Series DC Circuits
Basic Electricity
The brightness of a light bulb  or the power dissipated by any electrical load, for that matter  may be varied by inserting a variable resistance in the circuit, like this:

This method of electrical power control is not without its disadvantages, though. Consider an example where the circuit current is 5 amps, the variable resistance is 2 Ω, and the lamp drops 20 volts of voltage across its terminals. Calculate the power dissipated by the lamp, the power dissipated by the variable resistance, and the total power provided by the voltage source. Then, explain why this method of power control is not ideal.
A modern method of electrical power control involves inserting a fastoperating switch inline with an electrical load, to switch power on and off to it very rapidly over time. Usually, a solidstate device such as a transistor is used:

This circuit has been greatly simplified from that of a real, pulsecontrol power circuit. Just the transistor is shown (and not the “pulse” circuit which is needed to command it to turn on and off) for simplicity. All you need to be aware of is the fact that the transistor operates like a simple, singlepole singlethrow (SPST) switch, except that it is controlled by an electrical current rather than by a mechanical force, and that it is able to switch on and off millions of times per second without wear or fatigue.
If the transistor is pulsed on and off fast enough, power to the light bulb may be varied as smoothly as if controlled by a variable resistor. However, there is very little energy wasted when using a fastswitching transistor to control electrical power, unlike when a variable resistance is used for the same task. This mode of electrical power control is commonly referred to as PulseWidth Modulation, or PWM.
Explain why PWM power control is much more efficient than controlling load power by using a series resistance.
Don’t just sit there! Build something!! 
Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.
You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuitbuilding exercises, follow these steps:
 Carefully measure and record all component values prior to circuit construction.
 Draw the schematic diagram for the circuit to be analyzed.
 Carefully build this circuit on a breadboard or other convenient medium.
 Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements onebyone on the diagram.
 Mathematically analyze the circuit, solving for all values of voltage, current, etc.
 Carefully measure those quantities, to verify the accuracy of your analysis.
 If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction against the diagram, then carefully recalculate the values and remeasure.
Avoid very high and very low resistor values, to avoid measurement errors caused by meter “loading”. I recommend resistors between 1 kΩ and 100 kΩ, unless, of course, the purpose of the circuit is to illustrate the effects of meter loading!
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another timesaving technique is to reuse the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.
The formula for calculating total resistance of three seriesconnected resistors is as follows:

Algebraically manipulate this equation to solve for one of the series resistances (R_{1}) in terms of the other two series resistances (R_{2} and R_{3}) and the total resistance (R). In other words, write a formula that solves for R_{1} in terms of all the other variables.
A technician wants to energize a 24 volt motor, but lacks a 24 volt battery to do it with. Instead, she has access to several “power supply” units which convert 120 volt AC power from a power receptacle into lowvoltage DC power that is adjustable over a range of 0 to 15 volts. Each of these power supplies is a box with a power cord, voltage adjustment knob, and two output terminals for connection with the DC voltage it produces:

Draw a picture of how this technician might use power supplies to energize the 24 volt motor.
Suppose I connect two resistors in series with one another, like this:

How much electrical resistance would you expect an ohmmeter to indicate if it were connected across the combination of these two seriesconnected resistors?

Explain the reasoning behind your answer, and try to formulate a generalization for all combinations of series resistances.
Lightemitting diodes, or LEDs, are rugged and highly efficient sources of light. They are far more rugged and efficient than incandescent lamps, and they also have the ability to switch on and off much faster because there is no filament inside needing to heat or cool:

LEDs are low voltage devices, typically rated in the range of 1.5 to 2 volts DC maximum. Single diodes generally draw low currents as well, about 20 milliamps each. The problem is, how do you operate an LED from a typical electronic power source, which may output 24 volts DC or more?

The answer is to use a series dropping resistor:

Calculate the necessary resistance value and minimum power rating of a series dropping resistor for an LED rated at 1.7 volts and 20 mA, and a power supply voltage of 24 volts.
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 twowire cable, which is then connected to the same 110 volt source. Assuming that each conductor within the cable has an endtoend resistance of 3 ohms, how much power will the heater dissipate?

