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Current

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Current Current is defined as the flow of positive charge. Current is defined as the flow of positive charge. I = Q/t I = Q/t I: current in Amperes or Amps (A) I: current in Amperes or Amps (A) Q: charge in Coulombs (C) Q: charge in Coulombs (C) t: time in seconds (s) t: time in seconds (s)

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Electron Flow In a normal electrical circuit, it is the electrons that carry the charge. In a normal electrical circuit, it is the electrons that carry the charge. So if the electrons one way, which way does the current move? So if the electrons one way, which way does the current move?

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Problem How many electrons per hour flow past a point in a circuit if it bears 11.4 mA of direct current? How many electrons per hour flow past a point in a circuit if it bears 11.4 mA of direct current?

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EMF Cells convert chemical energy into electrical energy. Cells convert chemical energy into electrical energy. The potential difference (voltage) provided by a cell is called its electromotive force (or emf). The potential difference (voltage) provided by a cell is called its electromotive force (or emf). The emf of a cell is constant, until near the end of the cell ’ s useful lifetime. The emf of a cell is constant, until near the end of the cell ’ s useful lifetime. Misnomer: The emf is not really a force. Misnomer: The emf is not really a force.

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Battery A battery is composed of more than one cell in series. A battery is composed of more than one cell in series. The emf of a battery is the sum of the emf ’ s of the cells. The emf of a battery is the sum of the emf ’ s of the cells.

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Circuit Components VoltmeterAmmeter OhmmeterSwitch

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Problem If a typical AA cell has an emf of 1.5 V, how much emf do 4 AA cells provide? If a typical AA cell has an emf of 1.5 V, how much emf do 4 AA cells provide? Draw the battery composed of these 4 cells. Draw the battery composed of these 4 cells.

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Problem Draw a single loop circuit that contains a cell, a light bulb, and a switch. Draw a single loop circuit that contains a cell, a light bulb, and a switch. Put a voltmeter in the circuit so it reads the potential difference across the light bulb. Put a voltmeter in the circuit so it reads the potential difference across the light bulb.

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Series Series components are put together so that all the current must go through each one Series components are put together so that all the current must go through each one Bulbs in series all have the same current. Bulbs in series all have the same current.

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Parallel Parallel components are put together so that the current divides, and each component gets only a fraction of it. Parallel components are put together so that the current divides, and each component gets only a fraction of it. Bulbs in parallel do not have the same current. Their current is proportional to the resistance of the paths. Bulbs in parallel do not have the same current. Their current is proportional to the resistance of the paths.

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Problem Draw a circuit with a cell and two bulbs in series. Draw a circuit with a cell and two bulbs in series.

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Problem Draw a circuit having a cell and four bulbs. Exactly two of the bulbs must be in parallel. Draw a circuit having a cell and four bulbs. Exactly two of the bulbs must be in parallel.

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Mini Lab A Draw a circuit containing one cell, one bulb, and a switch. Create this with the available components. Measure the voltage across the cell and then across the bulb. What do you observe? Draw a circuit containing one cell, one bulb, and a switch. Create this with the available components. Measure the voltage across the cell and then across the bulb. What do you observe?

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Mini Lab B Draw a circuit containing a cells and two bulbs in series. Create this with the available components. Draw a circuit containing a cells and two bulbs in series. Create this with the available components. What do you observe happens to the other bulb’s brightness when you unscrew one of them? What do you observe happens to the other bulb’s brightness when you unscrew one of them? Measure the voltage across the battery and across each bulb. What do you observe? Measure the voltage across the battery and across each bulb. What do you observe?

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Mini Lab C Draw a circuit containing a cells and two bulbs in parallel. Create this with the available components. Draw a circuit containing a cells and two bulbs in parallel. Create this with the available components. What do you observe happens to the bulbs when you unscrew one bulb? What do you observe happens to the bulbs when you unscrew one bulb? Measure the voltage across the battery and across each bulb. What do you observe? Measure the voltage across the battery and across each bulb. What do you observe?

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Mini Lab D Draw a circuit containing a cells, three bulbs in series, and a switch which shorts the last bulb when it is closed. Create this with the available components. Draw a circuit containing a cells, three bulbs in series, and a switch which shorts the last bulb when it is closed. Create this with the available components. What do you observe happens to the bulbs when you close the switch? What do you observe happens to the bulbs when you close the switch?

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Ohm ’ s Law

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Resistors Resistors are devices put in circuits to reduce the current flow. Resistors are devices put in circuits to reduce the current flow. Resistors are built to provide a measured amount of “ resistance ” to electrical flow, and thus reduce the current. Resistors are built to provide a measured amount of “ resistance ” to electrical flow, and thus reduce the current.

