Introduction to Electric and Electronic Circuits PDF - University of Sheffield

Introduction to Electric and Electronic Circuits EEE123 Semester 1 AER125 Semester 1 EEE218 Semester 1 Department of Electronic & Electrical Engineering University of Sheffield, UK 1 2 ...

Introduction to Electric and Electronic Circuits EEE123 Semester 1 AER125 Semester 1 EEE218 Semester 1 Department of Electronic & Electrical Engineering University of Sheffield, UK 1 2  Contents 1. Basic Concepts......................................4 5.5 Phasor representation..............50 1.1 Charge................................................4 5.6 Phasor notation............................54 1.2 Potential Difference......................4 5.7 Complex Numbers Revision....55 1.3 Sources...............................................7 5.8 Complex numbers in AC circuits 1.4 Resistance.........................................8 56 1.5 Non-ideal Voltage Sources...... 17 6. First Order AC Circuits.................... 58 1.6 Current Sources........................... 20 6.1 First Order Series AC Circuits.58 2. Network Analysis.............................. 21 6.2 First Order Parallel AC Circuits 61 2.1 Superposition Theorem............ 21 7. Second Order AC Circuits............... 64 2.2 Kirchhoff’s Laws.......................... 24 7.1 Series resonant circuit..............64 2.3 Thévenin’s Theorem.................. 27 7.2 Parallel resonant circuit...........67 2.4 Norton’s Theorem...................... 29 8. Power Dissipation in AC Circuits. 72 3. Circuit Components.......................... 31 8.1 Root-Mean-Square (RMS)........72 3.1 Capacitance, C............................... 32 8.2 Power Factor.................................74 3.2 Capacitors in parallel................. 35 8.3 VA Rating........................................76 3.3 Capacitors in series.................... 35 9. Magnetism and Electromagnetics81 3.4 Inductance, L................................. 37 9.1 Magnetic Field..............................81 3.5 Inductors in series...................... 37 9.2 Magnetic Field due to an 3.6 Inductors in parallel.................. 38 Electric Current.............................................82 4. Transient Circuits............................. 40 9.3 Magnetic Circuits.........................83 4.1 Charging a capacitor through a 9.4 Electromagnetic Induction......88 resistor 40 9.5 Self Inductance.............................90 4.2 Current growth in an inductor. 42 9.6 Mutual Inductance......................92 5. AC Circuits Introduction................. 46 9.7 Transformers................................93 5.1 Terminology.................................. 46 10. Motors and Machines....................... 97 5.2 Pure resistance AC circuit........ 47 10.1 Force on a Current Carrying Conductor........................................................97 5.3 Pure capacitance AC circuit.... 48 10.2 DC Motors.......................................98 5.4 Pure inductance AC circuit...... 49 Appendix 1 – Useful Equations.......... 105 Course notes edited by Ken Mitchell, Adam Funnell and Jo Shien Ng. Revised August 2020. 3 Section 1: Basic Concepts EEE123 / AER125 / EEE218 1. Basic Concepts 1.1 Charge Electric charge is the basis for describing all electrical phenomena; the separation of charge gives rise to an electric field and the movement of charge is called an electric current. Usually we use the symbol Q to denote charge and the unit is the Coulomb (C). Electric charge exists in discrete quantities which are integral multiples of the charge on an electron: Negative charge on an electron = −1.6 × 10−19 C Positive charge on a proton = +1.6 × 10−19 C Movement of charge (electrons) gives rise to an electric current for which we use the symbol, I, and the unit Ampère or Amp (A). Multiples of the unit exist, commonly used ones are: 1mA (milli-amp) = 0.001 A 1μA (micro-amp) = 10-6 A 1kA (kilo-amp) = 1000 A The formal definition of the Ampère was updated in 2019 to use a precise, fixed measurement of the charge on an electron. Every electron, or proton, has a fixed charge of 1.602 176 634 × 10−19 Coulombs. 1 Coulomb is measured as the amount of charge passing a point in a circuit when a current of 1 Ampère flows for 1 second, or: where, Q is the charge (C), I is the current (A) and t is the time (s). Alternatively we can say that the current is the rate of flow of charge: Example A car battery is rated at 120 Amp-hour (A-h). What is the total charge it can supply assuming it is initially fully charged? 