rule holds true for the voltage across a portion of a network as determined by polarities, and it can be extended to networks with any number of sources.  The superposition principle is not applicable to power effects since the power loss in a resistor varies as the square (nonlinear) of the current or voltage.  Steps to be followed to apply the superposition theorem:  Select any one energy source.  Replace all the other energy sources by their internal series resistances for voltage sources. Their internal shunt resistances for current sources.  With only one energy source calculate the voltage drops or branch currents paying attention to the voltage polarities and current directions.  Repeat steps 1, 2 and 3 for each source individually.  Add algebraically the voltage drops or branch currents obtained due to the individual source to obtain the combined effect of all the sources.  Example network: R1 A R2 r + + R3 - - V1 V2 B Figure 1.31Superposition theorem network PiyushRupala, EE Department Basic Electrical Engineering (3110005) 19 1. D.C.Circuits Step-1 R1 A R2 V1 + - I1 R3 I2 r B Figure 1.32Superposition theorem network for step-1 Now apply Mesh analysis in loop  1, - I1R1 - I1R3  I2R3 - I1r  V1  0 Now apply Mesh analysis in loop  2, - I2R2 - I2R3  I1R3  0 Now , current flow from R3 branch is a lg ebric sum of I1 and I2 Step-2 R1 A R2 + I3 R3 I4 - r V2 B Figure 1.33Superposition theorem network for step-2 Now apply Mesh analysis in loop  1, - I3R1 - I3R3  I4 R3 - I3r  0 Now apply Mesh analysis in loop  2, - I4 R2 - V2 - I4 R3  I3R3  0 Now , current flow from R3 branch is a lg ebric sum of I3 and I4 Finally , current flow from R3 is a lg ebric sum of step  1 and step -2 1.13. Explain Thevenin’s theorem  Thevenin theorem is an analytical method used to change a complex circuit into a simple equivalent circuit consisting of a single resistance in series with a source voltage.  Thevenin’s can calculate the currents and voltages at any point in a circuit.  Thevenin’s Theorem states that “Any linear circuit containing several voltages and resistances can be replaced by just one single voltage in series with a single resistance PiyushRupala, EE Department Basic Electrical Engineering (3110005) 20 1. D.C.Circuits Connected across the load“.  In other words, it is possible to simplify any electrical circuit, no matter how complex, to an equivalent two-terminal circuit with just a single constant voltage source in series with a resistance (or impedance) connected to a load as shown below.  Thevenin’s Theorem is especially useful in the circuit analysis of power or battery systems and other interconnected resistive circuits where it will have an effect on the adjoining part of the circuit.  Thevenin’s equivalent circuit A RTH A A Linear Network + containing RL RL Several emf’s and - Resistance Eth B B Figure 1.34Thevenin’s equivalent circuit  As far as the load resistor RL is concerned, any complex “one-port” network consisting of multiple resistive circuit elements and energy sources can be replaced by one single equivalent resistance Rth and one single equivalent voltage Eth.  Rth is the thevenin resistance value looking back into the circuit and Eth is the Thevenin’s voltage (open circuit voltage) at the terminals.  Steps to be followed to apply the Thevenin’s theorem:  Remove the load resistor Rth or component concerned.  Find Rth by shorting all voltage sources or by open circuiting all the current sources.  Find Eth by the usual circuit analysis methods.  Find the current flowing through the load resistor Rth.  Example network: R1 R2 A r + R3 RL V1 - B Figure 1.35Thevenin’s theorem network PiyushRupala, EE Department Basic Electrical Engineering (3110005) 21 1. D.C.Circuits Step-1 R1 R2 A + V1 + - R3 Eth I1 I2 r - B Figure 1.36Thevenin’s theorem network (step-1) Now apply Mesh analysis in loop  1, - I1R1 - I1R3  I2R3 - I1r  V1  0 Now apply Mesh analysis in loop  2, - I2R2 - Eth - I2R3  I1R3  0 Loop - 2 is open that ' s way I2  0, So, Eth  I1R3 Eth  Thevenin equivalent voltage Rth  Thevenin equivalent Re sis tan ce RL  Load Re sis tan ce Step-2 R1 R2 A Rth     r+ R  R  + R  1 3 2 R3 Rth    r+ R1   R3   Rth     + R2     r+ R1   R3   r    B Figure 1.37Thevenin’s theorem network (step-2) Step-3 Rth IL Eth + IL  RL Rth  RL - Eth Figure 1.38Thevenin’s theorem network (step-3) PiyushRupala, EE Department Basic Electrical Engineering (3110005) 22 1. D.C.Circuits 1.14. Explain Norton’s theorem  Norton’s theorem is an analytical method used to change a complex circuit into a simple equivalent circuit consisting of a single resistance in parallel with a current source.  Norton’s Theorem states that “Any linear circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in parallel with a Single Resistor“.  