C h a p t e r CHAPTE NTODUCTON TO ELECTC CCUTS.0 NTODUCTON This chapter is explaining about the basic principle of electric circuits and its connections. The learning outcome for this chapter are the students should be able to explain clearly basic electrical quantities, types of electrical circuits, electrical power, electrical energy and solve related problems.. ELECTC Electric is an energy which cannot see but can be felt and be used by human on today and future. Electric energy can be created impact from action as friction, heat and electromagnetic field Electric energy can be change into other form of energy such as: a) Light energy - lamp b) Heat energy - ron c) Sound energy - adio d) Kinetic energy - Motor There are two types of electric which is the static electric and dynamic electric. a) Static electricity A situation where no electron movement in certain direction. b) Dynamic electricity A situation where got electron movement in certain direction... Electrical Quantities i. Electromotive force (e.m.f) Force or electric pressure that cause the flow of electrons or the flow of current in given circuiy. The example the source that produces electric energy are batteries and generator. Symbol : E Unit : olt()
C h a p t e r ii. Electrical charge There are two types of charge which is positive and negative charge. Electric charge is measured in Coulomb. Symbol : Q Unit : Coulomb(C) iii. Current The movement of the electric charge cause by free electrons movement. t flows from the positive terminal to the negative terminal. Symbol : Unit : Ampiar (A) iv. oltage The potential different between two points in a circuit. Symbol : Unit : olt() v. esistance t is the property of material by which it oppose the flow of current through it. Symbol : Unit : Ohm (Ω) vi. Conductor A material that allow electric current to flow easily. An example is copper and iron. vii.nsulator A material that does not allow or prevent the electrical current flow in normal condition. t has a lot of valence electrons but the valence electron are difficult to be free from is parent atom. For example rubber, glass, air. viii.semiconductor A material that has a conductance value between conductor and insulator. t has 4 valences electron and can be use to make electronic component. For examples silicon and germanium. xi.esistivity t is the characteristic of conductive material to opposition or decrease the current flows in it. Symbol : ρ (ho) Unit : Ohm meter ( ( Ω m)
C h a p t e r.. esistance The resistance of given material depends on the physical properties of the material. There are 4 factor that influence the value of resistance: i. Length of conductor, l The length of conductor is proportional to the resistance. The longer the length of the wire, the higher the resistance value. l ii. Surface area, A Area is inverse proportional to the resistance. As the resistance increase the cross section area of a conductor will decreases. A iii. esistivity esistivity is proportional to the resistances. Higher the resistance, higher the resistivity. ρ iv. Conductor Temperature, T The conductor temperature is proportional to the resistance. As the conductor temperature increase the value of resistance also increase. T Mathematically, formula for the resistance of a wire of length l and the cross section area A is as equation below: ρ l A (.) Where: A cross section area ( m ) ρ esistivity ( Ω m ) l Length (m) esistance ( Ω )
C h a p t e r 4 Example. Calculate the resistance of a.5 km length of aluminium wire. Given diameter wire 0mm and the resistivity of aluminium is 0.05 µω.m. Solution. 6 Been given, d 0x0 m, l.5x0 m, ρ 0.05x0 Ωm ρl Use equation,, A d 0x0 6 Where A π ( ) π ( ) 78.54x0 m 6 (0.05x0 )(.5 x0 ) 0.477Ω 6 78.54x0. ELECTC CCUT Electric circuit is a combination of conductor or cable which makes the current flow from voltage sources to electrical components or load.there are two types of electric circuit: i. Complete electric circuit ii. Non Complete electric circuit.. Complete electric circuit t is also called basic circuit or simple circuit as shown in Figure.. t is closed end connection that can make current go through completely which the current flow from source and back flow to sources again. The circuits must have voltage supply (), electric current () and resistance (). Figure. : Complete Electric Circuit.. Non Complete electric circuit t is a circuit without one of three component either voltage sources or load resistance. The current flow will never happen with perfect in non complete circuit. There are two types of non complete circuit: Open circuit and Short circuit
