good day everyone and welcome back last class we looked at some basic electrical concepts and laws of circuit analysis today we are going to apply those concepts and laws to analyze complex circuits we are going to focus on dc circuits and explore the different methods we can use to analyze and simplify them these different methods include the voltage and current divider rule mesh analysis nodal analysis and the superposition theorem then we will look at some theorems for reducing complex circuits namely thevenin's theorem and norton's theorem and finally we will look at the theorem for maximum power transfer from a circuit to its load before we begin i want to quickly recap ohm's law v is equal to ir and the power formula p is equal to v i this table shows what formula we would need to find the unknown variables given two known variables it's just mathematical manipulation and substitution of these two formulation to obtain the desired variable and i won't go into any further detail with this this is just here for your convenience we are going to analyze dc circuits but firstly what constitutes our dc circuit well dc values are constants meaning they do not vary with time and are unidirectional in a dc circuit the only active elements we have are dc voltage and current sources and the only passive element we will consider is the resistor the dc voltage can be produced by a battery or a dc generator so graphically the dc voltage is expressed as shown it is a constant value and does not change with time and hence is in one direction the same applies for dc currents since they do not change with time they can be expressed with both capital and common letters but you will usually see me use capital letters which i told you in the last lesson signifies average or constant values before i introduce the new methods we are going to review the voltage and current divider rules i'll just read from the points a voltage divider rule the voltage across a resistor within a loop is a fraction of the source voltage this fraction is the resistance of that resistor over the total resistance in the loop so for example the voltage across the resistor r1 vr1 is a fraction of the source voltage e this fraction is given by that resistor r1 divided by the sum of the resistances in the loop r1 plus r2 a similar analysis can be done to find the voltage across the resistor r2 the current divider rule the current through a resistor in a parallel branch is a fraction of the total current from the source this fraction is the equivalent resistance of the resistors in the other branches over the total resistances in the circuit remember it must be reduced to two branches before analysis so for example the current through the resistance rx ix is a fraction of the source current i t this fraction is given by the equivalent resistance of the resistors in the other branches rt divided by rx plus rt remember rt is resolved in parallel now these rules can only be applied to simple circuits as illustrated in this slide and the last we cannot use them in bigger more complex circuits and hence would need to employ other methods of analysis note a complex circuit would be one with multiple sources loops and or nodes of these methods is mesh analysis now a mesh is simply a selected loop in our complex circuit and this method is named as such because it establishes equations for voltages in selected loops of a complex circuit using kirchhoff's voltage law and expresses them in terms of currents and resistances we know from kojov's voltage law that the sum of voltages in a loop equals zero for example in the simple circuit shown for a clockwise flowing current the mesh equation expands to vs is equal to v1 plus v2 plus v3 each circulating current variable and direction must be specified for each selected loop so that we can obtain the corresponding voltage equation then using ohm's law to substitute for the voltage drops in terms of i and r the equation would solve for our specified currents in this and of course the number of equations we obtain need to be equal to the number of unknown current variables we we have in order to solve for them all so we have this complex circuit for example and we need to find the current through the 12 ohm resistor i now we cannot just use the divider rules to do this because the circuit is too complex it has two voltage sources influencing the current through the 12 ohm using using method analysis we first select our loops loop current variables and their directions i've chosen i want to flow clockwise in the left loop and i2 to float clockwise as well in the right loop based on these selected mesh currents the current truly 12 ohm i is equal to i1 minus i2 now we need to apply kojov's voltage law and ohms loaded in meshes to obtain the corresponding mesh equation before we can do that we must obtain the voltage directions across the components in the loops for mesh one we have the voltage direction as shown according to the polarity of the voltage source and the selected direction of i1 remember the voltage across the voltage sources are predetermined based on their polarities the voltage is in the direction of the positive terminal of the source but the voltage drops across resistors in the loops is based on the selected current direction the voltage drops i in the opposite direction to the current flowing through it now applying kuchovs voltage law and ohm's law to mesh one these voltage