120 A-h means that the battery can supply 1A for 120h, or 2A for 60h, or 30A for 4h, or 120A for 1h etc. It does not matter which one we assume in the calculation of charge as they all give the same answer. 1 120 60 60 = 4.32 × 105 C Assume I = 1A and t = 120 × 60 × 60 seconds: 30 4 60 60 = 4.32 × 105 C Check, if I = 30A and t = 4 × 60 × 60 seconds: 1.2 Potential Difference Normally in any material the number of positive and negative charges is equal so there is no net distribution of charge: + − − + + − + − − + − + + − − + − + − + 4 Section 1: Basic Concepts EEE123 / AER125 / EEE218 However, the charges can be separated by giving energy to the system: + + + + + + + + + + Electric Potential Field Difference − − − − − − − − − − There is now a potential difference between the positive and negative charges caused by their separation, and an electric field exists between the layers. I.e. work has been done in separating the charges which equals the energy given to the charges. The potential difference is measured in Volts (V). Normally voltages or potentials are measured relative to some reference value usually 0V (known as earth or ground). This is analogous to a tank of water situated above the ground as shown in the figure below: h1 h2 Flow of water h3 ground The potential energy due to gravity will cause the water to flow out of the pipe at a certain flow rate. The pressure, or difference in potential, that forces the water out of the pipe is directly related to the head of water (h1−h2) in such a way that the pressure is zero when h1=h2. Now if point h3, corresponding to the ground level is defined as zero potential then the pressure acting on the fluid in a pipe is actually the difference in potential energy ((h1−h3) − (h2−h3)) which has the same value as before. It is not therefore necessary to assign a precise energy value to h3 and in fact it would be extremely cumbersome to do so. Clearly it is the difference in potentials that we are interested in. For an electrical system: Voltage (potential difference) is analogous to the water pressure Current is analogous to the water flow So far we have seen that energy is needed to be input to our electrical system in order to separate charge and this has resulted in a potential difference. Expressing this in the form of an equation: where E is the energy input (Joules, J), V is the potential difference or voltage (Volts, V) and Q is the charge (Coulombs, C). Previously we have shown that: and so: (Joules) The power (the rate of doing work) is: (Watts) i.e. power is the product of the voltage and the current. 5 Section 1: Basic Concepts EEE123 / AER125 / EEE218 Example A car battery rated at 120A-h has a potential difference across its terminals of 12V. What is the total energy stored when it is fully charged. For how long could the battery supply the car’s radio (10W) and the headlights (200W)? What is the current drawn in each case? 12 120 3600 = 5.18 × 106 J or 5.18 MJ The total energy stored in the battery is: The radio is rated at 10W (or 10 J/s) hence the battery could supply it for: !.#$ #%& !.#$ #%' #% ()%% = 5.18 × 105 seconds or = 144 hours and current drawn is: #% * #+ = 0.83 A The headlights are rated at 200W (or 200 J/s) hence the battery could supply them for: !.#$ #%& +!6%% , -./0 +%% = 25900 seconds or ()%% = 7.2 hours 1 2345 and current drawn is: +%% , -./0 * #+ = 16.67 A Check: we could calculate the total charge in the battery (see previous example) as 4.32 × 105 C. Since the current is 16.67 A or C/s the time to discharge is 4.32 × 105/16.67 = 25900 s as before. Example An electric heater is required to heat 15 litres of water from 12°C to 100°C. What is the electrical energy consumed (a) in megajoules (b) in kilowatt-hours (kWh)? If it is required to heat the water in 10 minutes, what is the power rating of the heater? Assume the specific heat capacity of water is 4200 Jkg−1C−1and that the heater is 95% efficient. The mass of water is 15kg and the required temperature rise, θ, is 100 −12 = 88°C. Therefore energy 7899 9: ;H 12 Section 1: Basic Concepts EEE123 / AER125 / EEE218 CY F1 G W% >H We could obtain R0 from the first equation and substitute it into the second but this is not necessary: C+% F1 G W% 20H Rearranging: DdE \ ] F#_`^ +%Ha#b \c F#_%.%%B+) +%Hea#b > E^ D'^ `^ %.