# Electric Circuits. Concepts and Principles. Electric Circuits as Applied Physics. The Resistor

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1 . Electric ircuits oncepts and Principles Electric ircuits as Applied Physics Electric circuits are one of the most practical applications of our understanding of electric and magnetic fields. In general, an electric circuit is any device that consists of a closed path for charges to move (a current), a source of energy to drive the motion of the charges (a potential difference, or voltage, often in the form of a battery), and various circuit elements that can either convert (resistors) or store (capacitors and inductors) the energy supplied by the energy source. The study of circuits is incredibly broad, since there are limitless ways to combine these elements into an electric circuit. We will restrict ourselves to studying circuits with only a limited number of elements, and with a source that supplies a constant voltage 1. The esistor In general, a resistor is any device that converts electrical energy into another form of energy, often heat. For example, a fluorescent light bulb converts electrical energy into light (with about a 20% efficiency, the remaining energy is converted into heat) and an incandescent light bulb converts electrical energy very efficiently into heat (with only about 5% of the incident energy converted to light). Since a conversion of electrical energy takes place in these devices, they are resistors. In all resistors, the electric potential energy of the charges entering the device is larger than the electric potential energy of the charges exiting the device, because some of the potential energy has been converted to other forms. This decrease in potential energy is due to a decrease in electric potential between the two ends of the device and is directly proportional to the resistance of the device. 1 ircuits with constant voltage sources are referred to as D, or direct current, circuits. 1

2 The definition of resistance for a device is: where = i is the potential difference between the two ends of the device, often termed the voltage drop across the device, and i is the current that flows through the device. The unit of resistance, A, is defined as the ohm (Ω). The previous expression relates the resistance of a resistor to properties of the circuit it is part of. However, it is also sometimes useful to directly relate the resistance to the actual physical parameters of the device itself. For simple, passive resistors (basically blocks of material connected to a voltage source), resistance is defined as: where = ρl A ρ is the resistivity of the material from which the resistor is constructed, L is the length of the resistor in the direction of current flow, and A is the cross-sectional area of the resistor. esistivity can range from 0 for a perfect conductor to for a perfect insulator. One final note on the properties of resistors concerns their rate of energy conversion. Since electric potential is the electric potential energy per unit of charge, and current is the charge flowing through the device per second, the product of change in electric potential and current is the change in electric potential energy per second. Thus, the rate of energy conversion, or power, in a resistor is given by: P = i( ) 2

3 The apacitor A capacitor is a device that stores energy in the electric field between two closely spaced conducting surfaces. When connected to a voltage source, electric charge accumulates on the two surfaces but, since the conducting surfaces are separated by an insulator, the charges cannot travel from one surface to the other. The charges create an electric field in the space between the surfaces, and the two surfaces have a difference in electric potential Ε Once charged, if the capacitor is removed from the original circuit and connected to a second circuit it can act as a voltage source and drive its collected charge through the second circuit. When used in this way, the capacitor clearly acts as a temporary storehouse of energy. To determine the energy stored in a capacitor, we first need to define the capacitance of the capacitor. The capacitance of a capacitor is defined as: = Q where Q is the magnitude of the electric charge stored on either conducting surface, and is the potential difference between the surfaces. The unit of capacitance,, is defined as the farad (F). The amount of energy that can be stored on a capacitor is a function of both its capacitance and the potential difference between its surfaces. The relationship between stored energy and these parameters is: 1 U = ( ) 2 2 3

4 The Inductor An inductor is a device that stores energy in the magnetic field created when current passes through a coil of wire. When connected to a voltage source, current will flow though the inductor, establishing a magnetic field. B i If the voltage source is suddenly removed, current will continue to flow in the coil because of electromagnetic induction. This induced current will act to replace the disappearing source current. The energy needed to drive this current comes from the energy stored in the magnetic field, so in this case the inductor acts as a temporary storehouse of energy. To determine the energy stored in an inductor, we first need to define the inductance of the inductor. The inductance of the inductor is defined as: where Φ L = i Φ is the magnetic flux within the inductor, and i is the current flowing through the inductor. The unit of capacitance, Tm 2, is defined as the henry (H). A The amount of energy that can be stored in an inductor is a function of both its inductance and the current flowing through it. The relationship between stored energy and these parameters is: 1 U = Li 2 2 4

