A battery transforms chemical energy into electrical energy. Chemical reactions within the cell create a potential difference between the terminals

Transcription

1 D.C Electricity

2 Volta discovered that electricity could be created if dissimilar metals were connected by a conductive solution called an electrolyte. This is a simple electric cell. The Electric Battery

3 A battery transforms chemical energy into electrical energy. Chemical reactions within the cell create a potential difference between the terminals by slowly dissolving them. This potential difference can be maintained even if a current is kept flowing, until one or the other terminal is completely dissolved.

4 Several cells connected together make a battery, although now we refer to a single cell as a battery as well.

5 Electric Current Electric current is the rate of flow of charge through a conductor: Unit of electric current: the ampere(amp), A. 1 A = 1 C/s. Charge is not used up (destroyed) in a resistor. The charge that enters must equal the charge that leave. Conservation of Charge.

6 Example: 4.00x10-3 C of charge flowed through a light bulb in a time of 0.25 s. a) What was the average current through the bulb in this time? b) How many electrons passed through the bulb per second on average? Solution: a) = 4.00x10-3 /0.25= 1.6x10-2 A b) # electrons = 1.6x10-2 /1.6x10-19 = 1x10 17 electrons

7 A complete circuit is one where current can flow all the way around. Note that the schematic drawing doesn t look much like the physical circuit!

8 In order for current to flow, there must be a path from one battery terminal, through the circuit, and back to the other battery terminal. Only one of these circuits will work:

9 By convention, current is defined as flowing from + to -. Electrons (the mobile charge carriers) actually flow in the opposite direction, but not all currents consist of electrons (e.g. proton beam in particle accelerator). Voltage source

10 An ammeter is a device used to measure current in circuit. The entire current must pass through the device so it is connected in series. V I R

11 Example: Two ammeters (A 1 and A 2 ) are positioned in a circuit as shown. How would the measured values of current by each ammeter compare? A) A 1 is smaller B) A 2 is smaller C) They are equal A 1 A 2 I

12

13 Voltage Voltage is an electric potential difference. If there is a voltage there an electric field which causes a force on the charges. This potential difference forces electrons through the circuit. The voltage between positions A and B is the energy(or work) per unit charge to move the charge from A to B. Units of voltage: J/C or V(volt) Voltage is always measured between two locations. A voltmeter is therefore connected across something to measure it s voltage

14 For example to measure the voltage across a resistor (in parallel): V I R The voltage (energy per unit charge) V provided by the battery is equal to the energy per unit charge lost in the form of heat (and radiation) in the resistor (R).

15 Example: If J of energy is dissipated in a resistor when 1.25x10 20 electrons pass through it, what must be the voltage across the resistor? Solution: The total charge that passed through the resistor is: Q= (1.6x10-19 C)(1.25x10 20 )= 20 C The energy per unit charge (or voltage) is: V=E/Q= (100 J)/(20 C) = 5.00V

16 Ohm s Law: Experimentally, it is found that the current in a wire is proportional to the potential difference between its ends:

17 The ratio of voltage to current is the resistance: 1Ω=1J s/c 2 Ohm s Law The voltage is across a resistor is equal to the product of the resistance and the current through it. If one increases the voltage across a resistor the current should increase in a linear manner.

18 Example: If there is a voltage of 20.0 V across a 5.00 Ω resistor, what current flows through the resistor? Solution: I = V R = 20.0/5.00= 4.00 A

19 In many conductors, the resistance is independent of the voltage; this relationship is called Ohm s law. Materials that do not follow Ohm s law are called nonohmic. Unit of resistance: the ohm, Ω. 1 Ω = 1 V/A.

20 For any given material, the resistivity increases with temperature. When doing problems one generally assume the resistances remain constant. Semiconductors are complex materials, and may have resistivities that decrease with temperature.

21 Electric Power Power, as in kinematics, is the energy transformed by a device per unit time:

22 The unit of power is the watt, W. For ohmic devices, we can make the substitutions:

23 Example: A 20.0Ω resistor has a current of 0.50 A going through it. How much energy is dissipated in the resistor in a time of 5.00 minutes? Solution: P = I 2 R = (0.50) 2 (20)= 5.0 W E = Pt = (5.0)[(5)(60)]= 1500 J

24 What you pay for on your electric bill is not power, but energy the power consumption multiplied by the time. We have been measuring energy in joules, but the electric company measures it in kilowatthours, kwh.

25 Power in Household Circuits The wires used in homes to carry electricity have very low resistance. However, if the current is high enough, the power will increase and the wires can become hot enough to start a fire. To avoid this, we use fuses or circuit breakers, which disconnect when the current goes above a predetermined value.

26 Fuses are one-use items if they blow, the fuse is destroyed and must be replaced.

27 Circuit breakers, which are now much more common in homes than they once were, are switches that will open if the current is too high; they can then be reset.

28 DC Circuits

29 Kirchhoff's circuit laws The Current Law (Junction Rule) The current law arises from the basic principle that charge is conserved. The law states that the total current which enters a junction, must equal the total current which leaves the junction. In the circuit element above there are currents (conventional) entering and leaving the junction. Applying the law, one would obtain the relationship: I 1 +I 2 =I 3 If this condition were not true there would either be more less charge entering the junction compared to the charge leaving the junction. This would suggest that it is not in steady state.

