Electric current - Gaincoli, Ch 18, sections 1 - 6 & 8
1. Describe a simple model of electrical conduction in a metal.
· Students should be aware of the term drift velocity and of the interactions of conduction electrons with the lattice ions.
· When an electric field is applied to a metal, the forces that it exerts on the free electrons lead to a small net motion or drift in the direction of the force (think of it like the floor is tilted slightly). This is in addition to the normal random motion of the electrons.
· The electric field does work on the moving charges, and the resulting kinetic energy is transferred to the material by means of inelastic collisions with the ion cores, which vibrate about their equilibrium positions in the lattice. This energy transfer increases the average vibrational energy and therefore the temperature of the material.
· In general, the motion of free electrons in a metal consists of random motion with very large average speeds (106 ms-1) and a much slower drift speed in the direction of the electric field force.
2. Define electric current.
· Current is defined to be the amount of charge transferred per unit time.
· In metals, the moving charges are always electrons drifting by as a result of an electric field force.
· So, if a charge Dq flows through an area in a time Dt, then the current through the area, I, is given by I = Dq / Dt
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Current can also be expressed in terms of drift velocity, vD, as follows:
Suppose we have a homogeneous conductor with an electron density of n (= total number N / length L of conductor). An electric field is applied to the conductor and the electrons acquire a drift velocity, vD. In a time Dt the electrons move a distance vDDt. So, in time Dt all electrons that were within a distance of vDDt of some point in the conductor would flow past that point. To find this number, we multiply the electron density times the length vdDt. So, the number of electrons that flow past is nvDDt. This represents a total charge Dq of (nvDDt)q. To find the current we divide: I = Dq / Dt = (nvDDt)q / Dt = nqvD.
· Current is measured in the MKS system with the unit Ampere. 1 A = 1 Cs-1. The Ampere is a fundamental unit of the metric system. It is defined in relation to the amount of force between two long current-carrying wires that are 1 meter apart. 1 Amp of current produces a force of 1 Newton.
3. Define and apply the concept of resistance.
· Students should be aware that R = V/I is a general definition of resistance. It is not a statement of Ohm's law. Students should be familiar with the term resistor (any circuit device made to have a specific value of resistance).
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Resistance is a measure of opposition to flow in a conductor in the sense that high resistance results in low current and low resistance results in high current.
· More properly, resistance is determined by physical properties of the conductor such as the extent to which it has "free electrons" as well as its size and length.
· Free electrons will, in general, encounter more obstacles if they are forced to move through a long conductor as opposed to a short conductor. So, Resistance is directly related to L, the length of the conductor.
· It also follows that the thicker the wire is, the less resistance one will encounter since there is more open space through which electrons may move. So, Resistance is inversely related to A, the cross section area of the conductor.
· Resistance is measured in Ohms (W)
4. State Ohm's law.
· Ohm's Law predicts that the amount of current in a conductor is directly related to the amount of voltage (potential difference) across the conductor.
· In equation form, I = V / R
5. Compare ohmic and non-ohmic behaviour.
· Ohm's Law is an idealized relationship between current and electric potential. Most items obey this relationship for certain voltages, but in general do not show a linear relationship between current and voltage. These devices are called non-Ohmic (non-linear).
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Any device for which a linear relationship between I and V exists is called Ohmic.
· For example, a resistor in a circuit will show a linear relationship between I and V. A filament light bulb will not. This is because of the nature of the conductor. As the free electrons in the filament begin to drift and transfer KE to the lattice structure of the filament, the lattice ions begin to increase their vibrational motion and it heats up. As the vibrational motion of the lattice increases, the available space through which the free electrons can move decreases, effectively reducing the area of the conductor. As we have seen, as the area decreases the resistance increases. For most conductors, resistance is dependent on temperature and increases with temperature. This creates a non-linear relationship between I and V.
6. Derive and apply expressions for electrical power dissipation in resistors.
· Suppose a total amount of charge Dq is moved across some part of a conductor in a time Dt by a potential difference of V. The total amount of work done to move the charge is DW = (Dq)V. The power developed is defined as the work done per time, so power is found by P = DW / Dt = (Dq) V / Dt = IV (since Dq / Dt = I).
· So, power = IV. Using Ohm's Law to relate I to V and R, we can also write:
P = IV = (V/R)(V) = V2 / R
P = IV = I (IR) = I2 R.
