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Kirchoffs Laws Physics – 12.1.7 Kirchhoff’s laws Conservation of charge and energy in simple d.c. circuits. The relationships between currents, voltages and resistances in series and parallel circuits; questions will not be set which require the use of simultaneous equations to calculate currents or potential difference Mr Powell Kirchoffs Law I The “current law” states that at a junction all the currents should add up. I3 = I1 + I2 or I1 + I2 - I3 = 0 • currents towards a point is designated as positive • those away from a point as negative. • In other words the sum of all currents entering a junction must equal the sum of those leaving it. • Imagine it like water in a system of canals! Kirchoffs Law I Examples; If I1 = 0.1A, I2 = 0.2A, I3 = 0.3A If I1 = -0.1A, I2 = -0.2A, I3 = -0.3A If I1 = 2A, I2 = 3A, I3 = 5A There are some important multipliers for current: 1 microamp (1 A) = 1 x 10-6 A 1 milliamp (mA) = 1 x 10-3 A Also remember to make sure you work out current in Amps and time in seconds in your final answers! Kirchoffs Law I – Test it out… You can test out this theory by making a resistor network and connecting two power sources which you alter to change the currents. Kirchoffs Law I – Questions? Work out the currents and directions missing on these two junctions? 7A 3A Kirchoff Law I – Complex Questions Given the following circuit can you pick out how the current might behave using kirchoffs current laws. We are looking to find the current in the main branch or 0.75 resistor and also through the 1 resistor You may want to redraw the circuit then apply simple ideas of additive resistance and ratios to find out the currents? Kirchoff Law I - Answers This example is more simple than it looks. In fact you have one resistor on its own (0.75) Then the other three are in parallel with each other. With the 1 on one branch and the two 1.5 resistors on the other. You can then simply use resistance ratios to determine the current flow. The ratio for the parallel part is 1:3 so we say that the least current flows through the most resistive part. Hence: current through 1 resistor must be 0.75A. The current through main branch must be the sum of these i.e. 1A Kirchoff Law I - Answers In detail this means can rewrite the circuit and fill in the values for V and I as such… Energy (p.d.) is shared simply according to resistance. Kirchoffs Law II The “voltage law” states that the sum of e.m.f’s around a circuit or loop is equal to zero. i.e. all energy is transferred before you return back to the cell. In other words the sum of all voltage sources must equal the sum of all voltages dropped across resistances in the circuit, or part of circuit. Think of it like walking around a series of hills and returning back to point of origin – you are then at the same height! For more complex examples we must note the following rules; There is a potential rise whenever we go through a source of e.m.f from the – to the + side. There is a potential fall whenever we go through a resistance in the same direction as the flow of conventional current. i.e. + to NB. Both laws become obvious when you start applying them to problems. Just use these sheets as a reference point. Simple e.m.f example 1) If I = 100 mA what is that in amps? 100 mA = 0.1 A 2) What is the current in each resistor? 0.1 A 3) Work out the voltage across each resistor. V = IR so V1 = 0.1A x 30 = 3 V V2 = 0.1A x 40 = 4 V ; V3 = 0.1A x 50 = 5V 4) What is the total resistance? RT = 30 + 40+ 50 = 120 5) What is the battery voltage? V = 0.1A x 120 = 12V Complex Examples Hint1: This is complex and you must try and be consistent in your calculations in direction and which way the current is flowing or p.d. is lost! If we apply Kirchhoffs laws (previous slide) about current we can say that; I1 = I2 + I3 or 0 = I2 + I3 - I1 Hint2: This type of circuit is the top end of AS and you will not be asked to solve one using this Simultaneous Eq method unless they give you some values to help you out! If we apply Kirchhoffs laws (previous slide) about pd = 0 in closed loop we can say the following two things Starting at Point “P” and going clockwise around the left-hand loop; -I3R2 + E1 - I1R1 = 0 Starting at Point “P” and going clockwise around the right-hand loop; -E2 - I2R3 + I3R2 = 0 EXTENSION WORK R1 P I3 E1 I2 E2 R2 I1 R3 Hint: consider only E direction not I’s Complex Examples Starting at Point “P” and going clockwise around the left-hand loop; -I3R2 + E1 - I1R1 = 0 Starting at Point “P” and going clockwise around the right-hand loop; - E2 - I2R3 +I3R2 - = 0 p.d. is lowered with flow I1 R1 P I3 E1 p.d. is increased with flow I1 PD against flow of current I2 E2 R2 I1 p.d. down with flow I3 left loop R3 p.d. is up with flow I3 right loop PD against flow of current EXTENSION WORK Complex question…. Use the theory from the previous slide to answer the question below working out the current flow we have called I3. Hint use the technique shown on the previous slide. But try and reason it out yourself with the rules you have been given. This may take some time! Your answer should include the following; 1. 2. 3. 4. Reference to the rules of current flow & p.d Explanation as to why each contribution is + or – Equation for each loop Answer for I3 E1 E1 = 6V E2 = 2V R1=10 I1 R2=10 R3=2 Answer: I3 = 0.4A R1 P I3 I2 E2 R2 R3 Complex example.. Kirchoffs 1st Law; I1 = I2 + I3 This worked example relies on two equations found from two loops. Each defined for a separate power source. Kirchoffs 2nd law; Solve simultaneously to find I’s 30V = 20I3 + 5I1 Sum PD Loop AEDBA - Eqn 1 Sum PD Loop FEDCF 10V = 20I3 - 10I2 10V = 20I3 - 10(I1 - I3) 10V = 30I3 - 10I1 - Eqn 2 Add 2 x Eqn 1 + Eqn 2 70V = 70 I3 I3 = 1 A Substitute this in Eqn 1 EXTENSION WORK I1 = 2A so I2 = 1 A