C h a p t e r 5 i. Open circuit s the circuit without the load, so there will be no current flow occur. alue of resistant in this circuit is a higher. Figure. show a open circuit. Pull out load () Figure. : Open Circuit ii. Short circuit The connection at the load will short with a conductor which no resistance value as shown in Figure.. The current which go through is bigger. Normally if short circuit occur, the fuse will burnt. Short with a cable Figure. : Short circuit. OHM S LAW Ohm s law can be define as the current flowing through the electrical circuit is directly proportional to the potential difference across the circuit and inversely proportional to resistance of the circuit. f the value of resistance is constant and value of voltage increase so the value of current can be increase. Mathematically the equation for Ohm s law is as equation. below. (.) where: Current (A) oltage () esistance ( Ω )
C h a p t e r 6 The relationship between current and voltage is as shown using the graph at Figure.4. This is the situation for constant value of resistance and temperature. (volt) (constant) (Ampere) Figure.4 : Graph oltage () vurses Current () for constant resistance For the non constant or changing value of resistance, the relationship between voltage and current are non linear as graph shown in Figure.5 below. Figure.5 : Graph oltage () vurses Current () for non constant resistance Example. Calculate the current value if the resistance is 0Ω and the supply voltage is 5. Then, calculates the new current value if the resistance has been change to 0 kω. Solution. Been given, 5 i) 0 Ω, Base on Ohm s Law, 5. A 0 5 ii) 0k Ω, 5.5x0. 5mA 0x0
C h a p t e r 7.4 ELECTC POWE Electric power is a job can be done in one time unit. esistor dissipate energy in the form of heat. So power absorbed by the resistor is given by equation.. Symbol : P Unit : Watt (W) P (.) By using Ohm s law, equation.4. or, t can derive new equation for power as in P (.4) P Where, P Power (W), Current (A) esistance (Ω) and oltage ().4. Wattmeter Wattmeter is use to measure the value of the power that has been use. There are two coils in wattmeter. The coil connected parallel to the load is voltage coil and series with the load is current coil. The symbol for wattmeter is as shown in Figure.6(a) and.6(b) is the internal wattmeter connection. Figure.7 is the electric circuit connection using the wattmeter. W (a)meter Symbol (b)nternal Connection Figure.6 : Wattmeter
C h a p t e r 8 Current coil oltage coil Load () S Figure.7 : Electric Circuit Connection using Wattmeter.5 ELECTC ENEGY Electric energy is a product of power and time. The symbol for electric energy is T or E. Mathematically electric energy is expressed as equation.5. T Pt T t T t.5 T t where, T electric energy (kwh) P power (W) t time (s) voltage () current (A) resistance ( Ω ) Meter kilowatt-hour be used to measure total of electrical energy which be used by user. The symbol metre.8. The electric power can be converted to horse power where horse power 746 watt
C h a p t e r 9 kwh Figure.8 : Symbol Kilowatt Hour Metre The unit for electric energy is Kilowatt hour (kwh) or Joule (J). When the current flow, electron in the conductor will repell each other and it will produce heat and thus causing the cabel that is used heating up. Work is the energy absorbed to supply load kw for hour. Watt is the power used when A current flows between point that have volt potential. Units for work is Joule. This is equal to the energy produced to Coulomb charge flows by ohm resistance. Total energy used to flow A current for second by ohm resistant is called Joule. t is can be called as watt second, that is watt power used for second. n mathematical equation it can be shown as equation.6. Joule Watt second work (J) power (W) x time (s).6 Example. A toaster taking 5A current from 40 supply for 5 minutes. Calculate, i. Power used ii. Energy absorbed in kj Solution. Given: 5 A, 40 dan t 5 x 60 900s i. P ( 5)(40) 00W. ii. T Pt ( 00)(900) 080000W 080kWj 080 kj.6 ESSTO CCUT ANALYSS esistor can be connected in three different ways which are series, parallel and combination of series and parallel.
