Calculate the amount of voltage between points {\bf A} and {\bf B} in this circuit. You must sketch polarity marks (+ , $-$) on the schematic diagram to show the polarity of $V_{AB}$, as well as show all of your mathematical work!
$V_{\bf AB}$ = 9.198 volts, {\bf A} positive and {\bf B} negative.
The voltage between points A and B is the supply voltage (26 volts) minus the voltage drops across the 1k and parallel subnetwork resistors. Alternatively, one could calculate $V_{AB}$ by adding the voltage drops of the 1k and 270 ohm resistors.
The latter solution makes it easiest to see the polarity of $V_{AB}$: noting how the voltage drops across the 1k and 270 ohm resistors are additive, we see point A being the most positive and point B being the most negative.
Calculate the resistance between points {\bf A} and {\bf B} ($R_{AB}$) for the following resistor networks:
{\bf Figure 1:}
$R_{AB}$ = 500 $\Omega$
{\bf Figure 2:}
$R_{AB}$ = 750 $\Omega$
{\bf Figure 3:}
$R_{AB}$ = 1.511 k$\Omega$
{\bf Figure 4:}
$R_{AB}$ = 940 $\Omega$
{\bf Figure 5:}
$R_{AB}$ = 880 $\Omega$
{\bf Figure 6:}
$R_{AB}$ = 80.54 $\Omega$
Note that the circuit in figure 4 is a ``trick:’’ two of the resistors contribute absolutely nothing to $R_{AB}$! Be sure to discuss why this is with your students.
Discuss with your students how they approached each of these problems, and let the entire class participate in the reasoning process. The point of this question, like most of the questions in the Socratic Electronics project, is not merely to obtain the correct answers, but to stimulate understanding of {\it how} to solve problems such as these.
Use Kirchhoff’s Voltage Law to calculate the magnitude and polarity of the voltage across resistors R_2 and R_4 in this resistor network:
In your discussion, be sure to explore more than one ``loop’’ when using KVL. Not only does this demonstrate the arbitrary nature of your loop choice, but it also serves as a double-check for your work!
It is not necessary to know anything about series-parallel or even parallel circuits in order to solve for R_2’s or R_4’s voltage—all one needs to know is how to use Kirchhoff’s Voltage Law.
Imagine you are using a digital voltmeter to measure voltages between pairs of points in a circuit, following the sequence of steps shown in the following diagrams:
How much voltage would be registered by the voltmeter in each of the steps? Be sure to include the sign of the DC voltage measured (note the coloring of the voltmeter leads, with the red lead always on the first point denoted in the subscript: $V_{BA}$ = red lead on ``B’’ and black lead on ``A’‘):
{\bullet} $V_{BA} = $
{\bullet} $V_{DB} = $
{\bullet} $V_{FD} = $
{\bullet} $V_{AF} = $
What is the algebraic sum of these voltages?
{\bullet} $V_{BA} = +10.8$ volts
{\bullet} $V_{DB} = +25.2$ volts
{\bullet} $V_{FD} = -12.0$ volts
{\bullet} $V_{AF} = -24.0$ volts
Ask your students this question: ``Will the algebraic sum of voltage measurements ever be other than zero in a loop?’’ Ask them to explain {\it why} this is, as best they can.
Calculate the amount of voltage dropped across resistor R_2:
Also, note the direction of current through it and the polarity of the voltage drop across it.
$V_{R2} = 12.11 \hbox{ volts}$, positive on top and negative on bottom. If you follow conventional flow notation, this means current goes down through resistor R_2. The actual flow of electrons through R_2, however, is up.
Discuss with your students how they obtained their answers for this question. The reasoning and procedures are far more important than the actual answer itself.
Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.
By having students
Complete the table of values for this circuit:
A noteworthy feature of this circuit’s schematic is how the power supply connections are shown. Unlike many of my schematic diagrams, I do not show a ``battery’’ symbol here for a voltage source. Instead, I show power supply ``rail’’ symbols (flat line and a ground symbol). Let your students know that this is very common symbolism in modern schematics, and that is merely saves having to draw lines to a voltage source symbol (as well as the source symbol itself).
Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.
Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.
By having students
Suppose you were designing a circuit that required two LEDs for ``power on’’ indication. The power supply voltage is 15 volts, and each LED is rated at 1.6 volts and 20 mA. Calculate the dropping resistor sizes and power ratings:
After doing this, a co-worker looks at your circuit and suggests a modification. Why not use a single dropping resistor for both LEDs, economizing the number of components necessary?
