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• 2.Cool Fact: Electric companies don’t sell any electricity. After all, the electricity is wiggling equally back and forth (AC Alternating Current.) Even in a DC system, the electricity still must flow in a complete circle. So, the utility company takes back every bit of electricity that they send out! So why are they billing us? The smart electric companies say they’re billing us for energy, for electromagnetism in other words. Some of them have even changed their name to “Energy Company.”
• 3.Electricity and Circuits AC and DC charge and current voltage and potential difference emf resistor and resistance capacitor and capacitance inductor and inductance Conservation laws
• 4.Analogy: Pizza delivery: Pizzas are delivered to a party by scooters from the pizza shop. What is analogous with the scooters? the shop? the party? the number of scooters? the pizzas? the type of pizzas?
• 5.Analogy: Water pipe Pizza delivery Water is pumped to an apartment on the 10th floor. What is analogous with the water? The pump which provides the pressure? The height of the building? Pizzas are delivered to a party by scooters from the pizza shop. What is analogous with the scooters? The pizzas? The shop? The party? The number of scooters? apartment reservoir
• 6.Charge Q (C) positive and negative charge – protons and electrons in an atom 1e- = 1.609x10-19C = 1 proton 1C = 6.24x1018 electrons Current I (A) I = nq/t = Q/t Convention : flow of + charges rate of charge flowing past a given, measured in coulombs/second
• 7.Voltage or Potential difference DV (V) electric potential at a point - amount of electric potential energy (PE) a point charge at that point. the difference of the potential between any two points. It is the work per unit charge that has to be done to move the charge from one point to the other. Voltage = DV = D(PE) /q = W/q current flows from higher potential (positive terminal) to lower potential (negative terminal, taken to be at zero potential). Analogy : falling object. the free carriers drift toward the negative terminal due to an electric field. It is equal to the work that needs be done by the electric field to move the charge unit through the wire outside the source.
• 8.The zero position of the p.d. is arbitrary. The negative terminal of the battery is taken to be the 0 point. Voltage = DV = (VA – VB) When a metal wire is connected across the two terminals of a battery, the source places an electric field E across the conductor. Free electrons of the conductor are forced to drift toward the positive terminal under the influence of this field.
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• 10.+ve charges of conductor/wire gets “attracted” to the –ve end ie charge carrier moves from higher to lower potential  lose potential energy potential energy increases as charge moves across because work must be done to move +ve charge against the field  +ve side ends up with higher potential than -ve side
• 11.Emf ε (V) is the pd (voltage) at the ends of a source when there is no current present. act likes a charge pump and it does not depend on the resistance of the circuit. energy per unit charge which comes from a battery and is NOT a ”force”. It is equal to the energy supplied to the charge unit when it moves from one pole of the source to another, but through the source the charges gain energy going against the electric field in the battery
• 12.Emf ε (V)
• 13.EMF is the generated voltages from batteries and is essential for an electronic circuit to drive currents through the circuit. It act likes a charge pump and it does not depend on the resistance of the circuit. The value is equal to the energy supplied to the charge unit when it moves from one pole of the source to another, but through the source. This electrical energy is converted from other forms of energy like chemical. Voltage in the outer part of the circuitry is equal to the work that needs be done by the electric field to move the charge unit from one pole of the source to the other, but through the wire. The potential of the positive endpoint of the source is higher as compared to the negative points. The sum of the voltage drops in a circuit is equal to the terminal voltage according to Kirchoff’s law.
• 14.Terminal Voltage DV (V) electric potential difference between two terminals when current passes through it. equal to the EMF internal resistance is zero. Electric potential energy U (J) energy loss by a charge q moving from higher to lower potential of a resistor with p.d DV across it. energy gain from a battery as the charge moves against the electric field.
• 15.Resistance R (W) collision between the carrier charges and the lattice atoms of the conductor. energy is transferred to the atoms  an increase in the vibrational energy and temperature. energy loss by the charge during collision : W = qDV Ohm’s Law : DV = IR
• 16.Analogy (revisited): Water pipe Pizza delivery Water is pumped to an apartment on the 10th floor. What is supplying the energy? What is the flow of water analogous to? How to reduce resistance (increase water flow)? Pizzas are delivered to a party by scooters from the pizza shop. What is supplying the energy? How to increase energy of the charge/pizzas delivered? How to reduce resistance? apartment reservoir
• 17.Examples : How many electrons does a 1.5V battery drive through a 10W resistor in 1 second? Over the course of an 8 hour day, 3.8x104C of charge pass through a typical computer (presuming it is in use the entire time). Determine the current for such a computer Determine the amount of time that a LED night light (I=0.0042A) would have to be used before 1.0x106C of charge passes through them. Determine the amount of electrical energy (in J) used by a 42-inch LCD television (210 W) when operated for 3 hours.
