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Lecture 5 o Aim of the lecture More detail on Capacitance Ohms Law Capacitance Energy storage Dielectrics Ohms Law resistance Power dissipation o Main learning outcomes familiarity with Dielectrics Resistance Addition of Capacitors Resistors Power dissipation in resistors Dielectrics Material between plates oA dielectric is o an insulator o either polar o or non-polar No dielectric between plates C = e0A/d With a dielectric, then C = ere0A/d + - Polar Dielectric Electrons leave one plate The same number arrive on the other +ve o In a polar material molecules are polarised act like small dipoles orientate to align with E field E -ve [pure water is like this] Polar Dielectric Consider Gauss’ Law surface as shown net charge inside surface reduced +ve E charge = e0 -ve q-q’ = e0 ∫ E.dA But charge is now q-q’ where q is the charge without dielectric q’ is the charge due to polar molecules ∫ E.dA Polar Dielectric +ve q-q’ = e0 ∫ E.dA E -ve which gives that the electric field is E = (q-q’)/e0A compared with E0 = q/e0A with no field The electric field is weaker when a dielectric is present for the same applied voltage The quantity of charge q’ depends on E0, (often) which depends on q so q’ is proportional to q Polar Dielectric +ve so q’ = const q and E -ve E = const E0 define that 1/const is er with the result that C = ere0A/d The quantity of charge q’ depends on E0, (often) which depends on q so q’ is proportional to q Other geometries are possible cylindrical Sphere above a plane A non-polar dielectric is one where the molecules are non-polar o In this case the molecules are CAUSED to be polar by the electric field, they are Induced dipoles In practice the only difference is that the values of er are (usually) smaller than for a polar dielectric Adding Capacitance The net charge on these two plates is zero, it is just an equipotential line Add a second capacitor in series Adding Capacitance Overall effect is to double distance between plates but C = ere0A/d so capacitance is halved C Total capacitance = C/2 C Adding capacitors in series REDUCES the total capacitance More generally When adding capacitors in series 1 CT = 1 C1 + 1 C2 Example: capacitance with partial dielectric Capacitors Connected in Parallel o Adding a secondbut identical in parallel C = erecapacitor 0A/d doubles the area of the plates doubles charge stored for same voltage applied so capacitance is doubled More generally When adding capacitors in parallel CT = C1 + C2 CT = 1 CT = C1 + 1 C1 + C2 1 C2 PARALEL SERIES Energy Stored The energy stored is equal to the energy in the electric field between the plates. Q = CV Work done to move a small charge, dq from one plate to the other is dW = VdQ = VCdV So total energy, E is E = ∫dW = C∫VdV = ½CV2 Summary for Capacitors Q = CV E = ½CV2 CT = C1 + C2 1 1 1 CT = C1 C = ere0A/d + C2 For parallel addition For series addition for parallel plate capacitor e0 is 8.854×10−12 F m–1 er is typically between 1 and 10 Practical Information: Capacitors are labelled in a ‘funny’ way TheMore units than are always you mF wanted or pF to know! And this means the maximum voltage you can put across it is 100V examinable This not means 100k pF ie 100 x 103 x 10-12 F = 100nF but useful (it cant be mF because it would be too big – you just have to know this!) Final Warning: some capacitors are ‘polarised’ you MUST put the voltage across them the correct direction. This little + means that this terminal must be +ve compared with the other Tantalum bead capacitors can explode if connected the wrong way round!! 22u 35 means 22mF max volts = 35V More on Ohms Law and Resistance Recall that V = IR Where V is the voltage applied across a resistance, R and I is the current that flows. The resistance is analogous to the resistance of a pipe to the flow of water through it. o Electrons are made to drift in an electric field caused by an external voltage. o They loose energy in collisions with the fixed atoms o They therefore do not accelerate o They drift at constant speed Consider a resistance with a voltage across it. V Suppose the current that flows is Ia Current = 2Ia If we apply the same voltage across two such resistances connected in parallel, then the current doubles, so the resistance is inversely proportional to the area, A, of conductor If we put two in series, then we need a voltage V across each to drive the same current, Ia, so resistance is proportional to length, L Resistance = r L/A for many materials r is a constant called the ‘resistivity’ of the material and is very different for different conductors. V=IR o Any real circuit has resistance o Usually wires are a small resistance we ignore it, assume it is zero Represent the resistance with a RESISTOR o The wires that we draw joining parts of a circuit are Taken to have zero resistance Resistance is represented by o An ampmeter has zero resistance o A voltmeter has a very high resistance (infinite if perfect) More practical details! Here is how you can tell what a resistor value is: These resistors are 100W with 5% accuracy Resistors in series. Recall that for identical resistors, the resistance is proportional to the length. This generalises for resistors in series Rtotal = R1 + R2 + R3 + ….. For resistors in parallel, it is like increasing the area, so two in parallel gives half the resistance, and in general: Rtotal The energy transferred to the atoms when the electrons collide with them in a resistor is converted to heat Power = current x voltage P = IR but remember ohms law V=IR So P = I2R P = V2/R