Re-calculate the dropping resistor ratings (resistance {\it and} power) for the new design.
With two resistors: $R_1 = R_2 = 670 \> \Omega$, rated for at least 0.268 watts (1/2 watt would be a practical rating).
With one resistor: $R_1 = 335 \> \Omega$, rated for at least 0.536 watts (1 watt would be a practical rating).
If students are not yet familiar with the ``+V’’ symbol used to denote the positive power supply connection in this schematic, let them know that this is a very common practice in electronic notation, just as it is common to use the ground symbol as a power supply connection symbol.
The follow-up question is a very practical one, for it is seldom that you have the exact components on-hand to match the requirements of a circuit you are building. It is important to understand which way is safer to err (too large or too small) when doing ``as-built’’ design work.
Complete the table of values for this circuit:
Complete the table of values for this circuit:
Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.
Complete the table of values for this circuit:
Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.
Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.
By having students
Identify which of these components are connected directly in series with each other, and which are connected directly in parallel with each other:
Assume that the open wire ends are connection points to a power source. In circuits where ground symbols appear, consider ground as the other side of the power source.
{\bf Figure 1:}
R1 in series with SW1.
{\bf Figure 2:}
R1 in series with R2; R3 in parallel with R4.
{\bf Figure 3:}
R1 parallel with R2.
{\bf Figure 4:}
R1 parallel with R2.
{\bf Figure 5:}
L1 in series with C1.
{\bf Figure 6:}
R3 in parallel with R4.
Work with your students to clearly identify rules by which series and parallel connections may be identified. This is extremely important for students to grasp if they are to be successful analyzing series-parallel networks of any kind. The most common problems I encounter as an electronics instructor with reference to series-parallel are invariably related to students’ lack of ability to consistently distinguish series sub-networks and parallel sub-networks in series-parallel combination circuits.
Complete the table of values for this circuit:
Ask your students to identify components in this series-parallel circuit that are guaranteed to share the same voltage, and components that are guaranteed to share the same current, without reference to any calculations. This is a good exercise in identifying parallel and series interconnections, respectively.
Students often have difficulty formulating a method of solution: determining what steps to take to get from the given conditions to a final answer. While it is helpful at first for you (the instructor) to show them, it is bad for you to show them too often, lest they stop thinking for themselves and merely follow your lead. A teaching technique I have found very helpful is to have students come up to the board (alone or in teams) in front of class to write their problem-solving strategies for all the others to see. They don’t have to actually do the math, but rather outline the steps they would take, in the order they would take them.
By having students
When the 5 k$\Omega$ potentiometer in this circuit is set to its 0
{\bullet} At 0
{\bullet} At 25
{\bullet} At 50
{\bullet} At 75
{\bullet} At 100
Calculate what the output voltages will be if a 1 k$\Omega$ load resistor is connected between the ``$V_{out}$’’ terminal and ground:
{\bullet} At 0
{\bullet} At 25
{\bullet} At 50
{\bullet} At 75
{\bullet} At 100
{\bullet} At 0
{\bullet} At 25
{\bullet} At 50
{\bullet} At 75
{\bullet} At 100
This question is really nothing more than five loaded voltage divider problems packed into one! It is a very practical question, as potentiometers are very often used as variable voltage dividers, and students must realize the effects a load resistance will have on the characteristics of such dividers. Point out to them the extreme nonlinearity created by the inclusion of the load resistance.
Determine the voltages (with respect to ground) at points {\bf A} and {\bf B} in this circuit under four different conditions: both loads off, load 1 on (only), load 2 on (only), and both loads on:
$$\begin{array} {|l|l|} \hline Voltage & Both loads off & Load 1 on (only) & Load 2 on (only) & Both loads on \\ \hline $V_A$ & & & & \\ \hline $V_B$ & & & & \\ \hline \end{array}$$
$$\begin{array} {|l|l|} \hline Voltage & Both loads off & Load 1 on (only) & Load 2 on (only) & Both loads on \\ \hline $V_A$ & 26.4 volts & 26.3 volts & 22.4 volts & 22.3 volts \\ \hline $V_B$ & 5 volts & 4.46 volts & 4.23 volts & 3.78 volts \\ \hline \end{array}$$
Students will have to re-consider (and possible re-draw) the circuit for each loading condition, which is one of the major points of this question. The fact that a circuit can ``change’’ just by throwing a switch is an important concept for electronics students to grasp.