30 The Voltage Law This law arises from the basic principle of conservation of energy. The law states that the sum of all the voltage (potential difference) increases and decreases around a complete path must add to zero. Since you have returned to the same position the change in electric potential energy (or change in electric potential) must equal zero.

31 Resistors in Series and in Parallel A series connection has a single path from the battery, through each circuit element in turn, then back to the battery.

32 According to Kirchhoff s rules the current through each resistor is the same, and the sum of the voltage drops across the resistors equals the battery voltage.

33 From this we get the equivalent (or total) resistance (that single resistance that gives the same current in the circuit). The equivalent resistance in this case is a resistor that could replace the group of resistors and draw the same current from the battery with the same voltage across it.

34 Example: Series Circuit a) What is the total resistance of the circuit below? b) Label the direction of conventional current. c) What current leaves the voltage source? d) Determine the current through each resistor, and the voltage across each resistor. 28 V 2.00 Ω 1.00Ω 4.00 Ω

35 Solution: a) Req= 2+4+1= 7.0 Ω 28 V I 2.00 Ω b) See diagram Using the equivalent circuit, one can calculate the current leaving the battery. c) I = V R = 28/7= 4.00 A 1.00Ω 4.00 Ω 28 V I If one assumes the wires have no resistance, the voltage of the source must equal the voltage across the total resistance (Req). d) Req The current I ( 4A) is the same current through each resistance since the resistors are in series. Using ohm s law on can calculate the voltage across resistor. For the 2.00 Ω, 4.00 Ω, and 1.00 Ω resistors the voltages are 8.0V, 16 V, and 4.0 V respectively.

36 In the movie below a light bulb is connected to a battery (switch open). The voltage across the light bulb is initially 9.00 V. Assume the wires have near zero resistance, and the battery always produces a constant voltage. What will happen to the current and voltage when the switch is closed and resistance of the wires in increased?

37 Explanation of previous slide: With the switch open no current flows through the circuit and the voltage across the bulb must be zero. The current would start to flow when the switch is closed. The current drops and the voltage across the bulb must decrease if the resistance in the wires increases. The total voltage across the entire circuit would still remain at 9.00 V since the the battery is assumed to be a constant voltage source. Batteries actually have an internal resistance, so in reality the voltage across the circuit (terminal voltage) is not a constant as it depends on the current. We will examine this later.

38 A parallel connection splits the current; the voltage across each resistor is the same:

39 The total current is the sum of the currents across each resistor: This gives the reciprocal of the equivalent resistance:

40 Example: Parallel Circuit a) What is total resistance of the circuit? b) What is the voltage across each resistor and the current through each resistor? c) What current leaves the battery? 6.00 V 2.00Ω 6.00Ω

41 Solution: a) 1 R Eq = 1 R R 2 =1/2 +1/6= 4/6 Req = 1.50 Ω 6.00 V I b) According to Kirchhoff s voltage law, the voltage across each resistor must be equal ( V= 6.00 V). This is assuming zero resistance in the wires connecting the resistors. Using Ohm s Law one can calculate the current through each resistor ( I = V ). The currents through the 2.00 Ω and 6.00 Ω R resistors are 3.00 A, and 1.00 A respectively. c) One could determine the current leaving the battery by using ohm s law [ I = V ] for the equivalent circuit R above, or by using the junction (current) rule [ I = I 1 + I 2 ] for the original circuit. I= 3 +1 = 4.00 A I I 1 I Ω 2.00Ω 6.00Ω 6.00 V

42 Example: Combination Circuit a) In the circuit shown calculate the equivalent (total) resistance of the entire circuit. Reduce the circuit in steps. b) Calculate the current that leaves the battery. c) Find the voltage across each resistor and the current through each resistor d) What was the power dissipated in the 5Ω resistor? e) How much energy does the entire circuit use in 10 V= 24 V minutes? 4Ω 3Ω 1Ω 5Ω

43 Solution: a) 6Ω b) Using Ohm s Law: I= 24/6= 4.00 A c) 4Ω Voltage across the 4 Ω resistor: V 4 =IR= (4)(4) = 16 V One can now use Kirchhoff s voltage law to find the voltage across the 3 Ω resistor: V 3 =0, V 3 = 8V Now use Ohm s law to find the current through the 3 Ω resistor : I 1 =V/R= 2.67 A One can now use the current law to find the current through the 1 Ω and 5 Ω resistors. 4=2.67+I 2, I 2 = 1.33 A And finally one can use Ohm s law to find the voltages across the 1 Ω and 5 Ω resistors. V=IR, V 1 = 1.33 V, V 5 = 6.67 V I I 1 1Ω V= 24 V 3Ω I 2 5Ω d) P=VI= (6.66 )( 1.33) = 8.89 W e) P=VI, E=Pt= VIt = (24)(4)[(10)(60)] = 57.6 kj

44 Electric Hazards Even very small currents 10 to 100 ma can be dangerous, disrupting the nervous system. Larger currents may also cause burns. Household voltage can be lethal if you are wet and in good contact with the ground. Be careful!

45 A person receiving a shock has become part of a complete circuit.

46 Faulty wiring and improper grounding can be hazardous. Make sure electrical work is done by a professional.

47 The safest plugs are those with three prongs; they have a separate ground line. Here is an example of household wiring colors can vary, though! Be sure you know which is the hot wire before you do anything.