%%B+) = 91.3 °C 1.4.5 Connecting resistances in series Consider the simple circuit as shown in the figure below which consists of an ideal voltage source (VS = 10V) connected to a single load resistor (RL = 2Ω). I Battery terminals Source Load VL VS =10 V RL = 2 Ω In solving electrical circuits it is always worthwhile indicating the direction of the currents and voltages using arrows as shown in the diagram. The normal convention is to have a positive current flowing out of the positive terminal of the source. The increase in potential (voltage) across a source is always from negative to positive, and the increase in potential across a sink (load) is always in the opposite direction to the current. Since there is only one source and one sink in this circuit the source voltage is dropped across the 10 sink, hence: 0 , Rearranging Ohm’s law the current, I, flowing around the circuit may be found: *1 #% + = 5A 1 and the power dissipated in the load resistor may be calculated: *E #%E + C 5+ 2 = 50W or 10 5 = 50W or + = 50W Now try a circuit having two load resistors connected in series, such that the same current flows through both, as shown in the figure below: I V1 R1 = 2 Ω Source VS =10 V V2 R2 = 3 Ω C# and C+ We can apply Ohm’s law to each resistor in turn: # + #G + C# G C+ FC# G C+ H C/ but now the source voltage is dropped across the pair of resistors: 0 where RT is the total resistance. 13 Section 1: Basic Concepts EEE123 / AER125 / EEE218 This means we can replace the pair of resistors with a single, equivalent resistor, RT = 5Ω and we can find the current flowing around the circuit: *5 #% ! = 2A 4 Knowing the current we can calculate the voltage across each resistor and the power dissipated in C# 2 2 = 4V each. Voltage drop across resistor R1 = # Power dissipated in resistor R1 = # + C# 2+ 2 = 8W Voltage drop across resistor R2 = + C+ 2 3 = 6V Power dissipated in resistor R2 = + + C+ 2+ 3 = 12W Note: To find the power we could have used P = VI or P = V2/R, but we must use the voltage across the resistor of interest (not the total voltage – this is a very common student error!) For resistor R1: *DE BE # # 4 2 = 8W # + or = 8W D For resistor R2: *EE )E + + 6 2 = 12W + ( or = 12W E The total power dissipated in the two load resistors is 8 + 12 = 20W. This could have also been C/ 2+ 5 = 20W found by considering our equivalent load resistor: + In this example we have only considered two resistors in series, but the same method would apply to any number of resistors connected in series, i.e. with the same current flowing through each one. C/ C# G C+ G C( G⋅⋅⋅⋅ GCg The equivalent resistance RT for a group of resistors in series is: 1.4.6 Connecting resistances in parallel Let us now consider a circuit in which the load consists of two resistors in parallel having the same voltage, VL, across each one as shown in the figure below: I I1 I2 Source V1 V2 R1 = 2 Ω R2 = 4 Ω VS =10 V Clearly, the total current, I, splits into two paths with some of the current, I1, flowing through resistor, #G + R1, and the remainder the current, I2, flowing through resistor, R2, i.e.: Clearly the voltages across the resistors are equal to the supply voltage: 0 # + Applying Ohm’s law to each resistor: *D #% *E #% # + = 5A and + B = 2.5A D E Hence: *D *E # # # # # G + G 0 h G i 10 h+ G Bi = 7.5 A D E D E 14 Section 1: Basic Concepts EEE123 / AER125 / EEE218 We could replace the pair of resistors by a single equivalent resistor, RT: *5 # C/ * or 4 5 Combining this with the previous equation gives: # # # # # ( B G G C/ [ + B B ( or 4 D E Check, the total current is: *5 #% Q h i = 7.5 A (as before) 4 j # C# 5+ 2= 50W We can also calculate the power dissipated in each resistor: + Power dissipated in resistor R1 = # + C+ 2. 5+ 4= 25W + Power dissipated in resistor R2 = + Therefore the total power dissipated is 50 + 25 = 75W Check that the equivalent resistor gives the total power: B / + C/ 7. 