Another concept employed in this question is that of voltages specified at single points with an implied reference of ground. Note to students how each voltage was simply referenced by a single letter, either {\bf A} or {\bf B}. Of course there is no such thing as voltage at a single point in any circuit, so we need another point to reference, and that point is ground. This is {\it very} commonly seen in electronic circuits of all types, and is a good thing to be exposed to early on in one’s electronics education.
A much less obvious point of this question is to subtly introduce the concept of discrete states (loading conditions) available with a given number of boolean elements (switches). Given two load switches, there are four possible states of circuit loading, previewing binary states in digital circuits.
One of the resistors in this voltage divider circuit is failed open. Based on the voltage readings shown at each load, determine which one it is:
Resistor R1 has failed open. This is evident because only load \#1 is receiving any power; the other two loads are completely ``dead’‘.
Discuss with your students how they were able to predict R1 was the faulty resistor. Is there any particular clue in the diagram indicating R1 as the obvious problem?
One of the resistors in this voltage divider circuit is failed open. Based on the voltage readings shown at each load, determine which one it is:
Resistor R2 has failed open. We can tell this because load \#3 is receiving no power at all while load \#2 is being over-powered.
Discuss with your students how they were able to predict R2 was the faulty resistor. Is there any particular clue in the diagram indicating R2 as the obvious problem?
One of the resistors in this voltage divider circuit is failed (either open or shorted). Based on the voltage readings shown at each load, determine which one and what type of failure it is:
Resistor R1 has failed (partially) shorted. We can tell this because both loads \#2 and \#3 are being over-powered.
Discuss with your students how they were able to predict R1 was the faulty resistor. Is there any particular clue in the diagram indicating R1 as the obvious problem? Some students may suspect an open failure in resistor R3 could cause the same effects, but there is a definite way to tell that the problem can {\it only} come from a short in R1 (hint: analyze resistor R2).
Explain that not all ``shorted’’ failures are ``hard’’ in the sense of being direct metal-to-metal wire connections. Quite often, components will fail shorted in a ``softer’’ sense, meaning they still have some non-trivial amount of electrical resistance.
Old vacuum-tube based electronic circuits often required several different voltage levels for proper operation. An easy way to obtain these different power supply voltages was to take a single, high-voltage power supply circuit and ``divide’’ the total voltage into smaller divisions.
These voltage divider circuits also made provision for a small amount of ``wasted’’ current through the divider called a {\it bleeder} current, designed to discharge the high voltage output of the power supply quickly when it was turned off.
Design a high-voltage divider to provide the following loads with their necessary voltages, plus a ``bleeder’’ current of 5 mA (the amount of current going through resistor R4):
The key to calculating all resistor values is to determine how much voltage each one must drop and how much current each one must carry. The current question may be answered by applying Kirchhoff’s Current Law (KCL) to each of the nodes in the circuit, while the voltage question may be answered by calculating the voltage difference between each pair of supply lines to the tube circuit.
{\bullet} $R_1 =$ 3.25 k$\Omega$
{\bullet} $R_2 =$ 11 k$\Omega$
{\bullet} $R_3 =$ 3.67 k$\Omega$
{\bullet} $R_4 =$ 9 k$\Omega$
Be sure to ask your students {\it how} they obtained the solution to this problem. If no one was able to arrive at a solution, then present the following technique: simplify the problem (fewer resistors, perhaps) until the solution is obvious, then apply the same strategy you used to solve the obvious problem to the more complex versions of the problem, until you have solved the original problem in all its complexity.
Calculate the necessary value of $R$ to create a voltage drop of 4 volts between test points {\bf A} and {\bf B}:
$R$ = 2.667 k$\Omega$
In a parallel circuit, certain general principles may be stated with regard to quantities of voltage, current, resistance, and power. Complete these sentences, each one describing a fundamental principle of parallel circuits:
``In a parallel circuit, voltage . . .’‘
``In a parallel circuit, current . . .’‘
``In a parallel circuit, resistance . . .’‘
``In a parallel circuit, power . . .’‘
For each of these rules, explain {\it why} it is true.
``In a parallel circuit, voltage {\it is equal across all components}.’‘
This is true because a parallel circuit by definition is one where the constituent components all share the same two equipotential points.