C h a p t e r 0.6. Series Circuit Series circuit is refer to the connection of the resistor in the circuit. The resistors is connected from end to end as in Figure.9. Series analysis are going to determine total resistance, circuit current and total voltage. T T + - + - + - + n - n Figure.9: Series esistors Total resistance, T is the sum of all resistor which exist in the circuit. Equation.7 use to calculate total resistance. + + +... + T n (.7) Current through every resistor is equal to the total current, T as show in equation.8:... T n (.8) Total voltage, T is the sum of all voltage drops on every resistor as shown in equation.9 below. + + +... + T n (.9) oltage drop is the reduction of the voltage supply in every resistor. t can be calculate using Ohm s law and oltage divider law. oltage drop calculation using ohm s law are as Equation.0 below.
C h a p t e r n T T T T n (.0) Meanwhile, to calculate voltage value across every resistance in series circuit using voltage divider is as shown in equation. for three resistor connected in series and equation. is for two resistor connected in series. ( ) + + T ( ) T (.) + + (. ( ) + + T ( ) + ( ) + T T (.) Example.4 : efering to the circuit below determine : 5Ω 0 0Ω
C h a p t e r i). Total resistance, T ii). Current in the circuit, T iii). The voltage drop across each resistor. Solution.4: i). ii). Total resistance, T T + (5 + 0) 5 Ω Current in the circuit, T T 0 4.8 A 5 T iii). The voltage drop across each resistor. T (4.8)(5) 7 T (4.8)(0) 48.6. Parallel Circuit The parallel circuit is a connection of resistor which is against between each other. The resistors connected in parallel is shown in Figure.0. Parallel analysis are also going to determine total resistance, circuit current and total voltage. T + + + T - - - Figure.0: Parallel esistor The total parallel resistance can be calculate by using equation.: Or + + T T + + (.)
C h a p t e r The voltage across each parallel resistor is equal to the source voltage, T as shown in equation.4;... T n (.4) Total current, T for parallel circuit is equal to the sumation of all current from each branch. This is shown in equation.5 + + +... + T n (.5) The value of the current for each branch also can be determine by using the current divider s law as equation.6. T + + T - - Figure.: Parallel Circuit using esistors ( ) + and ( ) + T T (.6) Example.5: efering to the circuit below, calculate : T Ω 4Ω 40
C h a p t e r 4 i). Total resistance, T ii). Total current, T iii). Current and Solution.5: i). Total resistance, T T + ()(4) + 4. Ω ii). Total current, T T T T 40. 80 A iii). Current and, 40 0A 40 60A 4.6. Combination Circuit Most of electric circuits are the combination of series and parallel circuit. Both formula of series and parallel circuit will be used to determine the value of current, voltage and total resistance. Figure. is the example for combination circuit. T Figure. : Combination Series and Parallel esistors
C h a p t e r 5 Example.6 : By referring to Figure., a 0 source is connected across resistors. f 0Ω, 0Ω, 5Ω. Calculate a). Total resistance, T b). Total current, T c). Current and Solution.6: a). Total resistance, T (0 )(5 ) + 0 + 5 8.57Ω T + 8.57 + 0 8.57Ω b). Total current, T T 0 6.46 A 8. 57 T c). Current 5 ( ) T ( ) 6. 46 + 0 + 5 T (6.46.79).67 A.79 A.7 KCHOFF S LAW Kirchoff s law is used to solve more difficult electric circuit, for example the circuit which having more than one power supply. There are two types of Kirchoff law: a) Kirchoff s current law b) Kirchoff s voltage law.7. Kirchoff s Current Law Kirchoff s current law is also known as first order of Kirchoff s law. Kirchoff s current law stated that the algebraic sum of all the currents entering and leaving a node is equal. Therefore, the sum of the current into a node (total current in) is equal to the sum of the currents out of the node (total current out) as shown in Figure.. i i i Figure.: Flow of Current going n and Out the Node
C h a p t e r 6 n mathematic expression its can be stated as equation.7. + (.7).7. Kirchoff s oltage Law Kirchoff s voltage law also known as the second order of Kirchoff s law. This law stated that the sum of the voltage drop and voltage source around a closed path is equal to zero. + - T - + - + Figure.4: Close Path oltage This can be stated in mathematic expression like equation.8 T + + (.8) Example.7: Ω 6Ω Ω 5 0 By using Kirchoff s current law, find the current thought each branch in the circuit above.