5+ ( = 75W We would find that if we have a number of resistors in parallel, such that the voltage across them is 1 1 1 1 1 equal then the total resistance is obtained using: G G G⋅⋅⋅⋅ G C/ C# C+ C( Cg Example Find the total resistance of the following network of resistors, the current flowing in each resistor, the voltage across each resistor, and the power dissipated. I V1 I2 I3 I4 R1 Source 3Ω V2 R2 V3 R3 V4 R4 VS =10 V 7Ω 5Ω 10Ω The key to solving this problem is to identify which resistors are in parallel and which are in series. Looking at the circuit it will be seen that the voltages across R2, R3 and R4 are the same, therefore these resistors are in parallel and may be replaced by a single resistor RP, hence: # # C D D D D D D = 2.258Ω _ _ _ _ E j Q l ' D^ The circuit now reduces to: I V1 R1 3Ω Source VP RP VS =10 V 2.258Ω Since the same current, I, flows through the two resistors these are in series and can be replaced by C/ C# G C 3 G 2.258 = 5.258Ω a single resistor: 15 Section 1: Basic Concepts EEE123 / AER125 / EEE218 Using this value the total current flowing out of the battery and through R1 and RP can be found: *5 #% !.+!$ = 1.902 A 4 C/ 1.902+ 5.258 = 19 W and the total power dissipated in all the load resistors is: + / C# 1.902 3 = 5.706 V The voltage across the resistors is found using Ohm’s law: # C 1.902 2.258 = 4.294 V and: + ( B 5.706 G 4.294 = 10 V check: 0 # Now the current in R2, R3 and R4 and the power dissipated in each may be obtained: *m B.+6B + C+ 0.613+ 7 = 2.63 W + Current through R2 is + + T = 0.613 A and E *m B.+6B ( C( 0.859+ 5 = 3.69 W + Current through R3 is ( ! = 0.859 A and ( j *m B.+6B B CB 0.430+ 10 = 1.85 W + Current through R4 is B #% = 0.430 A and B Q # C# 1.902+ 3 = 10.85 W The power dissipated in R1 is: + # G G 0.613 G 0.859 G 0.430 = 1.902 A Check: + ( B G +G (G 10.85 G 2.63 G 3.69 G 1.85 = 19 W Total current is Total power is # B 1.4.7 Quick calculation for two resistors in parallel IT IT I1 I2 V R1 R2 ≡ V RP 1 1 1 C# G C+ Two resistors in parallel may be replaced by an equivalent resistor, RP, where: G C C# C+ C# C+ C# C+ hence: C C# G C+ C# C+ Find I1 and I2 in terms of the total current IT. Using Ohm’s law: # C# + C+ /C /c e C# G C+ from which I1 and I2 may be found: # 4 h D_ E i / _ E + 4 h D_ E i / D D_ E and D D E D E E D E 16 Section 1: Basic Concepts EEE123 / AER125 / EEE218 1.5 Non-ideal Voltage Sources So far in our analysis of circuits we have assumed ‘ideal’ voltage sources, that is to say they will always produce their rated voltage no matter what current is being drawn from them. In practice cells and batteries possess internal resistance from the materials and chemicals they are constructed from. This means the voltage at the terminals will now be a function of the current drawn and there will also be power dissipation within the battery. A visible example of this is the dimming of a car’s headlights when the engine is cranked on the starter motor. Let us consider the example we solved earlier, but now assume the battery has a 1Ω internal resistance as shown in the figure: RINT = 1Ω I Batter y VINT VS Load VTER VL 10 V RL = 2 Ω The current, I, flows through both the internal resistance, RINT, and the load resistor, RL, and so the C/ C g/ G C, 1 G 2 = 3Ω two resistors are in series: and the current is: *5 #% ( = 3.33 A 4 A current flowing through a resistor (energy sink) will dissipate power as heat. The energy dissipated in the load, which maybe a light bulb, heater etc. is useful energy, whereas the energy C, 3.33+ 2 = 22.2 W dissipated within the battery is waste energy as it simply heats up the battery itself. + Power dissipated in load (useful energy) is , Power dissipated in battery (waste energy) is g/ + C g/ 3.33+ 1 = 11.1 W The total power dissipated in our system is 22.2 + 11.1 = 33.3 W. C, 3.33 2 = 6.67 V The voltage across the terminals of the battery, VTER, is equal to the voltage across the load, VL: / , Alternatively we could have calculated the voltage drop across the internal resistance and subtracted 0 n C g/ 10 n 3.33 1 = 6.67 V this from VS: / VS is called the open-circuit, or no-load voltage, of the battery since if the load was disconnected no current would flow and hence VINT would be zero and VS would appear across the terminals. Often we use the symbol E (open-circuit emf) to depict this. This example has shown that to transfer 22.2W of power to our load requires a total power of 33.3W with 11.1W being dissipated in the battery itself as waste energy. Let us now look at the condition for maximum power transfer between the source and the load. 17 Section 1: Basic Concepts EEE123 / AER125 / EEE218 Consider the circuit shown in the figure below: I RINT Battery E RL The current, I, in the circuit is given by: C g/ G C, and the power transmitted to the load is: + + C, + + C, c e C, , C g/ G C, C +g/ G 2C g/ C, G C,+ C +g/ C, G 2C g/ G C, We are looking for the value of RL