``In a parallel circuit, current{\it s add to equal the total}.’‘
This is an expression of Kirchhoff’s Current Law (KCL), whereby the algebraic sum of all currents entering and exiting a node must be equal to zero.
``In a parallel circuit, resistance{\it s diminish to equal the total}.’‘
Each resistance in a parallel circuit provides another path for electric current. When resistances are connected in parallel, their combined total paths provide less opposition than any single path because the current is able to split up and proportionately follow these alternative paths.
``In a parallel circuit, power {\it dissipations add to equal the total}.’‘
This is an expression of the Conservation of Energy, which states energy cannot be created or destroyed. Anywhere power is dissipated in any load of a circuit, that power must be accounted for back at the source, no matter how those loads might be connected to each other.
Rules of series and parallel circuits are very important for students to comprehend. However, a trend I have noticed in many students is the habit of memorizing rather than understanding these rules. Students will work hard to memorize the rules without really comprehending {\it why} the rules are true, and therefore often fail to recall or apply the rules properly.
Explain, step by step, how to calculate the amount of current ($I$) that will go through each resistor in this series circuit, and also the current ($I$) supplied by the DC voltage source:
First we need to identify all the relevant principles for series circuits:
{\bullet} The algebraic sum of all voltages in the circuit will be equal to zero (Kirchhoff’s Voltage Law)
{\bullet} Current is common throughout a series circuit, because there is only one path for current in the entire circuit
{\bullet} Resistances add in series
We know the voltage of the source and the resistance of the three loads. However, we cannot simply apply Ohm’s Law at this point because the source voltage is not impressed entirely on any one of the loads—rather the source voltage will be split up proportionately amongst the three loads in accordance with KVL. It is important to always apply Ohm’s Law {\it in context}: $V = IR$ is true only if $V$, $I$, and $R$ apply to the same component or set of components. Here, the 36 volts of the source applies to all three resistors, not to any one resistor.
However, we may apply the principle of resistances adding in series to arrive at a total resistance value for the circuit, which we may then apply to total voltage to find total current. Adding up the three resistors’ values, we get a total resistance of $R_{total} = 1500 + 10000 + 4700 = 16200$ ohms. Total circuit current is then calculated as follows:
$$I = {V \over R} = {36 \hbox{ V} \over 16200 \> \Omega} = 2.222 \hbox{ mA}$$
It is helpful to annotate all calculated values on the circuit schematic for easy reference. The reason this is helpful is because it applies a context to the calculated value. Here we will sketch arrows (in the direction of conventional flow) to document the 2.222 mA circuit current, based on the relationship between voltage and current for {\it sources} (i.e. current exits the positive pole of a source because the source is driving that current):
Since this is a series circuit, we know that this value of current (2.222 milliamps) will be common through all components. Now that we know the current through each resistor and the resistance of each resistor, we may apply Ohm’s Law to each resistor individually as such:
$$V_{R1} = I R_1 = (2.222 \hbox{ mA}) (1500 \> \Omega) = 3.333 \hbox{ V}$$
$$V_{R2} = I R_2 = (2.222 \hbox{ mA}) (10000 \> \Omega) = 22.222 \hbox{ V}$$
$$V_{R3} = I R_3 = (2.222 \hbox{ mA}) (4700 \> \Omega) = 10.444 \hbox{ V}$$
\filbreak
Once again it is recommended to annotate the circuit schematic with these calculated values, for the sake of keeping all calculations in context. The polarity (+ , $-$) of each voltage is important to note as well, and we know this by the relationship between voltage and current for {\it loads} (i.e. the positive pole of a load is the one where conventional flow enters, because the voltage dropped by a load is opposing current):
As a final check of our work, we may sum these three resistors’ voltage drops to ensure they do indeed add up to equal the source voltage in accordance with KVL:
$$ 3.333 \hbox{ V} + 22.222 \hbox{ V} + 10.444 \hbox{ V} = 36 \hbox{ V}$$
Determine the amount of voltage dropped by each resistor in this circuit, if each resistor has a color code of Brn, Blk, Red, Gld (assume perfectly precise resistance values—0
Also, determine the following information about this circuit:
{\bullet} Current through each resistor
{\bullet} Power dissipated by each resistor
{\bullet} Ratio of each resistor’s voltage drop to battery voltage ($E_R \over E_{bat}$)
{\bullet} Ratio of each resistor’s resistance to the total circuit resistance ($R \over R_{total}$)
Voltage across each resistor = 1.5 V
Current through each resistor = 1.5 mA
Power dissipated by each resistor = 2.25 mW
Voltage ratio = $1 \over 3$
Resistance ratio = $1 \over 3$
When performing the mathematical analysis on this circuit, there is more than one possible sequence of steps to obtaining the solutions. Different students in your class may very well have different solution sequences, and it is a good thing to have students share their differing problem-solving techniques before the whole class.