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• 19.2 resistors in series The same delivery person delivers to the two parties, but for different pizza orders (types)
• 20.2 resistors in parallel Different delivery person delivers to two different parties, with different pizza orders.
• 21.Analogy (revisited): 2 resistors in series: Water goes from one wheel to the next, causing the water not to flow smoothly due to turbulence. So speed of the 2nd wheel decreases. Part of the energy is lost in the 1st wheel, and the rest is lost in the 2nd wheel.
• 22.Analogy (revisited): 2 resistors in parallel: Water flows smoothly for both wheels. But water must be poured quicker for for the two-wheel system (bottom) to have the same flow rate as the single-wheel (top) system. So boy/pump must work harder to turn both wheels together. Same amount of energy is given to both wheels.
• 23.Resistors in combination Resistors in series – conservation of energy Resistors in parallel – conservation of charge
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• 25.How to solve simple circuit problems: Draw diagram, use the rules for parallel and series to find the total external resistance Use e = IR + Ir to find the current I or the e from the battery Use Ohm’s law to find particular p.d and currents Any components is series carries the same current Any components in parallel has same p.d
• 26.Simple Circuits Conservation of energy :
• 27.Example:
• 28.Examples: R1 = R3 = R5 = 1W R2 = R4 = R6 = 2W
• 29.Example: Find : the total current drawn current through the R4 resistor voltage across the R2 resistor 3kW
• 30.Example: Find : Ibattery, VBC, I3Ω, VDI, VIL
• 31.Examples: Determine the following quantities for the two circuits : the equivalent resistance, the total current from the power supply, the current through each resistor, the voltage drop across each resistor, and the power dissipated in each resistor.
• 32.Example: The diagram below shows a circuit with one battery and 10 resistors. Determine the current through, the voltage drop across, and the power dissipated by each resistor.
• 33.Power P (W) power is lost through a resistor as work has to be done by charge against the resistance. obeys conservation of energy principle – power source will supply just enough energy to charges to overcome load resistors
• 34.
• 35.Capacitors & Capacitance two metal plates separated by a non-conducting substance, or dielectric. a passive two-terminal electrical component used to store energy electrostatically in an electric field, as opposed to batteries which store energy via chemical reactions. a capacitor can dump its entire charge in a tiny fraction of a second, while a battery would take minutes (or days) to completely discharge. when charged, a capacitor has zero nett charge and NO current will flow across it.
• 36.When you connect a capacitor to a battery : The plate on the capacitor that attaches to the negative terminal of the battery accepts electrons that the battery is producing. The plate on the capacitor that attaches to the positive terminal of the battery loses electrons to the battery. Once it's charged, the capacitor has the same voltage as the battery. Once charged, a capacitor will carry zero nett charge and NO current will flow across it. - - + +
• 37.Capacitors & Capacitance capacitance (F) - capacitor's storage potential defined as the ratio of charge ±Q on each conductor to the p.d. DV between them: - - + +
• 38.Capacitors in combination Capacitors in series – conservation of energy Capacitors in parallel – conservation of charge
• 39.Work/energy for a capacitor positive charges accumulate on one plate, and leave the other - making this plate -Q and the other +Q. an electric field is set up between the plates which opposes any further charge transfer. work must be done against this field in order to continue charging. This work becomes energy stored in the capacitor’s electric field.
• 40.Charging/discharging The moment the switch S is closed, the circuit will have current I flowing in it, with no potential difference across the capacitor C and the 𝜀 of the battery is equal to the potential drop across the resistor R. When the capacitor is fully charged, the capacitor has a potential drop of DVC = 𝜀 = Q/C across it, while the resistor will have no potential difference across it since there will be no more current will flow in the circuit.
• 41.
• 42.Charging/discharging When the switch S is closed, the circuit will have current I flowing in it, with potential difference across the capacitor C and resistor R equal to the 𝜀 of the battery. The current will be divided between R and C. When the capacitor is fully charged, the capacitor will have a potential drop of DVC = 𝜀 = Q/C across it, and the potential difference across R is also equal to 𝜀. All current will flow through the resistor.
• 43.Example: Given C1 = 2µF, C2 = 6µF, and C3 = 3µF, what is the total capacitance of the circuit in the figure above? What is the energy stored in the circuit if the battery provides 10V? How much charge is stored in C3?
• 44.Example: In the circuit given below, C1 = 60µF, C2 = 20µF, C3 = 9µF and C4 = 12µF. If the potential difference between points a and c, Vac= 120V find the charge of C2.
• 45. The switch is flipped to b. Vb = 12V, C = 10mF, R = 20W a.) What is the current through the resistor just AFTER the switch is thrown? b.) What is the charge on the capacitor just AFTER the switch is thrown? c.) What is the charge on the capacitor at at time t = 0.3 msec after the switch is thrown?