directions give us 32 from the voltage source minus a 4 i1 the voltage drop across the 4 ohm minus 12 i1 minus i2 the voltage drop across the 12v remember these voltage drops are negative because they are opposite in direction to the voltage source to the voltage across the source simplifying the equation we have 8 minus 4 i1 plus 3 i2 is equal to 0. for mesh 2 the selected current direction of i2 gives us these voltage directions across the source and resistors applying just voltage law and ohm's law gives the equation 12 i2 minus i1 the voltage drop across the 12 ohm plus s3 i2 the voltage drop across the 3 ohm plus the 6 volt voltage source remember when analyzing mesh 2 the reference current is i2 so the current truly 12 ohm is i2 minus i1 not i1 minus i2 as before note that all these voltages are positive in the equation because they are all in the same direction the selected current direction for i2 in this loop gives us voltage drops in the same direction as the voltage across the source simplifying the equation we have 5 i2 minus 4 i1 plus 2 is equal to 0. using the elimination technique we can subtract the second mesh equation from the first to obtain i2 as three amps then substituting i2 into the second mesh equation we can calculate i1 to be 4.25 amps now that we have our mesh currents we can find the current i through the 12 ohm which is 1.25 amps if the question asks we can also obtain the voltages across the resistors of the circuit by applying ohm's law as shown so note that specifying different mesh current currents in your analysis will lead to different mesh equations but once the analysis is done correctly you should obtain the same answer as i did i have here a more complicated network on the solutions for finding the current i zero i'll leave you guys to go through this one on your own to solidify what you've learned this again has a dependent current source which is outside the scope of the course and i won't bring for example but you can learn from the example nonetheless so the first thing you all have to do before the next class is to review this example another method of analysis that can be used on complex circuits is nodal analysis this method is named as such because it establishes equations for currents at selected nodes using kuchov's current law and expresses them in terms of voltages and resistances we covered nodes and how to identify them in the last class so for this method we need to select the nodes of the complex circuit and specify the current variables and directions at each of them we know for kojov's current law that the sum of the currents entering a loop equals zero for the sum of the currents entering the node equals the sum of those leaving the for example in the simple circuit shown we have one node and the nodal equation expands to i1 is equal to i2 plus i3 each branch current variable and its direction must be specified at each node so that we can obtain the corresponding current equation then using ohm's law to substitute for the currents in terms of v and r the equations would solve for our specified voltages in the circuit and again the number of equations we obtain need to be equal to the number of unknown voltage variables we have in order to solve for them all but here we have another complex circuit but we want to apply nodal analysis this time to find the voltages p1 and v2 the nodes are identified on labels for you already so let's apply nodal analysis at each node we first need to select the branch currents connected to the node at node 1 i selected the current in the 5 arm current source branch to be i1 and to flow in the same direction of the source and towards the node i also selected the currents in the 4 ohm and 2 ohm branches to the i2 and i3 respectively and to flow out of the node applying kujo's current law to this node we have i1 is equal to i2 plus i3 now using ohm's law and substituting in terms of v on r we have this equation let's look closely at this equation i1 is just the value of the current source in that branch 5 amps i2 is v1 minus v2 the voltage drop across the form divided by the 4 ohms similarly i3 is v1 minus the earth zero potential divided by v2 remember the first voltage is always the voltage at the node where the current leaves and the other is is the voltage at the node where the current enters at node two i could be current in the 10 amp current source branch to be i4 and to flow in the same direction of the source and towards the node i also selected the current in the 6 ohm to be i5 and to flow out of the node the currents in the 5 amp current source branch i2 and the 4 ohm branch sorry i1 and i2 were already defined for us applying kuchov's current law to this node we have i2 plus i4 is equal to i1 plus i5 now using ohm's law and substituting in terms of v on r we have this equation looking closely at this equation i4 is just the value of the current source in that branch 10 amps i2 is the same as before v1 minus v2 divided by 4 ohms similarly i5 is v2 minus the earth zero potential divided by the six ohms and we know i1 is the five amp current source simplifying the equations and using the elimination technique we can solve for v2 which is 20 volts then substituting in the equation one we calculated v1 to be 13.33 volts if the question asks we can also obtain currents to the resistors of the circuit by applying ohm's law as shown note again that specifying different branch currents in your analysis