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Transcript
Electricity and circuit 1 Electric Charge • Ordinary matter is made up of atoms which have positively charged nuclei and negatively charged electrons surrounding them. • The unit of electric charge is the coulomb • Charge is quantized as a multiple of the electron or proton charge: Helium atom (schematic) Showing two protons (red), two neutrons (green) and two electrons (yellow). 2 Potential Difference and Electric Current • Charges can “lose” potential energy by moving from a location at high potential (voltage) to a location at low potential. • Charges will continue to move as long as the potential difference (voltage) is maintained. • A sustained flow of electric charge due to the potential difference is called an electric current. • If 1 Coulomb of charge (6.25 x 1018 electrons) pass a point each second, the current is 1 Ampere. 3 Current … • Electric current (I): the number of coulombs of charge that pass by a certain point per second. • Currents flow through metal wires via the motion of electrons, which are negatively charged, BUT the direction of motion of the electrons in a circuit is always opposite to the direction of the current. When charged particles are exchanged between regions of space A and B, the electric current flowing from A to B is defined as q I t where Δq is the change in the total charge of region B over a period of time Δt. 4 Ions moving across a cell membrane • The figure shows ions, labeled with their charges, moving in or out through the membranes of three cells. If the ions all cross the membranes during the same interval of time, how would the currents into the cells compare with each other? Assuming that the rate of flow is constant – Cell A has positive current going into it because its charge is increased, i.e. has a positive value of q. – Cell B has the same current as cell A, because by losing one unit of negative charge it also ends up increasing its own total charge by one unit. – Cell C’s total charge is reduced by three units, so it has a large negative current going into it. – Cell D loses one unit of charge, so it has a small negative current into it. 5 Current… • In most DC electric circuits, it can be assumed that the resistance to current flow is a constant so that the current in the circuit is related to voltage and resistance by Ohm's law. 6 Number of electrons flowing through a light bulb • If a light bulb has 1.0 A flowing through it, how many electrons will pass through the filament in 1.0 s? – We are only calculating the number of electrons that flow, so we can ignore the positive and negative signs. Also, since the rate of flow is constant, we may use the definition of current as Δq/Δt . – Solving for Δq = I Δt gives a charge of 1.0 C flowing in this time interval. – The number of electrons is • number of electrons • • • = coulombs ×electrons/coulomb = coulombs/coulombs/electron = 1.0 C/e = 6.2 × 1018 7 Problem #1 • Let’s consider what happens when the nerve is stimulated to transmit information. • When the blob at the top (the cell body) is stimulated, it causes Na+ ions to rush into the top of the tail (axon). This electrical pulse will then travel down the axon, like a flame burning down from the end of a fuse, with the Na+ ions at each point first going out and then coming back in. If 1010 Na+ ions cross the cell membrane in 0.5 ms, what amount of current is created? • note: this mechanism is adopted to be a taste sensor 8 Electrical Resistance • Most materials offer some resistance to the flow of electric charges through them. – This is called electrical resistance. 9 Resistor… • A resistor is a two-terminal electrical or electronic component that resists an electric current by producing a voltage drop between its terminals in accordance with Ohm's law. – The electrical resistance is equal to the voltage drop across the resistor divided by the current that is flowing through the resistor. • Resistors are used as part of electrical networks and electronic circuits. 10 Resistance… • Resistance (R) of a conductor depends on: – Material (resistivity ) – Length (L)- longer conductors have more resistance. – Cross section (A)- thick wires have less resistance than thin wires – Temperature - higher temperature means more resistance for most conductors L R A gold = 2.24x10-8Ωm aluminium = 2.65x10-8Ωm nichrome = 100x10-8Ωm The resistance of fig 2 will be greater than that of fig 1 Object fig 3 will have a smaller resistance than fig 1 because the charged particles have less of it to get through. How about figure 4? 