which will give maximum power transfer. This will occur when o C +g/ the denominator in the above equation is minimum, i.e. when: p G 2C g/ G C, q 0 oC, C, C +g/ n + G1 0 C, C, C g/. Hence: This means that the condition for maximum power transfer between source and load is when the load resistance is equal to the internal resistance of the source. This is also known as resistance matching. Example The voltage across the terminals of a 12V car battery drops to 9V when supplying a current of 300A to the starter motor. Find the internal resistance of the battery and the load resistance for maximum power transfer to the load. How does the system efficiency vary with RL? RINT I VTER RL 12V First we need to find the internal resistance of the battery. When the current is 300A the terminal voltage, VTER, is 9V so 3V must be dropped across the internal resistance, hence: * r4 ( C g/ (%% = 0.01Ω C, C g/ = 0.01Ω For maximum power transfer to the load: Under these conditions the new current will be: #+ _ %.%#_%.%# = 600 A r4 1 C, 600+ 0.01 = 3.6 kW and the power in the load is: + , 18 Section 1: Basic Concepts EEE123 / AER125 / EEE218 However, since the internal resistance carries the same current and has the same value then there is also 3.6 kW dissipated in the internal resistance. (This is equivalent to a 3 bar electric fire heating up the battery!). Batteries are often specified in terms of Ampere-hours (A-h). If our battery is 60A-h it can deliver (fully charged) 60A for 1hour or 600A for 0.1 hour (6 minutes). (Unfortunately the electrolyte would probably boil after 30 seconds!). The efficiency of a system is defined as: s 100% 100% 100% t/ t/ t/ t/ , g t/ t/ t/ G , 00 0 , G g/ In the example we have just looked at, where RL and RINT are equal, then: ()%% s 1 100% 100% = 50% _ 1 ()%%_()%% r4 Alternatively: E %.%# s 1 100% 1 100% 1 100% 100%= 50% 1_ 1_ 1_ %.%#_%.%# E E r4 r4 r4 What happens to the efficiency when the load resistance is say 1Ω? #+ _ %.%#_# = 11.88 A r4 1 , + C, 11.88+ 1 = 141 W g/ + C g/ 11.88+ 0.01 = 1.41 W #B# s 1 100% 100% = 99% 1 _ r4 #B#_#.B# What happens if RL is zero i.e. the terminals of the battery are shorted out? The current is now only limited by the internal resistance of the battery and a current of 1200 A would flow. The power dissipated in the battery would be 14.4kW and the power in the load would be zero and hence the efficiency would be zero. This is likely to cause serious damage to the battery and should never be done. The condition for maximum power transfer to the load is when the load resistance is equal to the internal resistance but the efficiency is only 50%. We could repeat the above calculations for a range of loads and obtain the curves shown in the figure below. PL η PLMAX 100% 50% RL = RINT RL = RINT RL RL 19 Section 1: Basic Concepts EEE123 / AER125 / EEE218 1.6 Current Sources An ideal current source is a device which will generate a prescribed current independent of the circuit to which it is connected. To do so it must be able to generate an arbitrary voltage across is terminals. A more practical representation has an internal resistance connected in parallel with its terminals. On this course only limited use is made of circuits containing current sources. I I RINT Ideal current source Practical current source 20 Section 2: Network Analysis EEE123 / AER125 / EEE218 2. Network Analysis There are many methods available to use for solving electrical networks. On this course we will limit our study to the following methods: Superposition Theorem Kirchhoff’s Laws Thévenin and Norton equivalent circuits At this stage we will only consider networks comprising of resistors and d.c. sources. However the theorems and methods may be applied to more complex circuits containing inductance, capacitance and fed from a.c. or d.c. sources. A network comprises a number of branches or circuit elements connected together and considered as a unit. If the network contains no source of e.m.f. it is termed a PASSIVE network If the network does contain an e.m.f. source it is termed an ACTIVE network 2.1 Superposition Theorem “An emf. acting on any linear network produces the same effect whether it acts alone or in conjunction with other emfs.” (only linear networks will be considered on this course). What this means is that a network containing several emf sources may be analysed by considering the effect of each source in turn, with all other sources being represented by their internal resistances, then the resultant current is obtained from the algebraic sum of the currents from each source. This is best illustrated by means of an example. Example A circuit consists of two batteries, each having an internal resistance, and a single 10Ω load resistor as shown in the figure. What is the current in the 10Ω resistor? Battery A Battery B 6V 4V 10Ω 2Ω 3Ω Using the superposition theorem, first find the currents in each branch of the circuit when supplied by each source in turn, then add the results to find the total currents. 