C h a p t e r 7 Solution.7: A Ω 6Ω Ω 5 0 Kirchoff s Current Law: - ( + ) () Kirchoff s oltage Law: Loop : 5 + ( ) 6 6 ( 5 ) + 0 0 () Loop : 0 + ( 6 ) ( ) 0 () replace () into () : 0 + 6( ) ( + 8 0 ) 0 (4) Solve equations () and (4) by using Cramer rules. i) Make the matrix equation from equation () and (4) 6 5 8 0 6 ii) Find the value of determination, perhaps D 8 6 D ()(8) ( 6)() 0 8 iii). Find the value of determination for each currents, 5 6 ( 5)(8) ( 6)(0) 0 0 8
C h a p t e r 8 0 A D 0 D 5 0 + 0 0 0 0 0.5A From equation (); ( + ) (.5 + ). 5A Negative value (-ve) in the current shows the actual current direction is leading to resistance. EFEENCES Bakshi, A.. and Bakshi, U.A., 009, Circuit Theory st Edition, Technical Publications Pune, ndia Bakshi, A.. and Bakshi, U.A., 008, Circuit Analysis, Technical Publications Pune, ndia POBLEMS. Calculate the resistance of the aluminium with.5 km length, 0 mm diameter and 0.05 µωm resistivity. (0.477Ω). Calculate the resistance of the aluminium bar with 0m length, cross section area 8cm x cm and 0.069 µωm resistivity. (.6 x 0-4 Ω). The cuprum with 500cm and.75 µωmm resistivity. Calculate the diameter of the conductor when the resistor is.5 kω. (d.6 x 0-5 m) 4. The heating element with 50 Ω resistance, 50 cm length and 0.7 mm. Calculate the resistivity. (. x 0-5 Ω) 5. Calculate the resistance for m length copper,.5 mm diameter and 0.07 µωm resistivity. (0.98Ω)
C h a p t e r 9 6. The resistivity of the silinder aluminium conductor is 80 µωmm, mm radius and 0 Ω resistance. Calculate the length. (l.6 x 0 - m) 7. Calculate the resistance of the conductor with.5 m length,.6 m cross section area and 6. µωm resistivity. (5.8Ω) 8. Calculate the resistance of a zinc with 0.05 µωm resistivity and 0.5m diameter. (.7 x 0 - Ω) 9. Calculate the current flowing through the aluminium wire with a length of km and a diameter of 0 mm if the 5 supply voltage. The resistivity of the wire is 0.8 µωm. 0. efer to the figure below, calculate the current flow in the conductor. 00 ρ 5.5µΩm d 5 cm 00 km (.9A). Calculate the current in the aluminium coil with km length and 0 mm diameter when the supply is 5. The resistivity is 0.8 µω.m (4.5A). What is the current of a circuit that has and 0.5 ohm of resistance?. What is the voltage if current is 0.5 A [ampere] and resistance is 0.8 ohm? 4. What is the resistance of a circuit if voltage is and current is A. 5. What is the current of a circuit if resistance is ohm and voltage is 5. 6. What is the voltage of a circuit if resistance is 7 ohm and current is 0.5 A. 7. Calculate the energy in Joule and Kilowatthour for: a. A 60W lamp switched on for 8 hours b. A kw kettle switched on for 5 minutes 8. Calculate the current flow in the circuit with 0Ω resistence and 5 voltage supply. Then, calculate the current if the resistance is increasing to 0 kω. 9. Calculate the power losses when the current is 5mA through resistance 6kΩ. 0. Calculate the current flow by a Filament lamp with 40 and resistance 960Ω.. A cattle with resistance 40 Ω and current.4 A. Calculate the power. (P0.4w)