11 Resistor • Resistor is simply a cylinder of ohmic material with wires attached to the end or by the symbol: • Symbol of resistor: Resistor Variable Resistor Resistor symbols (US and Japan) Resistor Variable resistor Resistor symbols (Europe) 12 Ohm’s Law • For many conductors, current depends on: R V I – Voltage - more voltage, more current • Current is proportional to voltage – Resistance - more resistance, less current • Current is inversely proportional to resistance V I R Calculate the value of R from this graph 13 Example: resistor versus organic semiconductor thin film Kuwat Triyana et al., Current-Voltage Characteristics of Organic Semiconducting Copper– Phthalocyanine Deposited on an Interdigitated Au Electrode, (Indonesian Journal of Chemistry, 2006). 14 Voltage… • Voltage is electric potential energy per unit charge, measured in joules per coulomb ( = volts). • It is often referred to as "electric potential", which then must be distinguished from electric potential energy by noting that the "potential" is a "per-unit-charge" quantity. • Like mechanical potential energy, the zero of potential can be chosen at any point, so the difference in voltage is the quantity which is physically meaningful. • The difference in voltage measured when moving from point A to point B is equal to the work which would have to be done, per unit charge, against the electric field to move the charge from A to B. 15 Voltage… Used to calculate current in Ohm's law. Used to express conservation of energy around a circuit in the voltage law. Used to calculate the potential from a distribution of charges. Is generated by moving a wire in a magnetic field. 16 Series and parallel circuits 17 DC Circuit • The basic tools for solving D C circuit problems are Ohm's Law, the power relationship, the voltage law, and the current law. • The following configurations are typical. 18 Two Loop Circuits • A circuit with two loops and two sources is involved enough to illustrate circuit analysis techniques. • It may be analyzed by direct application of the voltage law and the current law, but some other approaches are also useful. • Given the voltages, current analysis may be carried out by: – Voltage and current laws – Superposition theorem – Thevenin's theorem – Norton's theorem 19 Kirchhoff’s Voltage Law (KVL) • The voltage changes around any closed loop must sum to zero. Parallel: VB= VR1= VR2 Series: VB= VR1+ VR2 • No matter what path you take through an electric circuit, if you return to your starting point you must measure the same voltage, constraining the net change around the loop to be zero. 20 Kirchhoff’s Current Law (KCL) • The electric current which flows into any junction in an electric circuit is equal to the current which flows out. 21 Multiple Sources • Two Loop Circuits: A circuit with two loops and two sources is involved enough to illustrate circuit analysis techniques. • It may be analyzed by direct application of the voltage law and the current law, but some other approaches are also useful, such as: superposition theorem, Thevenin’s theorem or Norton’s theorem. 22 Superposition: Two Loop Problem • To apply the superposition theorem to calculate the current through resistor in the two loop circuit shown, the individual current supplied by each battery is calculated with the other battery replaced by a short circuit. 23 One Loop Circuit with Two Voltage Sources • Using the following figure; a. Find current flowing through the circuit and its direction b. Calculate voltage difference between a and b 5Ω + 3V b 12 V + - I a 7Ω a. Assume that the direction of current and loop is the same. From KVL: ΣV + ΣIR = 0 -3 + 12 + I(5+7) = 0 Therefore, I = -9/12 = -0.75 A It means that the current is opposite direction of loop b. the voltage difference between point a and b: Vab= -3 V + (-0.75 A)(5 Ω) = -6.75 V, or Vab= -12 V - (-0.75 A)(7 Ω) = -6.75 V 24 Two Loop Circuits with Two Sources • Using the following figure, find current flowing through each resistor 3Ω A 4Ω V2 2I1 A I1 I2 10Ω 4Ω V2 10Ω + + 30 V 3Ω 2I1 - 30 V B B • Using KCL at point A: I1-I2+(2I1) = 0 • Using KVL: -30 + 3I1+4I2 = 0 (left loop), +10(2I1) - V2 + 4I2 = 0 (right loop) • From three equations we can find: I1 = 2 A and I2 = 6 A, V2 = 64 V 25 Two Loop Circuits with Three Sources 1Ω I1 + 2V 1Ω A I2 4Ω Loop 1 Loop 2 + + - I3 - 4V 2Ω B - 2V 2Ω • Using KVL of loop 1: -2V+I1(1)-I2(4)+4V+I1(2)= 0 3I1 – 4I2 = -2 • Using KVL of loop 2: -2V+I3(1)-I2(4)+4V+I3(2)= 0 3I3 – 4I2 = -2 • Using KCL at point A: Therefore, what are: I1, I2 and I3? I1 + I2 +I3 = 0 how about direction of current? 26