EGR 1980 Recitation Worksheet 3 Using Equations in Circuit Analysis H Griffith 1 February 2014 Introduction In lecture we reviewed techniques for solving linear equations in 1 variable Today we ll demonstrate the usefulness of these techniques in analyzing circuits whose loading networks are slightly more complex Circuit Analysis Kirchhoff s Laws Last recitation we completed our definitions of all necessary circuit quantities and their associated units In addition we defined Ohm s Law which is useful for describing all necessary quantities in a circuit consisting of only one load or equivalently describing a more complex network which has been specified by its equivalent resistance Today we ll examine two additional laws which are useful for solving slightly more complicated circuits The first law known as Kirchhoff s Voltage Law or KVL states that the sum of all voltage rises in a circuit loop is equivalent to the sum of all voltage drops in a circuit loop The second law known as Kirchhoff s Current Law KCL states that the sum of all currents entering a node in a circuit is equivalent to the sum of all circuits exiting a node For example let s consider the application of KVL to a simple series circuit consisting of two 1k loads connected in series and excited by an ideal 10V voltage source In order to apply KVL it is first necessary to draw an assumed direction of current flow in the complete circuit By convention you should always assume that current flow is directed from the positive terminal of the source towards the negative as was discussed in lecture this assumption is not physically correct as the charges which are truly mobile in a conductor are electrons According to the laws of electricity electrons would move in the opposite direction Therefore we say that true electron flow in a circuit is always opposite that of conventional current flow Once the assumed current has been drawn it is then beneficial to define voltage drops across each load in the circuit Physically you may think of these drops as the voltage that is required to push current through the opposition provided by each resistor Note that since there is only one path for the flow of electricity the current flow in each component must be identical This can only occur if the voltage is divided in a manner such that a greater portion of the source voltage is allocated to loads with greater resistance ie To push the same amount of current through a component with greater resistance a greater portion of the source voltage is required Therefore series circuits are often referred to as voltage dividers I V1 V2 Once you have defined all voltage drops across loads pick a starting point in the circuit typically chosen as the negative terminal of the voltage source and moving clockwise apply KVL by equating all voltage rises to all voltage drops Note that the two are distinguished by the polarity that is realized when moving across the component ie when you move across the voltage source in a clockwise fashion the sense of polarity is from to which corresponds to a rise In contrast when moving across a load the sense of polarity is to corresponding to a drop Doing this yields the following linear equation Note that although the equation is linear it consists of two unknown values By the principles of algebra we are unable to solve for two unknown values given only one equation Therefore it is necessary to use additional physical information in order to express the two unknowns in terms of one other common unknown value Note that by Ohm s Law both unknown voltage drops may be expressed as the product of an unknown current and a known resistance value Since the loads are arranged in series this unknown current is the same in both components Therefore upon substitution the number of unknowns is reduced to one ie Note that in the above expression the corresponding prefixes of the units have been converted to their equivalent orders of magnitude This is essential in order to ensure that the resulting quantity the series current is expressed in terms of its base units of Amperes We may now apply the principle of combining like terms in order to simplify the equation as follows Now we may solve for the desired unknown current by applying the multiplicative property of equations Namely we isolate the unknown current by dividing by its coefficient in the equation ie It should be noted that an identical result could have been achieved by applying the principle of equivalent circuits Namely in the original schematic the effect of two 1k loads in series could have been replaced by the equivalent resistance value of 2k Upon doing this the resulting network would be reduced to the simplistic form in which Ohm s Law could be applied directly thereby giving the same value for the circuit s current While KVL is useful for solving complex circuits involving series loops KCL is useful for solving complex circuits involving parallel branches For example consider the following circuit To perform our analysis we begin by first sketching an assumed direction of conventional current flow By convention current moves away from the positive terminal of the battery through the first resistor before reaching the node and splitting between the two possible parallel branches Note that we may enforce KCL at the point of branching in the network ie I1 I2 I3 As was the case in the previous example the above linear equation involves 3 unknowns and is thus unsolvable in its current form However as was done in the last case we may utilize other physical laws in order to express the three unknowns in terms of one common unknown Namely we can express the two currents through each branch I 1 and I2 in terms of the voltage across each branch and the branch s resistance Note that since the two loads are arranged in parallel the voltage across each is the same by definition ie We can now complete our approach if we are able to express the first current in terms of the unknown common voltage across the parallel branch This can be accomplished by thinking physically about the voltage drop across the first resistor R1 Note that by KVL this voltage drop is given as the difference of the source voltage and the common voltage across the parallel branch ie Thus As was demonstrated in class the solution to this equation is greatly simplified by eliminating fractions This is accomplished by multiplying both sides of the equation by the LCD of all fractions involved