C h a p t e r 0. A toaster with 5A current and 40 supply was on for 5 minutes. Calculate, i. Power used (00w) ii. Energy in kj (080kJ). A ice cooker with.45 kw power, 0 voltage. Calculate: i. Current (5A) ii. esistance (5.Ω) iii. Energy if the rice cooker is switch on for half an hour.(.75kwh) (60kJ) 4. Calculate the amount of current () in a circuit, given below: (4A) 5. Calculate the amount of resistance () in a circuit, given below (9Ω) 6. Calculate the amount of voltage supplied by a battery in a circuit given below. (4v)
C h a p t e r 7. Calculate: 5 A 0 Ω i. Potential different, v ii. Power, P iii. Electrical energy if the circuit switch on for hours (50v, 50w, 0.5kwh) 8. Based on the circuit diagram below, calculate; i. Total resistance. ii. Total voltage, T iii. oltage drop in the resistance, the voltage divider law..5a 8 Ω 6 Ω T 4 Ω (8 Ω, 7, 6 ) 9. Three () resistors connected in parallel to each value, 6Ω, 5Ω and 0Ω and are supplied with a 00. Calculate: i. Total resistance ii. Total current iii. The voltage across each resistor iv. Current through each resistor (.4 Ω, 4.7A, 00, 6.7A, 0A, 5A) 0. Define the First and Second order of Kirchoff s Law. Based on figure below, find ; i). Total esistance. ii). oltage,. ii). Current. and. iii). Total power and power dissipated in. 8Ω T 40 Ω 4Ω (.4Ω, 40v, 60A, 0A, 8.8kw, 50.4kw)
C h a p t e r. Based on figure below, the voltage across 7. Specify the following values : i). The current flow each resistor,, and 4 ii). The voltage across each resistor,, and 4 iii). Supply voltage, s 8Ω T S 6Ω 4 4Ω Ω (9A, A, 6A, 8v, 6v, 6v). Based on Figure below, calculate the current value of each branch and voltage drop in each resistor using Kirchoff s Law. 4Ω 5Ω 4 6 4. Based on figure below, calculate the voltage supply. i. 6 A Ω (.5A, 0.5A, A, v, 6v,.5v) 0Ω ii. 5 A 0 Ω (0A, 50A)
C h a p t e r 5. Based on figure below, calculate Total resistance, T, Total current, T, oltage across, and oltage across, i. 0Ω j 5v 0Ω ii. 5Ω j 9v 5Ω 5Ω 6. Based on figure below, calculate total resistance, T Ω Ω Ω j 4Ω 4Ω Ω Ω Ω Ω 7. Based on figure below, calculate total resistance, T referring to point AB. i. ii.
C h a p t e r 4 8. Based on the figure, calculate Total esistance, T, Total Current, T, Current, Current and Total Power, P T i. j j 0v 5Ω 0Ω ii. j j 00v Ω 8Ω 9. Based on the figure, calculate Total esistance, T, Total Current, T, oltage across,, oltage across, and Total Power, P T 0Ω 8Ω b 0 v 0Ω 4 Ω 40. Based on the figure, calculate Total esistance, T, Total Current, T and Total Power, P T j Ω 6 v Ω Ω
C h a p t e r 5 4. Based on the figure, calculate Total esistance, T, Total Current, T, oltage across esistor 0Ω, 0, oltage across resistor 0Ω, 0 and oltage across resistor 50Ω, 50 j 0Ω 0 v 0Ω 50Ω 4. Based on the figure, calculate Total esistance, T, Total Current, T, Current, Current, oltage across esistor 8Ω, and oltage across resistor 4Ω, j b v 8Ω 8Ω 4Ω 4. By using the Ohm s Law, Current Divider Law and oltage Divider Law calculate Total Current, T, Current, Current and oltage drop at each resistors. j 0Ω 5Ω 7Ω 0 v
C h a p t e r 6 44. By using Kirchoff Law calculate Current, Current and Current i. 6Ω Ω v Ω 5v ii. 8Ω 5Ω 0 v 5 v 7 v iii. Ω Ω 4 v v Ω iv. 4Ω 4Ω v Ω 6v
C h a p t e r 7 v. 00Ω 80Ω 4 v 47Ω 50 Ω v 47Ω