1. Mr Powell 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

2.  The “current law” states that at a junction all the currents should add up. I 3 = I 1 + I 2 or I 1 + I 2 - I 3 = 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

3. Examples; If I 1 = 0.1A, I 2 = 0.2A, I 3 = 0.3A If I 1 = -0.1A, I 2 = -0.2A, I 3 = -0.3A If I 1 = 2A, I 2 = 3A, I 3 = 5A Kirchoffs Law I 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!

4.  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 – Test it out…

5.  Work out the currents and directions missing on these two junctions? Kirchoffs Law I – Questions? 7A 3A

6. 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?

7. 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

8. 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.

9.  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. Kirchoffs Law II

10. 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 V 1 = 0.1A x 30  = 3 V V 2 = 0.1A x 40  = 4 V ; V 3 = 0.1A x 50  = 5V 4) What is the total resistance? R T = 30  + 40  + 50  = 120  5) What is the battery voltage? V = 0.1A x 120  = 12V

11. Complex Examples P E1E1 E2E2 R1R1 R2R2 I2I2 R3R3 I3I3 I1I1  If we apply Kirchhoffs laws (previous slide) about current we can say that; I 1 = I 2 + I 3 or 0 = I 2 + I 3 - I 1  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; -I 3 R 2 + E 1 - I 1 R 1 = 0  Starting at Point “P” and going clockwise around the right-hand loop; -E 2 - I 2 R 3 + I 3 R 2 = 0 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! 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! EXTENSION WORK

12. Complex Examples P E1E1 E2E2 R1R1 R2R2 I2I2 R3R3 I3I3 I1I1 Starting at Point “P” and going clockwise around the left-hand loop; -I 3 R 2 + E 1 - I 1 R 1 = 0 Starting at Point “P” and going clockwise around the right-hand loop; - E 2 - I 2 R 3 +I 3 R 2 - = 0 p.d. down with flow I 3 left loop p.d. is increased with flow I 1 p.d. is lowered with flow I 1 PD against flow of current p.d. is up with flow I 3 right loop Hint: consider only E direction not I’s EXTENSION WORK

13. Complex question…. P E1E1 E2E2 R1R1 R2R2 I2I2 R3R3 I3I3 I1I1  Use the theory from the previous slide to answer the question below working out the current flow we have called I 3.  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.Reference to the rules of current flow & p.d 2.Explanation as to why each contribution is + or – 3.Equation for each loop 4.Answer for I 3 E 1 = 6V E 2 = 2V R 1 =10  R 2 =10  R 3 =2  Answer: I 3 = 0.4A

14. Kirchoffs 1st Law; I 1 = I 2 + I 3 Kirchoffs 2nd law; Sum PD Loop AEDBA 30V = 20  I 3 + 5  I 1 - Eqn 1 Sum PD Loop FEDCF 10V = 20  I 3 - 10  I 2 10V = 20  I 3 - 10(I 1 - I 3 ) 10V = 30  I 3 - 10  I 1 - Eqn 2 Add 2 x Eqn 1 + Eqn 2 70V = 70  I 3 I 3 = 1 A Substitute this in Eqn 1 I 1 = 2A so I 2 = 1 A Complex example..  This worked example relies on two equations found from two loops. Each defined for a separate power source.  Solve simultaneously to find I’s EXTENSION WORK