As far as the load resistance, RL is concerned this single resistance, RN is the value of the resistance looking back into the network with all the current sources open circuited and IN is the short circuit current at the output terminals as shown below.  Norton’s equivalent circuit A A A Linear Network containing Several energy RL IN RN RL sources and Resistances B B Figure 1.39Norton’s theorem equivalent circuit  The value of this “constant current” is one which would flow if the two output terminals where shorted together while the Norton’s resistance would be measured looking back into the terminals.  The basic procedure for solving a circuit using Norton’s Theorem is as follows:  Remove the load resistor RL or component concerned.  Find RN by shorting all voltage sources or by open circuiting all the current sources.  Find IN by placing a shorting link on the output terminals A and B.  Find the current flowing through the load resistor RL.  Example network: R1 R2 A r + R3 RL - V1 B Figure 1.40Norton’s theorem network PiyushRupala, EE Department Basic Electrical Engineering (3110005) 23 1. D.C.Circuits Step-1 R1 R2 A V1 + - I1 R3 I2 IN r B Figure 1.41 Norton’s theorem network (step-1) Now apply Mesh analysis in loop  1, - I1R1 - I1R3  I2R3 - I1r  V1  0 Now apply Mesh analysis in loop  2, - I2R2 - I2R3  I1R3  0 Here I2  IN IN  Norton ' s equivalent current RN  Norton ' s equivalent Re sis tan ce RL  Load Re sis tan ce Step-2 R1 R2 A RN     r+ R  R  + R  1 3 2 R3    r+ R1   R3   RN RN     + R2     r+ R1   R3   r    B Figure 1.42 Norton’s theorem network (step-2) Step-3 IL RN IN RN RL IL  IN RN  RL Figure 1.43 Norton’s theorem network (step-3) PiyushRupala, EE Department Basic Electrical Engineering (3110005) 24 1. D.C.Circuits 1.15. Time domain analysis of first order RC circuit Charging of Capacitor Discharging of Capacitor R R + - - + VR VR + + + V V - - - VC C VC C - + Figure 1.44Charging of capacitor Figure 1.45Discharging of capacitor Apply KVL in circuit , Apply KVL in circuit , V - VR - Vc  0 0  VR  Vc V  VR  Vc 0  iR  Vc V  iR  Vc dq 0R  Vc dq dt V R  Vc d  CVc  dt 0R  Vc d  CVc  dt V R  Vc dV dt 0  RC c  Vc dV dt V  RC c  Vc dV dt Vc  -RC c dV dt V - Vc  RC c 1 -1 dt  Vc dVc   RC dt 1 1  V - Vc dVc   RC dt -t log Vc   K (i ) Multiply min us sign both the side RC When, t  0, Vc  V -1 -1  V - Vc dVc   RC dt log V   K (ii ) -t Solve equation (i ) and (ii ) log V - Vc  K (i ) RC -t log Vc    log V  When, t  0, Vc  0 RC log V   K -t (ii ) log Vc  - log V   RC Solve equation (i) and (ii)  Vc  -t -t log    log V - Vc    log V   V  RC RC -t -t  Vc  RC log V - Vc  - log V   V e RC   -t  V - Vc  -t log   Vc  Ve RC  V  RC -t  V - Vc  RC  V e   -t V  1-  c   e RC V  PiyushRupala, EE Department Basic Electrical Engineering (3110005) 25 1. D.C.Circuits -t Vc  V (1- e RC ) dq dq Also, i Also, i  dt dt d(CVc ) d(CVc ) i i dt dt d -t dV i  C (V (1- e RC )) i C c dt dt -t d -t d i  VC (1- e RC ) i  C (Ve RC ) dt dt -t   1  RC -t  -1 RC i  VC  0-  -  e  i  CV e   RC   RC -t V -t VC RC i  - e RC i e R RC -t V -t i  -Ime RC i  e RC R -t i  im e RC V λ 0.632 V vc vc 0.37 t λ t Figure 1.46Charging voltage of capacitor Figure 1.48Dicharging voltage of capacitor λ t O I -0.37 Im ic 0.37 -Im λ Figure 1.47Charging current of capacitor Figure 1.49Dicharging current of capacitor 1.16. Time domain analysis of first order RL circuit PiyushRupala, EE Department Basic Electrical Engineering (3110005) 26 1. D.C.Circuits Charging of Inductor Discharging of Inductor R R + - + - VR VR + + + + V - VL L V - VL L - - Figure 1.50Charging of inductor Figure 1.51Discharging of inductor From KVL, From KVL, di di -iR - L  0 V - iR - L  0 dt dt di di -iR  L V - iR  L dt dt di -R di dt   di   i L V - iR L 1 -R 1 1   di   dt  di   dt i L V - iR L  -R   -R -R di   dt  log  i     t  K (i ) V - iR L  L  -R  V  log V - iR     t  K (i ) When, t  0, i   L R When, t  0, i  0 V  log    K (ii ) log V   K (ii ) R Solve (i ) and (ii ) Solve equation (i ) and (ii )  -R  V   -R   log  i     t  log    log V - iR     t  log V   L R  L  V   -R   -R   log  i  - log      t  log V - iR  - log V     t R  L   L  i  -R  V - iR   -R   log   log     t    t  V   L   V   L  R  -R   V - iR   L t  i   - R t  e  V     e L  V   -R   R R  t 1-   i  e  L   -R  V  V  t i  e  L  R V  t   -R  i   1- e  L   V  λt R   i  e  R   t   -R  i  Im  1- e  L        i  Im 1- e  λt  PiyushRupala, EE Department Basic Electrical Engineering (3110005) 27 1. D.C.Circuits λ   t   -R  0.632 Im i  Im  1 - e  L       V  λt i e iL iL R 0.37 Im t λ t Figure 1.52Charging current of inductor Figure 1.53Dicharging current of inductor PiyushRupala, EE Department Basic Electrical Engineering (3110005) 28