· The three forms of the relation for power are identical. Choose the one that is most convenient for a particular situation. The form P = I2 R is often referred to as the power dissipated in a resistor and it represents the rate at which heat (often called Joule heating) is transferred to the conductor.
Electric circuits - Giancoli, Ch 19, sections 1 - 5
7. Define electromotive force.
· In an electric circuit there must be a device somewhere in the loop where charge travels "uphill" from lower to higher potential despite the fact that the electric field force is trying to push it from higher to lower potential.
· The influence that makes charge move from lower to higher potential is called electromotive force (emf or e).
· EMF is not a force but rather an energy per unit charge quantity like potential. The SI unit of EMF is the Volt. A battery with an EMF of 1.5 V does 1.5 J of work to raise 1 Coulomb of charge.
· An ideal source of EMF maintains a constant potential difference between its terminals, independent of the current through it. Quantitatively the emf of the source is equal to this potential difference.
8. Describe the concept of internal resistance.
· Real sources do not behave as described above, because charge moving through them encounters some resistance. This is called the internal resistance, r, of the source.
· If the source behaves according to Ohm's Law then the internal resistance is constant and the current that passes through the source has a drop in potential = Ir.
· In an open circuit situation, the potential difference Vab of the source = e because there is no current flowing. In the top diagram the voltmeter reads Vab = e
· In a closed circuit situation, the potential difference Vab of the source = e - Ir. In the bottom diagram, the voltmeter reads Vab = e - Ir. Since this is a short circuit we know that Vab = 0, which allows us to determine the short-circuit current I = e / r.
9. Derive and apply the equations for equivalent resistances of resistors in series and in parallel.
· For the case of resistors in series (one path for current to follow)
Þ We will define the equivalent resistance of the circuit, Req, so that V = IReq
Þ The current through each resistor has the same value. This is because of conservation of charge. For any segment in the circuit, the amount of charge coming in must equal the amount going out. So, I1 = I2 = I3 = I.
Þ According to Ohm's law V1 = I1R1, V2 = I2 R2 , and V3 = I3R3.
Þ The sum V1 + V2 + V3 = V. This is because of conservation of energy. V (the single voltage gain) must be equal to the sum of the voltage drops so that we don't gain or lose any energy.
Þ Putting it all together,
V1 + V2 + V3 = V
I1R1 + I2R2 + I3R3 = IReq
I(R1 + R2 + R3) = IReq
Req = R1 + R2 + R3
· For the case of resistors in parallel (more than one path for current to follow):
Þ We will define the equivalent resistance of the circuit so that V = IReq
Þ Looking at the diagram: V, V1, V2 and V3 are all connected directly to points a and b. This is because ideal wires have no resistance. We can then say V = V1 = V2 = V3 because of conservation of energy.
Þ By looking carefully at how the current splits and combines again, we can say that I = I1 + I2 + I3 because of conservation of charge. Coming up the right hand side of the circuit, I2 and I3 combine to the right of R2 and this joins up with I1 to the right of R1. This represents the total amount of current coming from the source and since we can't create or lose any charge, . . .
Þ Applying Ohm's Law, we can write:
I = I1 + I2 + I3
and since V = V1 = V2 = V3
10. Draw circuit diagrams.
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Students should be able to recognize and use the accepted circuit symbols included in the Physics Data Booklet.
11. Describe the use of ammeters and voltmeters.
· Students should be able to describe and draw the correct positioning of ideal ammeters and voltmeters in circuits. Students will not be required to know about shunts and multipliers.
· Because of the nature of the voltmeter -- ideally it has an infinite resistance, Voltmeters are placed in a circuit parallel to the device whose voltage is being measured. This is so that adding the meter does not change the flow of current in the circuit.
· Because of the nature of the ammeter -- ideally it has zero resistance, Ammeters are placed in a circuit in series with the device whose current draw is being measured. This is so that adding the meter does not change the flow of current in the circuit.
12. Solve problems involving series and parallel circuits.
· Students should appreciate that many circuit problems can be solved by regarding the circuit as a potential divider. Students should be aware that ammeters and voltmeters have their own resistance.
Problems: Ch 18 #3, 5, 9, 23, 27, 33, 36, 63 Ch 19 #3, 5, 7, 13, 16, 19, 69
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