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Circuit Components Resistor

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Resistance Resistance depends on resistivity and on geometry of the resistor. Resistance depends on resistivity and on geometry of the resistor. R = L/A R = L/A : resistivity (Ω m) : resistivity (Ω m) L: length of resistor (m) L: length of resistor (m) A: cross sectional area of resistor (m 2 ) A: cross sectional area of resistor (m 2 ) Unit of resistance: Ohms (Ω) Unit of resistance: Ohms (Ω)

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Problem What is the resistivity of a substance which has a resistance of 1000 Ω if the length of the material is 4.0 cm and its cross sectional area is 0.20 cm 2 ? What is the resistivity of a substance which has a resistance of 1000 Ω if the length of the material is 4.0 cm and its cross sectional area is 0.20 cm 2 ?

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iClicker Two concentric circular loops of radii b and 2b, made of the same type of wire, lie in the plane of the page, as shown above. The total resistance of the wire loop of radius b is R. What is the resistance of the wire loop of radius 2b ? (A) R/4 (B) R/2 (C) R (D) 2R (E) 4R

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Problem What is the resistance of a mile of copper wire if the diameter is 5.0 mm? (resistivity of copper is 1.72 x ) What is the resistance of a mile of copper wire if the diameter is 5.0 mm? (resistivity of copper is 1.72 x )

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iClicker The five resistors shown below have the lengths and cross ‑ sectional areas indicated and are made of material with the same resistivity. Which resistor has the least resistance?

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Ohm ’ s Law Resistance in a component in a circuit causes potential to drop according to the equation: Resistance in a component in a circuit causes potential to drop according to the equation: ΔV = IR ΔV = IR ΔV: potential drop (Volts) ΔV: potential drop (Volts) I: current (Amperes) I: current (Amperes) R: resistance (Ohms) R: resistance (Ohms)

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Ohmmeter Measures Resistance. Measures Resistance. Placed across a resistor when no current is flowing. Placed across a resistor when no current is flowing.

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Ammeter An ammeter measures current. An ammeter measures current. It is placed in the circuit in a series connection. It is placed in the circuit in a series connection. An ammeter has very low resistance, and does not contribute significantly to the total resistance of the circuit. An ammeter has very low resistance, and does not contribute significantly to the total resistance of the circuit.

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Power P = W/t P = W/t P = ΔE/Δt P = ΔE/Δt Units Units Watts Watts Joules/second Joules/second

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Power in Circuits P = IΔV P = IΔV P: power (W) P: power (W) I: current (A) I: current (A) ΔV: potential difference (V) ΔV: potential difference (V) P = I 2 R P = I 2 R P = (ΔV) 2 /R P = (ΔV) 2 /R

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iClicker The product 2 amperes x 2 volts x 2 seconds is equal to (A) 8 coulombs (B) 8 newtons (C) 8 joules (D) 8 calories (E) 8 newton ‑ amperes

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iClicker The circuit shown above left is made up of a variable resistor and a battery with negligible internal resistance. A graph of the power P dissipated in the resistor as a function of the current I supplied by the battery is given above right. What is the emf of the battery? (A) V (B) 0.67 V (C) 2.5 V (D) 6.25 V (E) 40 V

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Problem How much current flows through a 100-W light bulb connected to a 120 V DC power supply? What is the resistance of the bulb? How much current flows through a 100-W light bulb connected to a 120 V DC power supply? What is the resistance of the bulb?

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iClicker A certain coffeepot draws 4.0 A of current when it is operated on 120 V household lines. If electrical energy costs 10 cents per kilowatt ‑ hour, how much does it cost to operate the coffeepot for 2 hours? (A) 2.4 cents (B) 4.8 cents (C) 8.0 cents (D) 9.6 cents (E) 16 cents

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Resistors in Circuits Resistors can be placed in circuits in a variety of arrangements in order to control the current. Resistors can be placed in circuits in a variety of arrangements in order to control the current. The overall resistance of a specific grouping of resistors is referred to as the equivalent resistance. The overall resistance of a specific grouping of resistors is referred to as the equivalent resistance. The equivalent resistance in a circuit determine the current output from the battery. The equivalent resistance in a circuit determine the current output from the battery.

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Resistors in Circuits Arranging resistors in series increases the resistance and causes the current to be reduced. Arranging resistors in series increases the resistance and causes the current to be reduced. Arranging resistors in parallel reduces the resistance and causes the current to increase. Arranging resistors in parallel reduces the resistance and causes the current to increase.

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Resistors in Series

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Resistors in Parallel

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iClicker The electrical resistance of the part of the circuit shown between point X and point Y is (A) 4/3 (B) 2 (C) 2.75 (D) 4 (E) 6

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iClicker Which two arrangements of resistors shown above have the same resistance between the terminals? (A) I and II (B) I and IV (C) II and III (D) II and IV (E) III and IV

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iClicker A lamp, a voltmeter V, an ammeter A, and a battery with zero internal resistance are connected as shown above. Connecting another lamp in parallel with the first lamp as shown by the dashed lines would (A) increase the ammeter reading (B) decrease the ammeter reading (C) increase the voltmeter reading (D) decrease the voltmeter reading (E) produce no change in either meter reading

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Kirchoff ’ s Rules

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Problem Draw a circuit containing, in order (1) a 1.5 V cell, (2) a 100-Ω resistor, (3) a 330-Ω resistor in parallel with a 100-Ω resistor (4) a 560-Ω resistor, and (5) a switch. Draw a circuit containing, in order (1) a 1.5 V cell, (2) a 100-Ω resistor, (3) a 330-Ω resistor in parallel with a 100-Ω resistor (4) a 560-Ω resistor, and (5) a switch. Calculate the equivalent resistance. Calculate the equivalent resistance. Calculate the current through the cell. Calculate the current through the cell. Calculate the current through the 330-Ω resistor. Calculate the current through the 330-Ω resistor.