5 . Electric ircuits Analysis Tools esistors in ircuits The circuit at right represents a 12 car battery and two mismatched headlights, 1 = 1.9 Ω and 2 = 2.1 Ω. a. Determine the magnitude of the potential difference across and the current through each circuit component. b. If the battery has a total stored energy of 800 W hr, and produces a constant potential difference until discharged, how long will the bulbs stay lit? 1 2 The potential difference across the car battery is given as 12. This means that the electric potential in the wire coming out of the top of the battery is 12 larger than the potential in the wire coming from the bottom. Since each of the resistors are attached to these same two wires, the top of each resistor is 12 higher in potential than the bottom. Therefore the potential difference across each resistor is 12. When circuit elements are connected such that the elements all have the same potential difference, the elements are said to be in parallel. Since the potential difference across each resistor is known, we can use the definition of resistance to calculate the current through each branch of the circuit. Analyzing branch #1 yields 1 = i i i i = = 1.9Ω = 6.32A 5

6 and branch #2 i i i = i = = 2.1Ω = 5.71A The current that flows through 1 and the current that flows through 2 must also flow through both the top and bottom wires connected to the battery. To complete the mental image of a closed circuit of current, we will say the current flows through the battery as well, although this is not technically true. Therefore, the current that flows through the battery (the total current flowing in the circuit) is: i i i battery battery battery = i + i 1 2 = 6.32A A = 12.0A We can summarize this information in a simple table: across () i through (A) battery To determine how long the headlights will stay lit, we must calculate the total power of the circuit (the total amount of electrical energy converted per second). We can do this separately for each headlight and then add the results: P 1 P 1 P 1 = i ( ) 1 1 = (6.32A)(12 ) = 75.8W and 6

7 P P P = i ( ) 2 = (5.71A)(12 ) = 68.5W 2 so the total power of the circuit is: P total = W Therefore, the battery will last for 800W hrs = 144.3W 5.54hrs apacitors in ircuits The device at right represents a simplified camera flash circuit. With = 3 and = 100 Ω, find such that the flash reaches 80% of its final voltage in 1.0 s. The circuit above, termed an circuit, can best be analyzed by considering the changes in electric potential experienced by a hypothetical charge journeying around the circuit: as it passes through the battery the potential increases by, as it passes through the resistor the potential decreases by = i = i and as it passes through the capacitor the potential decreases by = Q = Q 7

8 Putting these changes in potential together results in: i Q = 0 = 0 Note that the total change in potential (and potential energy) must be zero since the energy given to the charge by the battery is partially converted by the resistor and partially stored by the capacitor. Solving the above equation is more difficult than it looks, since the charge on the capacitor depends on the current (since the current is what deposits the charge on the capacitor) and the current depends on the charge on the capacitor (since as the capacitor accumulates charge, the voltage across it grows, leaving less voltage to drive the current through the resistor). An equation of this type is referred to as a differential equation, and solving it is beyond the requirements for this course. However, we can still understand and make use of the solution nonetheless. If we assume that the capacitor is uncharged when the switch is first closed, then the solution for the current in the circuit as a function of time is: i( t) = e t / This equation says that initially, at t = 0 s, the current in the circuit is equal to the current that would result from a circuit containing a single resistor. This is because since the capacitor is initially uncharged there is no voltage across it, leaving all of the voltage across the resistor. As charge begins to accumulate on the capacitor, however, the voltage across it grows, leaving less and less voltage to drive current through the resistor. Ultimately, the capacitor will reach full charge, where all of the voltage is across the capacitor, leaving no voltage across and no current through the resistor. The formula above says all that in just a few simple symbols! We ve still not ready to answer the question, however. The question asks about the voltage across the capacitor. Since the voltage across the resistor can be expressed as: = i ( t) = i ( t) = ( e ( t) = e t / t / ) 8