An important aspect of this question is for students to observe the identical ratios (voltage versus resistance), and determine whether or not these ratios are equal by chance or equal by necessity. Ask your students, “What kind of evidence would prove these ratios were merely equal by chance?” Setting mathematics aside and viewing this circuit from a purely experimental point of view, ask your students what data could possibly prove these ratios to be equal by chance in this particular case? Hint: it would only take a single example to prove this!
Calculate the output voltages of these two voltage divider circuits ($V_A$ and $V_B$):
Now, calculate the voltage between points {\bf A} (red lead) and {\bf B} (black lead) ($V_{AB}$).
$V_A =$ + 65.28 V
$V_B =$ + 23.26 V
$V_{AB} =$ + 42.02 V (point {\bf A} being positive relative to point {\bf B})
Calculate both the maximum and the minimum amount of voltage obtainable from this potentiometer circuit (as measured between the wiper and ground):
$V_{max}$ = 3.85 volts
$V_{min}$ = 0.35 volts
Be sure to ask your students how they obtained their answers, not just what the answers are. There is more than one correct way to analyze this circuit!
Incidentally, there is nothing significant about the use of European schematic symbols in this question. I did this simply to provide students with more exposure to this schematic convention.
Suppose that an electric heater, which is nothing more than a large resistor, dissipates 500 watts of power when directly connected to a 110 volt source:
Now suppose that exact same heater is connected to one end of a long two-wire cable, which is then connected to the same 110 volt source. Assuming that each conductor within the cable has an end-to-end resistance of 3 ohms, how much power will the heater dissipate?
$P$ = 321.1 watts
The purpose of this question, besides providing a good problem-solving exercise for students, is to get them to realize one of the practical implications of power-line resistance.
Suppose an analog voltmeter has a range of 0 to 10 volts, and an internal resistance of exactly 100 k$\Omega$:
Show how a single resistor could be connected to this voltmeter to extend its range to {\it 0 to 50} volts. Calculate the resistance of this ``range’’ resistor, as well as its necessary power dissipation rating.
The basic problem here is how to make the voltmeter see 10 volts while it’s being connected to a source with a value of 50 volts. This will require a {\it series} resistor to drop the extra 40 volts:
A power dissipation rating of ${1 \over 8}$ watt would be more than sufficient for this application.
Voltmeter ranging is a very practical example of voltage divider circuitry.
Determine the voltages registered by a voltmeter between the following points in this circuit. Be sure to note whether the voltmeter’s indication will be a positive value or a negative value in each case:
{V_A =} {\underline{ 40pt}} (red lead on {\bf A}, black lead on ground)
{V_B =} {\underline{ 40pt}} (red lead on {\bf B}, black lead on ground)
{V_C =} {\underline{ 40pt}} (red lead on {\bf C}, black lead on ground)
{V_D =} {\underline{ 40pt}} (red lead on {\bf D}, black lead on ground)
{V_{AC} =} {\underline{ 40pt}} (red lead on {\bf A}, black lead on {\bf C})
{V_{DB} =} {\underline{ 40pt}} (red lead on {\bf D}, black lead on {\bf B})
{V_{BA} =} {\underline{ 40pt}} (red lead on {\bf B}, black lead on {\bf A})
{V_{BC} =}{\underline{ 40pt}} (red lead on {\bf B}, black lead on {\bf C})
{V_{CD} =} {\underline{ 40pt}} (red lead on {\bf C}, black lead on {\bf D})
{V_A =} {\underline{ +30 volts}} (red lead on {\bf A}, black lead on ground)
{V_B =} {\underline{ +3 volts}} (red lead on {\bf B}, black lead on ground)
{V_C =} {\underline{ +9 volts}} (red lead on {\bf C}, black lead on ground)
{V_D =} {\underline{ +15 volts}} (red lead on {\bf D}, black lead on ground)
{V_{AC} =} {\underline{ +21 volts}} (red lead on {\bf A}, black lead on {\bf C})
{V_{DB} =} {\underline{ -18 volts}} (red lead on {\bf D}, black lead on {\bf B})
{V_{BA} =} {\underline{ -27 volts}} (red lead on {\bf B}, black lead on {\bf A})
{V_{BC} =} {\underline{ -6 volts}} (red lead on {\bf B}, black lead on {\bf C})
{V_{CD} =} {\underline{ +24 volts}} (red lead on {\bf C}, black lead on {\bf D})
Calculate the amount of voltage between test points TP1 and TP3, and also the amount of voltage between test points TP2 and TP4:
$V_{TP1-TP3}$ = 40pt $V_{TP2-TP4}$ =
$V_{TP1-TP3}$ = 15.83 volts 20pt $V_{TP2-TP4}$ = 22.22 volts
Ask your students to explain how they obtained their answers. There is more than one correct way to answer this question!
