Electric Circuits Part 2: Kirchhoff s Rules Last modified: 31/07/2018
Contents Links Complex Circuits Applying Kirchhoff s Rules Example Circuit Labelling the currents Kirchhoff s First Rule Meaning Kirchhoff s Second Rule Meaning Path Direction Sign Convention for EMFs Sign Convention for Resistors Kirchhoff s Rules Recipe Extra Example
Complex Circuits Contents The basic problem to be solved in analyzing a circuit is to determine the currents flowing in the various parts of the circuit. In the last lecture, we looked at how for some circuits, this could be done by repeated application of the rules for series and parallel resistor combinations, to find a circuit containing a single EMF and a single resistance. A circuit that can be solved in this way is known as a simple circuit. However, not every circuit can be solved using this method. Such circuits are called complex circuits. For example, the circuit at right is complex. (Two or more EMF s that are not connected in series, as in this circuit, is a good indicator that a circuit is complex.)
Example Circuit Contents Solving simple circuits involved using Ohm s Law (Remember the series and parallel formulas are derived using Ohm s Law). Complex circuits are also solved using Ohm s Law, but the method is a little more complicated and requires the use of Kirchhoff s Rules. To demonstrate the method, we will determine the currents flowing in the following circuit. There is nothing at all special about this example - any circuit can be solved using the same steps. 3 Ω 6 V 4 Ω 3 V The first step is to assign labels to the different currents.
Labelling the Currents Contents Looking at the circuit, we should realize that there are three different currents flowing - we ll call them i 1, i 2 and i 3. Each current flows in a separate branch of the circuit between two junctions. But each of these currents has two possible directions - there are eight (!!) different ways to label the currents: Which is the correct choice? It doesn t matter. Any choice will do. If after solving for a current we obtain a negative answer, this will tell us that the actual direction is opposite to the one we have chosen.
Contents Let s choose one of these possibilities for our example: 3 Ω 6 V 4 Ω 3 V i 1 i 3 i 2 We need to solve for three different currents. However, the currents are not independent. Looking at the junction at the bottom of the circuit, the current flowing in breaks into two parts. Therefore i 3 = i 1 + i 2. We have just used Kirchhoff s First Rule.
Kirchhoff s First Rule Contents At any junction (or node) in a circuit, the sum of currents flowing in equals the sum of currents flowing out. i = i out in For example: i 1 i 3 i 4 i 1 + i 2 = i 3 + i 4 i 2
Kirchhoff s First Rule: Meaning Contents Where does this rule come from? It is really just a consequence of the conservation of charge - charge does not suddenly appear or disappear. This requires that the flow of charge into a point - the node - must equal the flow of charge out, as the amount of charge cannot change.
Contents Kirchhoff s first rule alllows us to reduce the number of variables we need to solve for. In our example circuit, there are three currents, expressed in terms of only two variables. 3 Ω i 1 i 2 6 V 4 Ω 3 V i 1 + i 2 To solve for the two variables i 1 and i 2, we need two equations involving them. These equations are obtained by using Kirchhoff s Second Rule.
Kirchhoff s Second Rule Contents For any path around a circuit loop, the sum of gains in potential (i.e. EMFs) equals the sum of losses across resistors (given by Ohm s Law). E = ir For example: R 1 Around the blue path, E 1 + E 2 = i 1 R 1 + i 2 R 2 i 1 E 1 E 2 i 2 R 2
Kirchhoff s Second Rule: Meaning Contents Where does this rule come from? Again it is nothing really new. The rule comes from the conservation of energy - since we start and finish at the same point in the circuit, the total change in potential (i.e. energy) over the loop must be zero. The potential has changed along the way, so the total increase (the EMFs) must be equal to the total decrease (resistors) in potential.
Path Direction and Signs Contents Note that in this example, the blue path was taken in the clockwise direction. Why? No reason. The path is completely imaginary - just a tool to create the equations involving currents - so could have been taken anti-clockwise instead: R 1 i 1 E 1 E 2 Around the blue path: E 1 E 2 = i 1 R 1 i 2 R 2 i 2 R 2 The equation obtained is the same as before, but multiplied on both sides by 1. Why?
Sign Convention for EMFs Contents If our path takes us the wrong way through an emf (i.e. through the positive end first) then we are travelling from high potential to low - a loss instead of a gain. The emf must be then counted with a minus sign. emf counts as positive: path emf counts as negative: path The direction of the current through the emf is irrelevant to the sign in the equation.
Sign Convention for Resistors Contents Similar logic will apply to resistors in the circuit. Current flows from high potential to low, so if our path direction is opposite the current as labelled, then we are travelling from low potential to high - a gain instead of a loss. The ir term must be counted with a minus sign. ir counts as positive: ir counts as negative: i path i path
Contents Back to our example circuit, and we choose two loops: i 1 i 2 3 Ω 6 V loop A 4 Ωloop B 3 V i 1 + i 2 i 1 i 2 Applying K2 to the two loops gives: E = IR loop A: + 6 = 2i 1 + 4(i 1 + i 2 ) = 6i 1 + 4i 2 loop B: + 3 = 3i 2 + 4(i 1 + i 2 ) = 4i 1 + 7i 2
Contents We ve already seen that taking either of these loops in the opposite direction, would give the same equations. The directions used were good choices beacause they minimized the number of negative signs coming from going the wrong way through an emf or resistor. In fact for these loops, no negatives were required. It is not always possible to avoid them completely, but it is a good idea to try to minimize the number. Other loops are possible - for instance the loop around the outside of the circuit: i 1 i 2 3 Ω loop C 6 V 4 Ω 3 V i 1 + i 2 i 1 i 2
Contents Applying K2 to this new loop gives: E = IR loop C: 3 + 6 = 2i 1 3i 2 The previous loop equations were: loop A: 6 = 6i 1 + 4i 2 loop B: 3 = 4i 1 + 7i 2 Looking closely, we should notice that the loop C equation is formed from the loop A equation minus the loop B equation. If we remember our maths, then we know that the solutions to any two of these equations will be the same. We can choose any two different loops to use with Kirchhoff s second rule.