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iClicker In the diagrams above, resistors R 1 and R 2 are shown in two different connections to the same source of emf that has no internal resistance. How does the power dissipated by the resistors in these two cases compare? (A) It is greater for the series connection. (B) It is greater for the parallel connection. (C) It is the same for both connections (D) It is different for each connection, but one must know the values of R1 and R2 to know which is greater (E) It is different for each connection, but one must know the value of to know which is greater.

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Junction Rule Kirchoff ’ s 1st rule is also called the “ junction rule ”. Kirchoff ’ s 1st rule is also called the “ junction rule ”. The sum of the currents entering a junction equals the sum of the currents leaving the junction. The sum of the currents entering a junction equals the sum of the currents leaving the junction. This rule is based upon conservation of charge. This rule is based upon conservation of charge. Amount of current flowing in any direction is proportional to the rersistance along the path Amount of current flowing in any direction is proportional to the rersistance along the path

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iClicker When there is a steady current in the circuit, the amount of charge passing a point per unit of time is (A) the same everywhere in the circuit (B) greater at point X than at point Y (C) greater in the 1 resistor than in the 2 resistor (D) greater in the 1 resistor than in the 3 resistor (E) greater in the 2 resistor than in the 3 resistor

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Loop Rule Kirchoff ’ s 2nd rule is also referred to as the “ loop rule ”. Kirchoff ’ s 2nd rule is also referred to as the “ loop rule ”. The net change in electrical potential in going around one complete loop in a circuit is equal to zero. The net change in electrical potential in going around one complete loop in a circuit is equal to zero.

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iClicker Kirchhoff ’ s loop rule for circuit analysis is an expression of which of the following? (A) Conservation of charge (B) Conservation of energy (C) Ampere's law (D) Faraday's law (E) Ohm's law

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Ohm ’ s Law Graph Make a table of current and resistance data and graph the data such that voltage is the slope of a best-fit line. Make a table of current and resistance data and graph the data such that voltage is the slope of a best-fit line. Wire a circuit with a cell and one or more resistors. Wire a circuit with a cell and one or more resistors. Calculate and record the resistance. Calculate and record the resistance. Measure and record the corresponding current. Do this 5 times without duplicating your resistance values. Measure and record the corresponding current. Do this 5 times without duplicating your resistance values. Rearrange the equation ΔV = IR so that ΔV is the slope of a linear equation. Rearrange the equation ΔV = IR so that ΔV is the slope of a linear equation. Construct a graph from your data that corresponds to this rearranged equation. Construct a graph from your data that corresponds to this rearranged equation. Calculate and clearly report the slope of the line. How does this compare to the emf of 1.5 V for a D-cell? Calculate and clearly report the slope of the line. How does this compare to the emf of 1.5 V for a D-cell?

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Terminal Voltage When a current is drawn from a battery, the voltage across its terminals drops below its rated EMF. When a current is drawn from a battery, the voltage across its terminals drops below its rated EMF. The chemical reactions in the battery cannot supply charge fast enough to maintain the full EMF. The chemical reactions in the battery cannot supply charge fast enough to maintain the full EMF. Thus the battery is said to have an internal resistance, designated r. Thus the battery is said to have an internal resistance, designated r.

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Terminal Voltage and EMF A real battery is then modeled as if it were a perfect emf, ε. in series with a resistor r. A real battery is then modeled as if it were a perfect emf, ε. in series with a resistor r. Terminal voltage V ab Terminal voltage V ab When no current is drawn from the battery, the terminal voltage equals the emf. When no current is drawn from the battery, the terminal voltage equals the emf. When a current I flows from the battery, there is an internal drop in voltage equal to Ir, thus the terminal voltage (actual voltage delivered) is V ab = ε - Ir When a current I flows from the battery, there is an internal drop in voltage equal to Ir, thus the terminal voltage (actual voltage delivered) is V ab = ε - Ir

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iClicker In the circuit shown above, what is the value of the potential difference between points X and Y if the 6 ‑ volt battery has no internal resistance? (A) 1 V (B) 2 V (C) 3 V ( D) 4 V (E) 6V

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iClicker An immersion heater of resistance R converts electrical energy into thermal energy that is transferred to the liquid in which the heater is immersed. If the current in the heater is I, the thermal energy transferred to the liquid in time t is (A) Irt (B) I2Rt (C) IR2t (D) IRt2 (E) IR/t

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