9 the voltage across the capacitor is the amount of the source voltage that remains : = e = (1 e = = 0 t / t / This function shows that after a long time (t ), the voltage across the capacitor will equal the voltage of the source. ) Therefore, = (1 e 0.8(3) = (3)(1 e 0.8 = 1 e e 1/100 1/100 = 0.2 t / 1 = ln(0.2) 100 = 6.21 mf Thus, a 6.21 mf capacitor will reach 80% of its final voltage in 1.0 s. ) 1/100 ) 9

10 Inductors in ircuits The device at right represents a simplified electromagnet. With = 100 and = 15 Ω, find L such that the current reaches 5.0 A in 0.5 s. L The circuit above, termed an L circuit, can best be analyzed by considering the changes in electric potential experienced by a hypothetical charge journeying around the circuit: as it passes through the battery the potential increases by, as it passes through the resistor the potential decreases by = i = i and as it passes through the inductor the potential changes by L = Φ = Since by Faraday s Law of Induction, Φ i Li Φ ε = t ( Li) ε = t i ε = L t The emf induced by the inductor is the potential drop across it, so L i = L t 10

11 Putting these changes in potential together results in: i i L t = 0 because the total change in potential (and potential energy) must be zero since the energy given to the charge by the battery is partially converted by the resistor and partially stored by the inductor. As with the circuit, this is a differential equation. The solution for the current in the circuit as a function of time is: i( t) = (1 e t / L This equation says that initially, at t = 0 s, the current in the circuit is zero. This is because before the switch is closed, no current flows through the inductor so no magnetic flux exists within the inductor. When you close the switch, you are trying to change the magnetic flux in the inductor, and by Faraday s Law this attempted change in flux will induce an emf to oppose it. This opposing emf will keep the current from initially flowing. However, ultimately, the inductor will get used to the new flux and the opposing emf will diminish. Once the opposing emf completely disappears, the entire voltage of the battery will go toward driving current thought the resistor and the circuit will begin to behave like a circuit with a single resistor. This is why as t the current in the circuit is simply. Again, the formula says all that much more concisely than I can! ) We could use this expression for current to determine the time-dependence of any other circuit parameter, but since the question asks about the current directly, i( t) = (1 e = (1 e = 1 e e 7.5 / L = 0.25 t / L 7.5 / L 7.5 = ln(0.25) L L = 5.4 H ) (0.5)(15) / L Therefore, if the electromagnet has an inductance of 5.4 H, it will take 0.5 s for the current to rise to 5.0 A. ) 11

12 . Electric ircuits Activities 12

13 The left block below has front face dimensions of 10 cm by 4 cm, with a depth of 3 cm. The right block is made of the same material and is exactly one-half as wide, with front face dimensions of 5 cm by 4 cm, with a depth of 3 cm F A D B E ank the electrical resistance along each of the hypothetical current paths. Largest Smallest The ranking can not be determined based on the information provided. Explain the reason for your ranking: 13

14 Each of the circuits below consists of identical batteries and resistors. All of the switches are closed at the same time. A B D E F a. ank the circuits on the basis of their total resistance. Largest Smallest The ranking can not be determined based on the information provided. b. ank the circuits on the basis of the elapsed time before the battery dies. Largest Smallest The ranking can not be determined based on the information provided. Explain the reason for your ranking: 14

15 Each of the circuits below consists of identical batteries and resistors. All of the switches are closed at the same time. A B D E F a. ank the circuits on the basis of the current through the resistor labeled. Largest Smallest The ranking can not be determined based on the information provided. b. ank the circuits on the basis of the magnitude of the potential difference across the resistor labeled. Largest Smallest The ranking can not be determined based on the information provided. Explain the reason for your ranking: 15

16 Each of the circuits below consists of identical batteries, resistors, and capacitors. All of the switches are closed at the same time. ank each circuit on the basis of the time needed for the positive plate of the capacitor to reach 50% of full charge. A B D E F Largest Smallest The ranking can not be determined based on the information provided. Explain the reason for your ranking: 16

17 Each of the circuits below consists of identical batteries, resistors, and capacitors. All of the switches are closed at the same time. ank each circuit on the basis of the final charge on the positive plate of the capacitor. A B D E F Largest Smallest The ranking can not be determined based on the information provided. Explain the reason for your ranking: 17