Contents The solutions to these equations are (check this for yourself!): i 1 = 15 13 A and i 2 = 3 13 A, i 3 = i 1 + i 2 = 12 13 A. The negative sign of the solution for i 2 tells us that the real current is in the oppposite direction to our original guess, while the other two assumed directions were correct: 3 Ω 15 13 A 3 13 A 6 V 4 Ω 3 V 12 13 A
Contents Any circuit of EMFs and resistors can be solved using Kirchhoff s Rules, though the maths very quickly gets complicated. For example, look at the circuit at right: One of many ways of labelling the currents in this circuit is shown. Here there are three variables describing six different currrents, so we would need to solve a system of 3 equations, obtained using three different circuit loops. i 1 i 2 i 1 i 2 i 1 i 3 i 2 i 3 i 3
Kirchhoff s Rules Recipe Contents Every Kirchhoff problem is solved in the same way, using the following steps. There are no tricks, just longer and shorter maths equations to solve. 1. Label (with direction) every current in the circuit. (Exam questions will often have some currents already labelled, if so you should use these) 2. Use Kirchoff s First Rule to reduce the number of variables describing the currents to N. (Exam questions will not have N > 2) 3. Use Kirchhoff s Second Rule to construct N equations involving these N variables. Any N different circuit loops will do. 4. Solve these N equations. 5. Quickly redraw the circuit, showing the values and correct directions of all currents.
Example Contents (a) Calculate the magnitude and direction of the currents flowing in each of the resistors in the circuit at right. (b) Calculate the total power loss in the circuit. 10 V 5 V 4 Ω 1 Ω The first thing to notice is that there are two EMFs in this circuit. This tells us that it will be necessary to use Kirchhoff s Rules. However, this doesn t mean that we should completely forget about series and parallel resistors.
Contents The two resistors in the top left of the circuit are in parallel. 1 Ω 10 V 5 V 4 Ω 10 V 5 V 4 Ω 1 Ω and the two 1 Ω resistors are in series. 1 Ω Real world problems will often be like this - use parallel and series first, until we get stuck, then use Kirchhoff s Rules. 10 V 5 V 4 Ω
Contents First Step: Label the currents. Note that this is just one of many ways to do this. In an exam it is important to clearly show the currents you have chosen. Second Step: Use Kirchhoff s First Rule. Kirchhoff s First Rule allows us to express one of the currents in terms of the other two. Again there are multiple equally correct ways to do this, so it is important to clearly show your choice. i 10 V 5 V 4 Ω i i 10 V 5 V 4 Ω I + i I I
Contents Third Step: Choose loops to use with Kirchhoff s Second Rule. Again there are a few different possibilities, so clearly label the loops (with direction) that you are using. i 10 V 5 V A 4 ΩB i + I I Fourth Step: Use Kirchhoff s Second Rule to Obtain Equations. E = IR loop A +10 = 2i + 4(i + I) = 6i + 4I loop B +5 = 2I + 4(i + I) = 6I + 4i Fifth Step: Solve the Equations Solving these two equations gives: I = 0.5 A and i = 2 A
Contents Sixth Step: Work Out the Currents I = 0.5 A and i = 2 A means that the third current is i = i+i = 1.5 A. Note that I being negative means that the direction of the current is actually opposite to that we initially assumed. Finally, as we did in the last lecture, we can calculate the currents flowing in the series and parallel combinations. 2 A 0.5 A 10 V 5 V 4 Ω 1.5 A 1 A 1 A 10 V 0.5 A 1.5 A 4 Ω 5 V 1 Ω 2 A
Contents Calculate Power Loss We are also asked to find the total power lost in the resistors in the circuit. We should remember the formula for the power loss in a resistor from last lecture: P = i 2 R We could use this formula for each of the resistors and the currents we have just calculated, and add them together. The smarter method is to use the equivalent circuit at right. (The power loss in the equivalent circuit will be the same - this is what equivalent means!) 10 V 2 A 0.5 A 5 V 4 Ω 1.5 A Total Power Lost = 2 2 2 + 0.5 2 2 + 1.5 2 4 = 17.5 W
Contents Alternatively, we could look at the power being input by the EMF s - energy conservation tells us that the power input must equal the power output. The power input of an EMF is given by the same formula as for a resistor: P = VI Power Input = EI = 10 2 + 5 ( 0.5) = 17.5 W 10 V 5 V 2 A 4 Ω 0.5 A The 5 V EMF is taking energy away from the circuit, i.e. it is being charged. The 10 V EMF adds 20 W to the circuit. 17.5 W is dissipated in the five resistors, while the remaining 2.5 W is lost to the 5 V EMF.
Contents Kirchhoff s Rules can be used to solve for the currents in any circuit involving only resistors and EMFs. This will always require solving N equations with N variables, with N increasing as the circuit becomes more complicated. For N > 2, this is becomes very tedious to do by hand, so usually computers are used. Slightly modified versions of Kirchhoff s Rules are used to analyze even more complicated circuits, possibly involving AC currents or capacitors and other circuit elements (diodes, transistors etc).