In a series circuit, certain general principles may be stated with regard to quantities of voltage, current, resistance, and power. Complete these sentences, each one describing a fundamental principle of series circuits:
``In a series circuit, voltage . . .’‘
``In a series circuit, current . . .’‘
``In a series circuit, resistance . . .’‘
``In a series circuit, power . . .’‘
For each of these rules, explain {\it why} it is true.
``In a series circuit, voltage {\it drops add to equal the total}.’‘
This is an expression of Kirchhoff’s Voltage Law (KVL), whereby the algebraic sum of all voltages in any loop must be equal to zero.
``In a series circuit, current {\it is equal through all components}.’‘
This is true because a series circuit by definition has only one path for current to travel. Since charge carriers must move in unison or not at all (a consequence of the Conservation of Charge, whereby electric charges cannot be created or destroyed), the current measured at any one point in a series circuit must be the same as the current measured at any other point in that same circuit, at any given time.
``In a series circuit, resistance{\it s add to equal the total}.’‘
Each resistance in a series circuit acts to oppose electric current. When resistances are connected in series, their oppositions combine to form a greater total opposition because then same current must travel through every resistance.
``In a series circuit, power {\it dissipations add to equal the total}.’‘
This is an expression of the Conservation of Energy, which states energy cannot be created or destroyed. Anywhere power is dissipated in any load of a circuit, that power must be accounted for back at the source, no matter how those loads might be connected to each other.
Rules of series and parallel circuits are very important for students to comprehend. However, a trend I have noticed in many students is the habit of memorizing rather than understanding these rules. Students will work hard to memorize the rules without really comprehending {\it why} the rules are true, and therefore often fail to recall or apply the rules properly.
Use Kirchhoff’s Current Law to calculate the magnitude and direction of the current through resistor R_4 in this resistor network:
It is not necessary to know anything about series-parallel or even parallel circuits in order to solve the R_4’s current—all one needs to know is how to use Kirchhoff’s Current Law.
Explain, step by step, how to calculate the amount of current ($I$) that will go through each resistor in this parallel circuit, and also the current ($I$) supplied by the DC voltage source:
First we need to identify all the relevant principles for series circuits:
{\bullet} The algebraic sum of all currents at a node will be equal to zero (Kirchhoff’s Current Law)
{\bullet} Voltage is common throughout a parallel circuit, because every component shares the same two equipotential points
{\bullet} Resistances diminish in parallel
Following from the rule that voltage is common throughout a parallel circuit, we may conclude that each of the three resistors sees 36 volts from the source. Thus, we may immediately apply Ohm’s Law to calculate current through each of the resistors, knowing the voltage across each resistor and the resistance of each resistor:
$$I_{R1} = {V \over R_1} = {36 \hbox{ V} \over 1500 \> \Omega} = 24 \hbox{ mA}$$
$$I_{R2} = {V \over R_2} = {36 \hbox{ V} \over 10000 \> \Omega} = 3.6 \hbox{ mA}$$
$$I_{R3} = {V \over R_3} = {36 \hbox{ V} \over 4700 \> \Omega} = 7.660 \hbox{ mA}$$
It is helpful to annotate all calculated values on the circuit schematic for easy reference. The reason this is helpful is because it applies a context to the calculated value. Here we will sketch arrows (in the direction of conventional flow) to document all three resistor currents, based on the relationship between voltage and current for {\it loads} (i.e. current enters the positive pole of a load because the load is opposing that current):
\filbreak
From here we may apply KCL to calculate current values at the each node, knowing that every milliamp leaving a node must be matched by a milliamp of current entering the node. Current entering the upper-right node, therefore, will be the sum of the two currents exiting that node. The same thing happens at the lower-right node, where two currents entering that node merge to form a larger current exiting:
$$I = 3.6 \hbox{ mA} + 7.660 \hbox{ mA} = 11.260 \hbox{ mA}$$
Once again we will document this calculated value on the circuit schematic to maintain its context:
Applying KCL to the upper-left and lower-left nodes, and annotating the schematic once again:
$$I = 24 \hbox{ mA} + 11.260 \hbox{ mA} = 35.260 \hbox{ mA}$$
With the arrows showing this 35.260 mA current, we can see it passes straight out of (and back in to) the 36 volt source, which means this is our total current value for the parallel circuit.