18 Each of the circuits below consists of identical batteries, resistors, and capacitors. All of the switches are closed at the same time. A B D E F a. ank each circuit on the basis of the current through the battery just after the switch is closed. Largest Smallest The ranking can not be determined based on the information provided. Explain the reason for your ranking: b. ank each circuit on the basis of the current through the battery long after the switch is closed. Largest Smallest The ranking can not be determined based on the information provided. Explain the reason for your ranking: 18

19 Each of the circuits below consists of identical batteries, resistors, and inductors. All of the switches are closed at the same time. A B D E F a. ank each circuit on the basis of the current through the battery just after the switch is closed. Largest Smallest The ranking can not be determined based on the information provided. Explain the reason for your ranking: b. ank each circuit on the basis of the current through the battery long after the switch is closed. Largest Smallest The ranking can not be determined based on the information provided. Explain the reason for your ranking: 19

20 The cylindrical wire used to form a light bulb filament has radius 3.7 µm and length 1.7 cm. a. If the wire is made of tungsten, what is the resistance of the filament? Tungsten has a resistivity of 5.25 x 10-8 Ωm. b. If the light bulb is connected to a 12 battery, what is the power converted by the light bulb? c. If the battery has a total stored charge of 0.5 A hr, and produces a constant potential difference until discharged, how long will the light bulb light? d. What is the total energy converted by the light bulb? 20

21 The rectangular block of iron at right has front face dimensions of 10 cm by 4 cm, with a depth of 3 cm. y x a. Find the resistance of the block along each of the three coordinate directions. Iron has a resistivity of 9.68 x 10-8 Ωm. z b. If the block is connected to a 12 battery although the direction with the least resistance, what is the current through the block? c. If the battery has a total stored energy of 2.5 W hr, and produces a constant potential difference until discharged, how long will this current flow? 21

22 The circuit at right represents a Halloween decoration with light-up eyes and a spooky sound. The 6 battery provides current for the 1 = 12 Ω bulb and the 2 = 8 Ω speaker. 1 2 a. Determine the magnitude of the potential difference across and the current through each circuit component. across i through battery 1 2 b. If the battery has a total stored energy of 2.5 W hr, and produces a constant potential difference until discharged, how long will the decoration function? c. How much total energy is converted by the bulb? 22

23 The circuit at right represents a Halloween decoration with light-up eyes and a spooky sound. The 6 battery provides current for the 1 = 20 Ω bulb and the 2 = 8 Ω speaker. 1 2 a. Determine the magnitude of the potential difference across and the current through each circuit component. across i through battery 1 2 b. If the battery has a total stored charge of 0.5 A hr, and produces a constant potential difference until discharged, how long will the decoration function? c. What percentage of the total energy is converted by the speaker? 23

24 The circuit at right represents a 12 car battery and two headlights of 0.7 Ω each. 1 2 a. Determine the magnitude of the potential difference across and the current through each circuit component. across i through battery 1 2 b. If the battery has a total stored charge of 120 A hr, and produces a constant potential difference until discharged, how long will the bulbs stay lit? c. How much total energy is converted by bulb #1? 24

25 The circuit at right represents a 12 car battery and two mismatched headlights, 1 = 0.7 Ω and 2 = 1.1 Ω. 1 2 a. Determine the magnitude of the potential difference across and the current through each circuit component. across i through battery 1 2 b. If the battery has a total stored energy of 600 W hr, and produces a constant potential difference until discharged, how long will the bulbs stay lit? c. What percentage of the total energy is converted by bulb #1? 25

26 The circuit at right consists of a 24 battery and three resistors, 1 = 8 Ω, 2 = 4 Ω, and 3 = 6 Ω. Determine the magnitude of the potential difference across and the current through each circuit component across i through battery

27 The circuit at right consists of a 24 battery and three resistors, 1 = 8 Ω, 2 = 4 Ω, and 3 = 6 Ω. Determine the magnitude of the potential difference across and the current through each circuit component across i through battery

