Voltage Divider Circuits
AC Electric Circuits
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.
We know that the current in a series circuit may be calculated with this formula:

We also know that the voltage dropped across any single resistor in a series circuit may be calculated with this formula:

Combine these two formulae into one, in such a way that the I variable is eliminated, leaving only E_{R} expressed in terms of E_{total}, R_{total}, and R.
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% error):

Also, determine the following information about this circuit:
 Current through each resistor
 Power dissipated by each resistor
 Ratio of each resistor’s voltage drop to battery voltage ([(E_{R})/(E_{bat})])
 Ratio of each resistor’s resistance to the total circuit resistance ([R/(R_{total})])
Calculate the voltage dropped by each of these resistors, given a battery voltage of 9 volts. The resistor color codes are as follows (assume 0% error on all resistor values):
 R_{1} = Brn, Grn, Red, Gld
 R_{2} = Yel, Vio, Org, Gld
 R_{3} = Red, Grn, Red, Gld
 R_{4} = Wht, Blk, Red, Gld

Now, recalculate all resistor voltage drops for a scenario where the total voltage is different:

Many manufacturing processes are electrochemical in nature, meaning that electricity is used to promote or force chemical reactions to occur. One such industry is aluminum smelting, where large amounts of DC current (typically several hundred thousand amperes!) is used to turn alumina (Al_{2}O_{3}) into pure metallic aluminum (Al):

The alumina/electrolyte mixture is a molten bath of chemicals, lighter than pure aluminum itself. Molecules of pure aluminum precipitate out of this mix and settle at the bottom of the “pot” where the molten metal is periodically pumped out for further refining and processing. Fresh alumina powder is periodically dropped into the top of the pot to replenish what is converted into aluminum metal.
Although the amount of current necessary to smelt aluminum in this manner is huge, the voltage drop across each pot is only about 4 volts. In order to keep the voltage and current levels reasonable in a large smelting facility, many of these pots are connected in series, where they act somewhat like resistors (being energy loads rather than energy sources):

A typical “potline” might have 240 pots connected in series. With a voltage drop of just over 4 volts apiece, the total voltage powering this huge series circuit averages around 1000 volts DC:

With this level of voltage in use, electrical safety is a serious consideration! To ensure the safety of personnel if they must perform work around a pot, the system is equipped with a “movable ground,” consisting of a large switch on wheels that may be connected to the steel frame of the shelter (with concrete pilings penetrating deep into the soil) and to the desired pot. Assuming a voltage drop of exactly 4.2 volts across each pot, note what effect the ground’s position has on the voltages around the circuit measured with respect to ground:

Determine the voltages (with respect to earth ground) for each of the points (dots) in the following schematic diagram, for the ground location shown:

Draw an equivalent schematic diagram for this circuit, then calculate the voltage dropped by each of these resistors, given a battery voltage of 9 volts. The resistor color codes are as follows (assume 0% error on all resistor values):
 R_{1} = Brn, Grn, Red, Gld
 R_{2} = Yel, Vio, Org, Gld
 R_{3} = Red, Grn, Red, Gld
 R_{4} = Wht, Blk, Red, Gld
 R_{5} = Brn, Blk, Org, Gld

Compare the voltage dropped across R1, R2, R3, and R4, with and without R5 in the circuit. What general conclusions may be drawn from these voltage figures?
The formula for calculating voltage across a resistor in a series circuit is as follows:

In a simpleseries circuit with one voltage source and three resistors, we may rewrite this formula to be more specific:

Suppose we have such a series circuit with a source voltage of 15 volts, and resistor values of R_{1} = 1 kΩ and R_{2} = 8.1 kΩ. Algebraically manipulate this formula to solve for R_{3} in terms of all the other variables, then determine the necessary resistance value of R_{3} to obtain a 0.2 volt drop across resistor R_{1}.
As adjustable devices, potentiometers may be set at a number of different positions. It is often helpful to express the position of a potentiometer’s wiper as a fraction of full travel: a number between 0 and 1, inclusive. Here are several pictorial examples of this, with the variable m designating this travel value (the choice of which alphabetical character to use for this variable is arbitrary):

Using an algebraic variable to represent potentiometer position allows us to write equations describing the outputs of voltage divider circuits employing potentiometers. Note the following examples:

Algebraically manipulate these four equations so as to solve for m in each case. This will yield equations telling you where to set each potentiometer to obtain a desired output voltage given the input voltage and all resistance values (m = …).
When the 5 kΩ potentiometer in this circuit is set to its 0%, 25%, 50%, 75%, and 100% positions, the following output voltages are obtained (measured with respect to ground, of course):

 At 0% setting, V_{out} = 0 V
 At 25% setting, V_{out} = 2.5 V
 At 50% setting, V_{out} = 5 V
 At 75% setting, V_{out} = 7.5 V
 At 100% setting, V_{out} = 10 V
Calculate what the output voltages will be if a 1 kΩ load resistor is connected between the “V_{out}” terminal and ground:

 At 0% setting, V_{out} =
 At 25% setting, V_{out} =
 At 50% setting, V_{out} =
 At 75% setting, V_{out} =
 At 100% setting, V_{out} =
Calculate both the total resistance of this voltage divider circuit (as ßeen” from the perspective of the 25 volt source) and its output voltage (as measured from the V_{out} terminal to ground):

Note that the two 5 kΩ potentiometers are set to their 80% positions (m = 0.8), while the 100 kΩ potentiometer is set exactly to midposition (50%, or m = 0.5).
Calculate both the total resistance of this voltage divider circuit (as ßeen” from the perspective of the 25 volt source) and its output voltage (as measured from the V_{out} terminal to ground):

Note that the upper 5 kΩ potentiometer is set to its 20% position (m = 0.2), while the lower 5 kