Divider Circuits and Kirchhoff’s Laws - Review Sheet | PHYS 3036, Study notes of Physics

Download Divider Circuits and Kirchhoff’s Laws - Review Sheet | PHYS 3036 and more Study notes Physics in PDF only on Docsity! Chapter 6 DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS Contents 6.1 Voltage divider circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 6.2 Kirchhoff’s Voltage Law (KVL) . . . . . . . . . . . . . . . . . . . . . . . 173 6.3 Current divider circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 6.4 Kirchhoff’s Current Law (KCL) . . . . . . . . . . . . . . . . . . . . . . . 187 6.5 Contributors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 6.1 Voltage divider circuits Let’s analyze a simple series circuit, determining the voltage drops across individual resistors: + - R1 R2 R3 5 kΩ 7.5 kΩ 10 kΩ45 V 165 166 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS E I R Volts Amps Ohms R1 R2 R3 Total 5k 10k 7.5k 45 From the given values of individual resistances, we can determine a total circuit resistance, knowing that resistances add in series: E I R Volts Amps Ohms R1 R2 R3 Total 5k 10k 7.5k 22.5k 45 From here, we can use Ohm’s Law (I=E/R) to determine the total current, which we know will be the same as each resistor current, currents being equal in all parts of a series circuit: E I R Volts Amps Ohms R1 R2 R3 Total 5k 10k 7.5k 45 22.5k 2m 2m 2m 2m Now, knowing that the circuit current is 2 mA, we can use Ohm’s Law (E=IR) to calculate voltage across each resistor: E I R Volts Amps Ohms R1 R2 R3 Total 5k 10k 7.5k 45 22.5k 2m 2m 2m 2m 10 20 15 It should be apparent that the voltage drop across each resistor is proportional to its resistance, given that the current is the same through all resistors. Notice how the voltage across R2 is double that of the voltage across R1, just as the resistance of R2 is double that of R1. If we were to change the total voltage, we would find this proportionality of voltage drops remains constant: E I R Volts Amps Ohms R1 R2 R3 Total 5k 10k 7.5k 22.5k 8m 8m 8m 8m 40 80 60 180 6.1. VOLTAGE DIVIDER CIRCUITS 169 1 2 wiper contact Potentiometer The wiper contact is the left-facing arrow symbol drawn in the middle of the vertical resistor element. As it is moved up, it contacts the resistive strip closer to terminal 1 and further away from terminal 2, lowering resistance to terminal 1 and raising resistance to terminal 2. As it is moved down, the opposite effect results. The resistance as measured between terminals 1 and 2 is constant for any wiper position. 1 2 less resistance more resistance 1 2 less resistance more resistance Shown here are internal illustrations of two potentiometer types, rotary and linear: Resistive strip Wiper Terminals Rotary potentiometer construction 170 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS Resistive strip Wiper Terminals Linear potentiometer construction Some linear potentiometers are actuated by straight-line motion of a lever or slide button. Others, like the one depicted in the previous illustration, are actuated by a turn-screw for fine adjustment ability. The latter units are sometimes referred to as trimpots, because they work well for applications requiring a variable resistance to be ”trimmed” to some precise value. It should be noted that not all linear potentiometers have the same terminal assignments as shown in this illustration. With some, the wiper terminal is in the middle, between the two end terminals. The following photograph shows a real, rotary potentiometer with exposed wiper and slidewire for easy viewing. The shaft which moves the wiper has been turned almost fully clockwise so that the wiper is nearly touching the left terminal end of the slidewire: Here is the same potentiometer with the wiper shaft moved almost to the full-counterclockwise position, so that the wiper is near the other extreme end of travel: 6.1. VOLTAGE DIVIDER CIRCUITS 171 If a constant voltage is applied between the outer terminals (across the length of the slidewire), the wiper position will tap off a fraction of the applied voltage, measurable between the wiper contact and either of the other two terminals. The fractional value depends entirely on the physical position of the wiper: less voltagemore voltage Using a potentiometer as a variable voltage divider Just like the fixed voltage divider, the potentiometer’s voltage division ratio is strictly a function of resistance and not of the magnitude of applied voltage. In other words, if the potentiometer knob or lever is moved to the 50 percent (exact center) position, the voltage dropped between wiper and either outside terminal would be exactly 1/2 of the applied voltage, no matter what that voltage happens to be, or what the end-to-end resistance of the potentiometer is. In other words, a potentiometer functions as a variable voltage divider where the voltage division ratio is set by wiper position. This application of the potentiometer is a very useful means of obtaining a variable voltage from a fixed-voltage source such as a battery. If a circuit you’re building requires a certain amount of voltage that is less than the value of an available battery’s voltage, you may connect the outer terminals of a potentiometer across that battery and ”dial up” whatever voltage you need between the potentiometer wiper and one of the outer terminals for use in your circuit: 174 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS implied, for positive readings in digital meter displays. However, for this lesson the polarity of the voltage reading is very important and so I will show positive numbers explicitly: E2-1 = +45 V When a voltage is specified with a double subscript (the characters ”2-1” in the notation ”E2−1”), it means the voltage at the first point (2) as measured in reference to the second point (1). A voltage specified as ”Ecg” would mean the voltage as indicated by a digital meter with the red test lead on point ”c” and the black test lead on point ”g”: the voltage at ”c” in reference to ”g”. COMA V V A A OFF . . . . . . cd Ecd The meaning of RedBlack If we were to take that same voltmeter and measure the voltage drop across each resistor, stepping around the circuit in a clockwise direction with the red test lead of our meter on the point ahead and the black test lead on the point behind, we would obtain the following readings: E3-2 = -10 V E4-3 = -20 V E1-4 = -15 V 6.2. KIRCHHOFF’S VOLTAGE LAW (KVL) 175 + - 1 2 3 4 + + + - - - R1 R2 R3 5 kΩ 10 kΩ 7.5 k Ω 45 VV ΩCOMA V Ω COMA V Ω COMA V Ω COMA E2-1 E3-2 E4-3 E1-4 +45 -10 -20 -15 We should already be familiar with the general principle for series circuits stating that individual voltage drops add up to the total applied voltage, but measuring voltage drops in this manner and paying attention to the polarity (mathematical sign) of the readings reveals another facet of this principle: that the voltages measured as such all add up to zero: -10 V -20 V -15 V +45 V 0 V + voltage from point to point 12 voltage from point to point voltage from point to point voltage from point to point 23 34 41 E2-1 = E3-2 = E4-3 = E1-4 = This principle is known as Kirchhoff’s Voltage Law (discovered in 1847 by Gustav R. Kirchhoff, a German physicist), and it can be stated as such: ”The algebraic sum of all voltages in a loop must equal zero” By algebraic, I mean accounting for signs (polarities) as well as magnitudes. By loop, I mean any path traced from one point in a circuit around to other points in that circuit, and finally back to the initial point. In the above example the loop was formed by following points in this order: 1-2-3-4-1. It doesn’t matter which point we start at or which direction we proceed in tracing the loop; the voltage sum will still equal zero. To demonstrate, we can tally up the voltages in loop 3-2-1-4-3 of the same circuit: 176 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS 0 V + voltage from point to point voltage from point to point voltage from point to point voltage from point to point +10 V -45 V +15 V +20 V 32 21 14 43 E2-3 = E1-2 = E4-1 = E3-4 = This may make more sense if we re-draw our example series circuit so that all components are represented in a straight line: + 12 3 4 + + +- - - 2 - current current R1 R2 R3 45 V 5 kΩ 10 kΩ 7.5 kΩ It’s still the same series circuit, just with the components arranged in a different form. Notice the polarities of the resistor voltage drops with respect to the battery: the battery’s voltage is negative on the left and positive on the right, whereas all the resistor voltage drops are oriented the other way: positive on the left and negative on the right. This is because the resistors are resisting the flow of electrons being pushed by the battery. In other words, the ”push” exerted by the resistors against the flow of electrons must be in a direction opposite the source of electromotive force. Here we see what a digital voltmeter would indicate across each component in this circuit, black lead on the left and red lead on the right, as laid out in horizontal fashion: + 12 3 4 + + +- - - 2 - current R1 R2 R3 45 V5 kΩ 10 kΩ 7.5 kΩ -10 V -20 V -15 V +45 V V Ω COMA -10 V Ω COMA V Ω COMA V Ω COMA -20 -15 +45 E3-2 E4-3 E1-4 E2-1 If we were to take that same voltmeter and read voltage across combinations of components, starting with only R1 on the left and progressing across the whole string of components, we will see how the voltages add algebraically (to zero): 6.2. KIRCHHOFF’S VOLTAGE LAW (KVL) 179 + - + - - + + - - + + - + - 5 V 8 V 3 V 11 V 8 V 10 V 2 V Try any order of steps from any terminal in the above diagram, stepping around back to the original terminal, and you’ll find that the algebraic sum of the voltages always equals zero. Furthermore, the ”loop” we trace for KVL doesn’t even have to be a real current path in the closed-circuit sense of the word. All we have to do to comply with KVL is to begin and end at the same point in the circuit, tallying voltage drops and polarities as we go between the next and the last point. Consider this absurd example, tracing ”loop” 2-3-6-3-2 in the same parallel resistor circuit: + - + - + - + - 1 2 3 4 5678 R1 R2 R36 V 180 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS 0 V + voltage from point to point voltage from point to point 230 V voltage from point to point voltage from point to point 6 6 2 -6 V 0 V +6 V 3 3 3 E3-2 = E6-3 = E3-6 = E2-3 = E2-2 = KVL can be used to determine an unknown voltage in a complex circuit, where all other voltages around a particular ”loop” are known. Take the following complex circuit (actually two series circuits joined by a single wire at the bottom) as an example: 1 2 3 4 5 6 7 8 9 10 + - + - + - + -+ - + - 35 V 15 V 20 V 13 V 12 V 25 V To make the problem simpler, I’ve omitted resistance values and simply given voltage drops across each resistor. The two series circuits share a common wire between them (wire 7-8-9-10), making voltage measurements between the two circuits possible. If we wanted to determine the voltage between points 4 and 3, we could set up a KVL equation with the voltage between those points as the unknown: E4-3 + E9-4 + E8-9 + E3-8 = 0 E4-3 + 12 + 0 + 20 = 0 E4-3 + 32 = 0 E4-3 = -32 V 6.2. KIRCHHOFF’S VOLTAGE LAW (KVL) 181 1 2 3 4 5 6 7 8 9 10 + - + - + - + -+ - + - 35 V 15 V 20 V 13 V 12 V 25 V Measuring voltage from point 4 to point 3 (unknown amount) V Ω COMA E4-3 ??? 1 2 3 4 5 6 7 8 9 10 + - + - + - + -+ - + - 35 V 15 V 20 V 13 V 12 V 25 V V Ω COMA Measuring voltage from point 9 to point 4 (+12 volts) E4-3 + 12 +12 184 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS 6.3 Current divider circuits Let’s analyze a simple parallel circuit, determining the branch currents through individual resistors: + - + - + - + - R1 R2 R3 1 kΩ 3 kΩ 2 kΩ 6 V Knowing that voltages across all components in a parallel circuit are the same, we can fill in our voltage/current/resistance table with 6 volts across the top row: E I R Volts Amps Ohms R1 R2 R3 Total 6 6 6 6 1k 3k 2k Using Ohm’s Law (I=E/R) we can calculate each branch current: E I R Volts Amps Ohms R1 R2 R3 Total 6 6 6 6 1k 3k 2k 6m 2m 3m Knowing that branch currents add up in parallel circuits to equal the total current, we can arrive at total current by summing 6 mA, 2 mA, and 3 mA: E I R Volts Amps Ohms R1 R2 R3 Total 6 6 6 6 1k 3k 2k 6m 2m 3m 11m The final step, of course, is to figure total resistance. This can be done with Ohm’s Law (R=E/I) in the ”total” column, or with the parallel resistance formula from individual resistances. Either way, we’ll get the same answer: 6.3. CURRENT DIVIDER CIRCUITS 185 E I R Volts Amps Ohms R1 R2 R3 Total 6 6 6 6 1k 3k 2k 6m 2m 3m 11m 545.45 Once again, it should be apparent that the current through each resistor is related to its resistance, given that the voltage across all resistors is the same. Rather than being directly proportional, the relationship here is one of inverse proportion. For example, the current through R1 is half as much as the current through R3, which has twice the resistance of R1. If we were to change the supply voltage of this circuit, we find that (surprise!) these proportional ratios do not change: E I R Volts Amps Ohms R1 R2 R3 Total 1k 3k 2k 545.45 24 24 24 24 24m 8m 12m 44m The current through R1 is still exactly twice that of R2, despite the fact that the source volt- age has changed. The proportionality between different branch currents is strictly a function of resistance. Also reminiscent of voltage dividers is the fact that branch currents are fixed proportions of the total current. Despite the fourfold increase in supply voltage, the ratio between any branch current and the total current remains unchanged: = = = = = = IR1 Itotal Itotal 11 mA 11 mA 44 mA 44 mA IR2 6 mA 24 mA 2 mA 8 mA 0.54545 0.18182 = Itotal 11 mA = 44 mA = IR3 3 mA 12 mA 0.27273 For this reason a parallel circuit is often called a current divider for its ability to proportion – or divide – the total current into fractional parts. With a little bit of algebra, we can derive a formula for determining parallel resistor current given nothing more than total current, individual resistance, and total resistance: 186 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS Current through any resistor En Rn In = Voltage in a parallel circuit Etotal = En = Itotal Rtotal Substituting . . . Itotal Rtotal for En in the first equation . . . Current through any parallel resistor In = Rn Itotal Rtotal . . . or . . . In = Itotal Rn Rtotal The ratio of total resistance to individual resistance is the same ratio as individual (branch) current to total current. This is known as the current divider formula, and it is a short-cut method for determining branch currents in a parallel circuit when the total current is known. Using the original parallel circuit as an example, we can re-calculate the branch currents using this formula, if we start by knowing the total current and total resistance: IR1 = 545.45 Ω 11 mA 1 kΩ = 6 mA 11 mA 545.45 Ω = 11 mA 545.45 Ω = IR2 = IR3 = 3 kΩ 2 mA 2 kΩ 3 mA If you take the time to compare the two divider formulae, you’ll see that they are remarkably similar. Notice, however, that the ratio in the voltage divider formula is Rn (individual resistance) divided by RTotal, and how the ratio in the current divider formula is RTotal divided by Rn: 6.5. CONTRIBUTORS 189 general relationship as such: Iexiting = Ientering Mr. Kirchhoff decided to express it in a slightly different form (though mathematically equiva- lent), calling it Kirchhoff’s Current Law (KCL): Ientering + (-Iexiting) = 0 Summarized in a phrase, Kirchhoff’s Current Law reads as such: ”The algebraic sum of all currents entering and exiting a node must equal zero” That is, if we assign a mathematical sign (polarity) to each current, denoting whether they enter (+) or exit (-) a node, we can add them together to arrive at a total of zero, guaranteed. Taking our example node (number 3), we can determine the magnitude of the current exiting from the left by setting up a KCL equation with that current as the unknown value: I2 + I3 + I = 0 2 mA + 3 mA + I = 0 . . . solving for I . . . I = -2 mA - 3 mA I = -5 mA The negative (-) sign on the value of 5 milliamps tells us that the current is exiting the node, as opposed to the 2 milliamp and 3 milliamp currents, which must were both positive (and therefore entering the node). Whether negative or positive denotes current entering or exiting is entirely arbitrary, so long as they are opposite signs for opposite directions and we stay consistent in our notation, KCL will work. Together, Kirchhoff’s Voltage and Current Laws are a formidable pair of tools useful in analyzing electric circuits. Their usefulness will become all the more apparent in a later chapter (”Network Analysis”), but suffice it to say that these Laws deserve to be memorized by the electronics student every bit as much as Ohm’s Law. • REVIEW: • Kirchhoff’s Current Law (KCL): ”The algebraic sum of all currents entering and exiting a node must equal zero” 6.5 Contributors Contributors to this chapter are listed in chronological order of their contributions, from most recent to first. See Appendix 2 (Contributor List) for dates and contact information. 190 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS Jason Starck (June 2000): HTML document formatting, which led to a much better-looking second edition. Ron LaPlante (October 1998): helped create ”table” method of series and parallel circuit analysis.