28 Assuming the capacitor is initially uncharged and the switch is closed at time t = 0 s, the current through the resistor in the simple circuit at right can be modeled by: i ( t) = e Based on this result, determine the following as functions of time. t a. The potential difference across the resistor: b. The potential difference across the capacitor: c. The charge on the capacitor: d. The energy stored in the capacitor: e. The energy converted by the resistor: 28

29 The device at right represents a simplified camera flash circuit. With = 3 and = 5000 µf, find such that the flash has 12 m stored after 1.2 s. 29

30 The device at right represents a simplified camera flash circuit. With = 3 and = 147 Ω, find such that the flash reaches 95% of full charge in 1.5 s. 30

31 The device at right represents a simplified camera flash circuit. With = 3 and = 147 Ω, find such that the flash has 75% of its maximum energy stored after 1.9 s. 31

32 The device at right represents a simplified camera flash circuit. With = 3 and = 140 Ω, find such that the flash has 15 mj stored after 1.2 s. 32

33 The capacitor is initially uncharged. a. What is the current through 1 and the current through 2 immediately after the switch is first closed? b. What is the current through 1 and the current through 2 long after the switch is closed? c. What is the current through 1 and the current through 2 immediately after the switch is opened (after being closed a long time)?

34 The capacitor is initially uncharged. a. What is the potential difference across 1 and the potential difference across 2 immediately after the switch is first closed? b. What is the potential difference across 1 and the potential difference across 2 long after the switch is closed? c. What is the potential difference across 1 and the potential difference across 2 immediately after the switch is opened (after being closed a long time)?

35 The capacitor is initially uncharged. a. What is the potential difference across 1 and the potential difference across 2 immediately after the switch is first closed? b. What is the potential difference across 1 and the potential difference across 2 long after the switch is closed? c. What is the potential difference across 1 and the potential difference across 2 immediately after the switch is opened (after being closed a long time)?

36 The capacitor is initially uncharged. a. What is the current through 1 and the current through 2 immediately after the switch is first closed? b. What is the current through 1 and the current through 2 long after the switch is closed? c. What is the current through 1 and the current through 2 immediately after the switch is opened (after being closed a long time)?

37 Assuming the switch is closed at time t = 0 s, the current through the inductor in the simple L circuit at right can be modeled by: t L il ( t) = (1 e ) Based on this result, determine the following as functions of time. L a. The potential difference across the resistor: b. The potential difference across the inductor: c. The energy stored in the inductor: d. The energy converted by the resistor: 37

38 The device at right represents a simplified electromagnet. With = 40 and L = 15 H, find such that the current (and magnetic field) reaches 80% of its final value in 2.5 s. L 38

39 The device at right represents a simplified electromagnet. With = 40 and = 14.7 Ω, find L such that the current reaches 2.5 A in 4.0 s. L 39

40 The device at right represents a simplified electromagnet. With = 40 and L = 8.3 H, find such that the magnet has 75% of its maximum energy stored after 1.9 s. L 40

41 The device at right represents a simplified electromagnet. With = 40 and = 10 Ω, find L such that the magnet has 100 J of energy stored after 0.3 s. L 41

42 a. What is the current through 1 and the current through 2 immediately after the switch is first closed? b. What is the current through 1 and the current through 2 long after the switch is closed? c. What is the current through 1 and the current through 2 immediately after the switch is opened (after being closed a long time)? 1 2 L 42

43 a. What is the potential difference across 1 and the potential difference across 2 immediately after the switch is first closed? b. What is the potential difference across 1 and the potential difference across 2 long after the switch is closed? c. What is the potential difference across 1 and the potential difference across 2 immediately after the switch is opened (after being closed a long time)? 1 2 L 43

44 a. What is the potential difference across 1 and the potential difference across 2 immediately after the switch is first closed? b. What is the potential difference across 1 and the potential difference across 2 long after the switch is closed? c. What is the potential difference across 1 and the potential difference across 2 immediately after the switch is opened (after being closed a long time)? 2 1 L 44

45 a. What is the current through 1 and the current through 2 immediately after the switch is first closed? b. What is the current through 1 and the current through 2 long after the switch is closed? c. What is the current through 1 and the current through 2 immediately after the switch is opened (after being closed a long time)? 2 1 L 45

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