Calculate the total amount of current that the battery must supply to this parallel circuit:
Now, using Ohm’s Law, calculate total resistance ($R_{total}$) from total (source) voltage $V_{total}$ and total (source) current $I_{total}$.
$I_{total} = 40.0 \hbox{ mA}$
$R_{total} = 250 \> \Omega$
While some students seem able to immediately grasp the concept of parallel resistances diminishing in (total) value, it is worthwhile to approach it from an Ohm’s Law perspective as well to give other students a more formal rationale for this effect.
Complete the table of values for this circuit:
Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.
Complete the table of values for this circuit:
Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.
The circuit shown here is commonly referred to as a {\it current divider}. Calculate the voltage dropped across each resistor, the current drawn by each resistor, and the total amount of electrical resistance ``seen’’ by the 9-volt battery:
{\bullet} Current through the 2 k$\Omega$ resistor =
{\bullet} Current through the 3 k$\Omega$ resistor =
{\bullet} Current through the 5 k$\Omega$ resistor =
{\bullet} Voltage across each resistor =
{\bullet} $R_{total}$ =
{\bullet} Current through the 2 k$\Omega$ resistor = 4.5 mA
{\bullet} Current through the 3 k$\Omega$ resistor = 3 mA
{\bullet} Current through the 5 k$\Omega$ resistor = 1.8 mA
{\bullet} Voltage across each resistor = 9 volts
{\bullet} $R_{total}$ = 967.74 $\Omega$
Some students may find the diagram hard to follow, and so they will find the task of analysis helped by drawing an equivalent schematic diagram for this circuit, with all terminal points labeled. I recommend you not suggest this solution immediately, but rather challenge your students to think of problem-solving techniques on their own. Surely, someone in the class will have thought of doing this, and the impact of such a suggestion coming from a peer is greater than if it came from you, the instructor.
Be sure to ask your students this question: ``Why is this type of circuit commonly called a {\it current divider}?’‘
Examine these two variable-resistance ({\it rheostat}) networks, each one with a large-range potentiometer and a small-range potentiometer:
For each network, determine which pot is the {\it coarse} adjustment and which pot is the {\it fine} adjustment for total network resistance, and explain your reasoning.
{\bf Series network}
100k = Coarse adjustment ; 5k = Fine adjustment
{\bf Parallel network}
5k = Coarse adjustment ; 100k = Fine adjustment
General principles to keep in mind here are that series resistances {\it add} while parallel resistances {\it diminish}. The total resistance of a series network is always greater than any of its constituent resistances, and so the largest resistance in a series network tends to dominate. The total resistance of a parallel network is always less than any of its constituent resistances, and so the least resistance in a parallel network tends to dominate.
Identify which of these components are connected directly in series with each other, and which are connected directly in parallel with each other:
Assume that the open wire ends are connection points to a power source.
{\bf Figure 1:}
R2 in parallel with R3.
{\bf Figure 2:}
R1 in series with R2.
{\bf Figure 3:}
R2 in series with R3.
{\bf Figure 4:}
R1 in series with R2; R3 in series with R4.
{\bf Figure 5:}
R1 in parallel with R3; R2 in parallel with R4.
{\bf Figure 6:}
R1 in series with R2.
Work with your students to clearly identify rules by which series and parallel connections may be identified. This is extremely important for students to grasp if they are to be successful analyzing series-parallel networks of any kind. The most common problems I encounter as an electronics instructor with reference to series-parallel are invariably related to students’ lack of ability to consistently distinguish series sub-networks and parallel sub-networks in series-parallel combination circuits.
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by Siemens AG