will lead to different nodal equations but once the analysis is done correctly you should obtain the same answers i obtained the last method of analysis we will learn today is superposition it is based on the superposition principle applied to electrical circuits this states that the voltage across or current through an element in a linear circuit is the algebraic sum of the voltages across or currents through that element due to each independent source acting alone so according to this principle we can turn off all independent sources except one source find the desired voltage or current due to that active source using basic laws then repeat this step for each of the other independent sources then finally we can compute the final answer by adding algebraically all the contributions due to the independent sources let's solidify this like before with an example we have a complex circuit and we would like to find a voltage across the 12 ohm resistor v considering the current source only we remove the voltage source from the original circuit to do this we short circuit the voltage source so that the voltage across its terminal is now zero we now have a simplified circuit on which we can use the current divider theorem to find the current to the 12 ohm resistor and then ohm's law to find the voltage across it 6 volts this is the voltage across the 12 ohm resistor due to the current source only now considering the voltage source only we remove the current source from the original circuit to do this we open circuit the current source so that the current through it is nose zero we are therefore left with this simplified circuit on which we can use the voltage divider theorem to find the voltage across the 12 ohm resistor 24 volts this is the voltage across the 12 ohm resistor due to the voltage source only now the final answer is the contribution of both sources so we add the voltages we obtain due to both separately to get 30 volts most students find this to be the easiest of the three options of methods of analysis onto equivalent theorems any circuit with sources dependent on or independent and resistors can be replaced by an equivalent circuit containing a single source and a single resistor these theorems imply that we can replace arbitrarily complicated networks with simple networks for purposes of analysis so if complex circuits are to be connected to loads or other circuits it would be easier for analysis if we simplify them for example if we have a complex circuit for our phone charger it can be connected to a phone which has another complex circuitry simplifying these circuits would be beneficial for us to apply the laws and methods we've learned to calculate and predict the current sound voltages at their two minutes these complex circuits can be simplified into two forms thevenin's theorem states that a linear two-terminal circuit with independent and or dependent sources can be replaced by an equivalent circuit consisting of a voltage source vth in series with a resistor rth the vth is the open circuit voltage at the terminals across the voltage across the open terminals and rth is the input or equivalent resistance at the terminals when the independent sources are turned off so the circuit may contain multiple active and passive components within this box but when we apply seven and theorem we can simplify it to just one equivalent circuit a voltage source in series with our equivalent resistor the value of this voltage source vth is given by the open circuit voltage of the original complex circuit this means to find this voltage value we need to apply a method of analysis on the complex the value of the equivalent resistance rth is calculated by turning off all the sources in the complex circuit and simplifying the resistances to find the equivalent resistance between the load terminals the second form a complex circuit can be simplified into is given by norton's theorem it states that a linear two-terminal circuit with one or more independent sources can be replaced by an equivalent circuit consisting a current source i n in parallel with a resistor where i n is the short circuit current through the terminals and rn is the input or equivalent resistance at the terminals when the independent sources are turned off it is the same as rth so when we apply norton's theorem we can simplify the complex circuit into just one equivalent current source and an equivalent resistance in parallel with this source the value of this current source i n is given by the short circuit current of the original complex circuit this means to find this current value we need to apply a method of analysis on the complex circuit the value of the equivalent resistance rn is calculated in the same way as per thevenin's theorem by turning off all the sources and finding the equivalent resistance between the load terminals these two theorems are related to each other any circuit with a voltage source in series with a resistor ied theft in an equivalent circuit is in turn equivalent to a current source in parallel with that same resistor the norton equivalent circuit this is called source transformation so if you found both equivalents you would realize that the thevenin voltage source value divided by the thevenin resistance value would be equal to the northern current source value similarly the norton current source value multiplied by its naught and resistance value will give you the thevenin voltage source