The circuit shown here is commonly referred to as a voltage divider. Calculate the voltage dropped across the following pairs of terminals, the current through each resistor, and the total amount of electrical resistance ßeen” by the 9volt battery:

 • Voltage between terminals 2 and 3 =
 • Voltage between terminals 4 and 5 =
 • Voltage between terminals 6 and 7 =
 • Voltage between terminals 6 and 8 =
 • Voltage between terminals 4 and 8 =
 • Voltage between terminals 2 and 8 =
 • Current through each resistor =
 • R_{total} =
Can you think of any practical applications for a circuit such as this?
What will happen in this circuit as the switches are sequentially turned on, starting with switch number 1 and ending with switch number 3?

Describe how the successive closure of these three switches will impact:
 • The total amount of circuit resistance “seen” by the battery
 • The total amount of current drawn from the battery
 • The current through each resistor
 • The voltage drop across each resistor
Also, provide a safetyrelated reason for the existence of the fourth resistor in this circuit, on the lefthand side of the circuit (not bypassed by any switch).
In a series circuit, certain general rules may be stated with regard to quantities of voltage, current, resistance, and power. Express these rules, using your own words:
“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 why it is true.
Predict how all test point voltages (measured between each test point and ground) in this circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

 • Resistor R_{1} fails open:
 • Resistor R_{2} fails open:
 • Resistor R_{3} fails open:
 • Solder bridge (short) past resistor R_{2}:
For each of these conditions, explain why the resulting effects will occur.
A student is troubleshooting a tworesistor voltage divider circuit, using a table to keep track of his test measurements and conclusions. The table lists all components and wires in the circuit so that the student may document their known status with each successive measurement:

Measurement taken Battery Wire +/1 R_{1} Wire 2/3 R_{2} Wire 4/
Prior to beginning troubleshooting, the student is told there is no voltage across R_{2}. Thus, the very first entry into the table looks like this:
Measurement taken Battery Wire +/1 R_{1} Wire 2/3 R_{2} Wire 4/
V_{R2} = 0 V
Based on this data, the student then determines possible faults which could cause this to happen, marking each possibility in the table using letters as symbols. The assumption here is that there is only one fault in the circuit, and that it is either a complete break (open) or a direct short:
Measurement taken Battery Wire +/1 R_{1} Wire 2/3 R_{2} Wire 4/
V_{R2} = 0 V O O O O S O
“O” symbolizes a possible “open” fault, while “S” symbolizes a possible “shorted” fault.
Next, the student measures between terminals 1 and 4, obtaining a full 6 volt reading. This is documented on the table as well, along with some updated conclusions regarding the status of all wires and components:
Measurement taken Battery Wire +/1 R_{1} Wire 2/3 R_{2} Wire 4/
V_{R2} = 0 V O O O O S O
V_{1−4} = 6 V OK OK O O S OK
After this, the student measures between terminals 1 and 2 (across resistor R_{1}), and gets a reading of 0 volts. Complete the table based on this last piece of data:
Measurement taken Battery Wire +/1 R_{1} Wire 2/3 R_{2} Wire 4/
V_{R2} = 0 V O O O O S O
V_{1−4} = 6 V OK OK O O S OK
V_{R1} = 0 V
This voltage divider circuit has a problem: there is no voltage output between terminals 7 and 8.

A technician has taken several measurements with a voltmeter, documenting them chronologically from top to bottom in the farleft column:
Measurement Batt (+)/1 R_{1} 2/3 R_{2} 4/5 R_{3} 6/7 R_{4} 8/()
V_{1−8} = 6 V
V_{1−5} = 0 V
V_{5−7} = 6 V
Fill in all cells of this table with one of three different symbols, representing the status of each component or wire (numbers separated by a slash indicate the wire connecting those terminals):
 • O for an “open” fault
 • S for an “short” fault
 • OK for no fault
You are to assume that there is only one fault in this circuit, and that it is either a complete break (open) or a direct short (zero resistance). After completing the table, assess whether or not the exact fault may be known from the data recorded thus far. If not, suggest the next logical voltage measurement to take.
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