21 Section 2: Network Analysis EEE123 / AER125 / EEE218 First consider the effect of battery A alone. Battery B is represented by its internal resistance alone. Let the currents in the branches be I1, I2 and I3 respectively. (Note it is usual to chose the direction of the current to flow out of the positive terminal of the battery). The circuit then becomes: Battery A I1 C I2 I3 6V 10Ω 3Ω 2Ω D It can be seen that the 10Ω and 3Ω resistor are in parallel so the resistance between points C and D 1 30 is: CA 2.31[ 1 1 13 10 G 3 The circuit can then be drawn as: Battery A I1 C 6V RCD 2Ω D C/ CA G 2 4.31[ RCD is then in series with the 2Ω resistance of battery A which gives a total circuit resistance of: 6 and the current flowing out of battery A, I1, may be found: 1.393K # C/ 4.31 Once we have found I1, the current that flows through battery A and our equivalent resistor RCD, we 3 can then work back to find the currents, I2 and I3: 0.321K + # 10 G 3 10 1.071K ( # 10 G 3 (Note the direction of the currents as indicated by the arrows in the diagram. We need to know this when we come to sum them with the currents from battery B). 22 Section 2: Network Analysis EEE123 / AER125 / EEE218 Now consider the effect of battery B alone. Battery A is represented by its internal resistance alone. Let the currents in the branches be I4, I5 and I6, so the circuit becomes: C’ I4 Battery B I6 I5 4V 2Ω 10Ω 3Ω D’ It can be seen that now it is the 10Ω and 2Ω resistor which are in parallel so the resistance between 1 20 points C’ and D’ is: CA′ ′ 1.667[ 1 1 12 10 G 2 C/′ CA′ ′ G 3 4.667[ RC’D’ is then in series with the 3Ω resistance of battery B which gives a total circuit resistance of: 4 and the current flowing out of battery B, I4, may be found: 0.857K v B C/′ 4.667 Once we have found I4, the current that flows through battery B and our equivalent resistor RC’D’, we 2 can then work back to find the currents, I5 and I6: 0.143K ! B 10 G 2 10 0.714K ) B 10 G 2 We can now superimpose these two sets of results to find the resultant current in each branch of the circuit. (Take care with the directions). For the 10Ω resistor in the central branch, battery A gave the contribution of I2 = 0.321A(↓) flowing downwards and battery B gave the contribution of I5 = 0.143A(↓) again flowing downwards. Hence #%w F↓H +G ! 0.321 G 0.143 = 0.464A(↓) flowing downwards. the total current is: For the branch containing the battery A we have I1 (↑)flowing upwards and I6(↓) flowing downwards F↓H )n # 0.714 n 1.393 = −0.678A flowing downwards (or +0.678A flowing upwards) (i.e. in the opposite direction), hence: Likewise if we follow the same procedure for battery B we have I3 = 1.072(↓) flowing downwards v F↓H (n B 1.071 n 0.857 = 0.214A(↓) flowing downwards (or −0.214A flowing upwards) and I4 = 0.857A(↑) flowing upwards. Hence the current in the branch containing battery B is: What this indicates is that we have a positive current flowing into the positive terminal of battery B (IB is in the downward direction) and battery B is in fact being charged by battery A. 23 Section 2: Network Analysis EEE123 / AER125 / EEE218 The general method for solving circuits by superposition is to consider the effect of each source in turn. (Other voltage sources are short circuited, current sources are open-circuited) and then to sum the components taking care with the directions of the currents. 2.2 Kirchhoff’s Laws Before embarking on Kirchhoff’s laws it is first necessary to define some terms. Consider the network shown below: BRANCH NODE NODE LOOP NODE BRANCH ─ Part of a circuit connecting two nodes NODE ─ The point where two or more branches meet LOOP - Closed path formed by connecting branches 2.2.1 Kirchhoff’s 1st Law The algebraic sum of all the instantaneous currents entering any node is zero at all times. For example: i1 i2 i4 i3