volume so finding a thought and no are not an equivalent circuit requires essentially the same process as finding a stubborn equivalent this simplifying process can be difficult to understand so let's break it down into steps step one pick a good breaking point in the circuit remember you cannot split a dependent source on its control variable so the breaking point of the circuit must be at its output terminals where the load is connected if you are given a circuit with a load inserted simply remove the load resistor and then proceed with the analysis step two if you are finding the seven and equivalent circuit the value of the seven voltage source is the open circuit voltage across the terminals so after having removed the load resistor you must use a method of analysis to calculate the voltage across the opened load terminals of the original circuit if you are finding the norton's equivalent circuit the value of the norton's current source is the short circuit current to the load terminal therefore after removing the load resistor from the original circuit short circuit the output terminals and use a method of analysis to calculate the current through this short circuit since the original circuits are usually complex the methods of analysis you use would either be mesh nodal or superposition [Music] step three find the cement or norton resistance if there are only independent sources in the circuit remove all power sources in the original circuit voltage sources are shorted and current sources are open to remove them and calculate the total resistance between the open connection points if they are both independent and dependent sources in the circuit we can't use this method we have to compute rth from both the open circuit divided by i short circuit from the last step remember we will not explore depend on sources in this course so you could ignore the parts that refer to dependent sources they are just there for the completeness of the lecture for the final step we simply draw the equivalent circuit for thevenin's the complex circuit is now replaced with or simplified to a voltage source of value the open circuit we calculated in step two and series with the calculated equivalent resistance for northerns the complex circuit is now replaced with or simplified to a current source of value i short circuit we calculated in step two and parallel with the calculated equivalent resistance the load resistor reattaches between the two open points of the equivalent circuit at the end of the analysis remember only steps 2 and 4 differ for sevenin and nota so you are presented with several options of methods of analysis the equivalent theorems and their steps i would like you to apply them to this circuit to find the thevenin's equivalent circuit i'll also like you to apply to this one to find the norton's equivalency during the tutorial class i will go through these solutions in detail but i urge you all to attempt it on your own this topic requires a lot of practice and you need to identify what your misconceptions are and where your confusion lies you will inevitably make mistakes doing circuit analysis so i suggest you make those mistakes before you enter the exam so after having reduced our circuit a typical analysis done is to determine the most power this circuit can produce or transfer to a load or another circuit maximum power transfer occurs when the resistive value of the load is equal in value to that of the source's internal resistance so when rl the resistance of the load is equal to rth in this case the source's internal resistance you have maximum power dissipated in the loop this statement can be proven through experimentation here we have a seven and equivalent circuit supplying a variable load resistor where the internal resistance is 25 ohms and the load is variable between 0 to 100 ohms we also have the supply voltage 100 volts now we are going to do this the long way using load resistances in increments of 5 ohms we will find the value of rl at which the source provides maximum power so incrementing the value of the load and calculating the power transferred to or dissipated by the load we will do this for each increment i only showed one sample calculation when the load resistance is 5 ohms the current would then be calculated using ohm's law the source voltage divided by the internal resistance plus the load gives 3.33 amps then we can use the current to calculate the power dissipated or transferred to the loop i squared rl which gives 55.6 watts now this calculation was done for the subsequent increments of load resistances the results for each increments were tabulated as shown and a graph of power versus load resistance was what from the graph we can see the peak or maximum power occurs at 25 ohms which is the value of the internal resistance rs therefore this proves that the maximum power is dissipated in the load when the load resistance is equal to the internal resistance value 25 ohms the last task i have for you all is to try this same experiment on the norton equivalent circuit shown find the value of rl at which the source provides maximum power so i've given you all a lot to do for next class so please make every effort to attempt them any difficulties will be addressed as we go through the solutions in the tutorial class again if you all have any questions about the theory presented in this lecture feel free to ask in the next class or on the class forum you