• 46.Example: For the circuit shown, points A and B are at the same potential. Determine … the potential drops and currents across each resistor the potential drops across each capacitor the charge stored in each capacitor the capacitance of the unknown capacitor
• 47.
• 48.Complex Circuits For more complex circuits, Kirchhoff’s Rules applies. The rules can be used for both DC and AC circuits. The criteria for the rules is that the circuit must be in steady-state. There are two rules : Kirchhoff's first rule, Kirchhoff's point rule, or Kirchhoff's junction rule (or nodal rule). Kirchhoff's second rule, Kirchhoff's loop (or mesh) rule.
• 49.K1: Due to conservation of charges principle. what goes in must come out. At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node,
• 50.K1: Due to conservation of charges principle. At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node, or: The algebraic sum of currents in a network of conductors meeting at a point is zero.
• 51.K2 : Due to conservation of energy principle. You can only spend what you have been given the sum of the emfs in any closed loop is equal to the sum of the potential drops in that loop.
• 52.K2 : Due to conservation of energy principle. The directed sum of the electrical potential differences (voltage) around any closed network is zero, ie. the sum of the emfs in any closed loop is equivalent to the sum of the potential drops in that loop. P.d. (voltage) is defined as the energy per unit charge. The total amount of energy gained per unit charge must equal the amount of energy lost per unit charge, ie. the total work done by the charges as the current flows through the resistors is equal to the energy gain by the charges from the electric fields as the charges flow through the power sources.
• 53.
• 54.How to solve complex circuit problems: Draw diagram, use the rules for parallel and series to find the external resistance, if any. Label the directions of emf’s (in the direction of the energy gain by the charge). Label the directions for the currents across all resistors. If direction is unknown, assume any one. For each loop, choose a positive loop direction. Solve by applying : K1 to the nodes, apply K2 to the loops.
• 55.Example: Find : I1, I2, I3
• 56.Example: Find : I1, I2, I3, Vab, Vhe, Vad
• 57.Example: Determine : the current through each of the other resistors, the voltage of the battery on the left, and the power delivered to the circuit by the 20V battery
• 58.Example: Find the unknown currents I1 and I2. Find the unknown battery voltage V2. Compare VAB, VCD and VEF.
• 59.Example: Find : 𝜀 , IR1, VR5 I = 3A
• 60.Summary Simple circuit : DV = SIR = 𝜀 – Ir , DV = Q/C Potential energy of charge : U = qDV Components in : series : I1 = I2 = … : parallel : DV1 = DV2 = … Power lost : 𝑃=𝐼2𝑅= ∆𝑉2 𝑅 =𝐼∆𝑉 Energy stored : 𝑈= 𝑄2 2𝐶 = 𝑐∆𝑉2 2 = 𝑄∆𝑉 2 Complex circuit : Kirchhoff Rule 1 : SIin = SIout : Kirchhoff Rule 2 : S𝜀 = SIR
• 61.of Circuits
• 62.Concept Test Charge flows through a light bulb. Suppose a wire is connected across the bulb as shown. When the wire is connected, 1. all the charge continues to flow through the bulb. 2. half the charge flows through the wire, the other half continues through the bulb. 3. all the charge flows through the wire. When a wire is connected across B as shown, bulb A 1. burns more brightly. 2. burns as brightly. 3. burns more dimly. 4. goes out. short circuit Bulb B will go out and total resistance in the circuit will decrease.
• 63.Concept Test The three light bulbs in the circuit all have the same resistance. Given that brightness is proportional to power dissipated, the brightness of bulbs B and C together, compared with the brightness of bulb A, is 1. twice as much. 2. the same. 3. half as much. If the four light bulbs in the figure are identical, which circuit puts out more light? 1. I. 2. Both emit same amount of light. 3. II. Potential difference across B and C is equal to that across A. Given that P=V2/R, the power loss by B and C will be half that by A. Total resistance in I is smaller and so the total current will be larger in the circuit and hence, more power will be loss.
• 64.Concept Test If the four light bulbs in the figure are identical, which circuit puts out more light? 1. I. 2. The two emit the same amount of light. 3. II.
• 65.Concept Test The light bulbs in the circuit are identical. When the switch is closed, 1. both go out. 2. the intensity of light bulb A increases. 3. the intensity of light bulb A decreases. 4. the intensity of light bulb B increases. 5. the intensity of light bulb B decreases. 6. some combination of 1–5 occurs. 7. nothing changes. The potential difference across the switch is zero. The potential diff across the bulbs are still equal.
• 66.Concept Test An ammeter A is connected between points a and b in the circuit below, in which the four resistors are identical. The current through the ammeter is 1. I/2. 2. I/4. 3. zero. 4. need more information The resistors are identical and so the potential diff across a and b is zero.