9Voltage and CurrentVoltage “V” is the potential difference between two points. (Imagine as height) Current “I” is the rate of flow of charged particles. (Imagine as water flow)
10What are resistors and conductors? An applied voltage across an object causes an electric field across the material. The electric field accelerates any “free” electrons in the material. This motion is the electric current. Electrons collide with atoms which slows them down increasing resistance to the current.
11What factors determine resistance? The greater the length of an object the more resistance it will have. The greater the cross sectional area the less resistance it will have. Resistance = Resistivity * Length / Area
12What properties affect resistivity? Some materials (e.g. copper) have lots of free electrons and have low resistivity, (good conductors). Others (e.g. glass) have almost no free electrons. Semiconductors (e.g. silicon) have modest numbers In a superconductor the electrons don’t ever collide with the atoms so the resistivity is zero.
13Some real resistorsCommercial resistors are often carbon or metal film. 6 inches of HB pencil (Carbon) is about 16 Ohms If we connect 6V across it we get 1 volt per inch If we connect 12V across it we get 2V per inch (and it catches fire)
14I = V / R Ohms Law Amps = Volts / Ohms For an ideal resistor “R” The current “I” increases with applied voltage “V” (Electromotive force)The greater the resistance “R” to the current the less current I flows.I = V / RAmps = Volts / Ohms
15Resistors in series R = R1 + R2 Our resistors both resist the current. Its like one longer resistor (or pencil). We add the resistances R1 and R2 so:R = R1 + R2I = V / (R1+R2)
17The potential dividerJust like in our pencil the voltage will distribute itself proportional to the resistance.E.G if R1 is twice R2 then 1/3 of the voltage will be across R2.So V will be 4 volts.
18The potential divider V2 = V1 * R2 / (R1+R2) (We can prove this from Ohms law)I = V1/(R1+R2)I = V2/R2
19Resistors in parallelOur resistors both carry current so its like one thicker resistor. We add the currents so I = V / R1 + V / R2From Ohms law we have I = V / RSo: / R = 1 / R1 + 1 / R2Or: R = (R1 * R2) / (R1 + R2)
20A simple networkFirst we combine the parallel resistors. Using R = (R1 * R2) / (R1 + R2)So R = 2000*2000/( ) = 1KWe now have a potential divider with 1K at the top and our combined 1K at the bottom.Using V2 = V1 * R2 / (R1+R2)So V = 12Volts * 1000 / ( )So V = 6 Volts.
21Kirchoff’s lawsSometimes you can (and should) calculate what you want just using the principles of:Resistors in series.Resistors in parallel.Potential dividers.However for network analysis you need Kirchoff’s laws.
22Kirchoff's 2 Laws We have already started using these. Current is conserved. So if you add up all the current into a connection (a node) it will add to zero.Voltage measurements are consistent. So if in a loop you add up the voltage differences between successive nodes the result will be zero.There are some pitfalls so we need to look in detail.
23Current is conservedTo apply this rule we mark an arrow on every link in a circuit and label the links I1, I2 etc.You regard current in the direction of your arrow as positive, otherwise its negative.(It doesn’t matter which way you mark the arrows)
24Current is conservedFor each node we can write an equation. In this case its:0 = I1 + I2 - I3 - I4using positive for arrows pointing to the node and negative if they point away.Alternatively I1 + I2 = I3 + I4
25Voltage measurements are consistent For each loop we can write an equation. Since voltages add we can work round the loop:0 = (V2-V1) + (V3-V2) +(V4-V3)+ (V1-V4)Mathematically this is true. (But that doesn’t prove Kirchoff's law)
26Direction of voltagesIf we mark a current arrow I on a component (e.g. a resistor) then for V= I * R we must regard the tail of the arrow as positive.So in this exampleV2-V3 = I * R
27Previous example using Kirchoff’s laws I2+I3=I1(V-0)/2K+(V-0)/2K= (12-V)/1KV/2+V/2= 12-VV=12-V2*V=12V=6 volts
28Using Kirchoff’s current law again I2+I3=I1(V-3)/2K+(V-0)/2K= (12-V)/1K(V-3)/2+V/2= 12-VV-3+V=24-2*V4*V=27V=6.75 volts
29The current law for multiple nodes We write an equation for each node. Assuming V0 and V2 are the supply:(At V1) I1+I2+I3=0(At V3) I4+I5 -I3=0Ohms law gives equations for I e.g.I2 = (V0 – V1)/R Note not V1-V0 !!Solve using simultaneous equations.7 unknowns: V1, V3, I1, I2, I3, I4, I57 equations: 2 above plus 5 Ohms Law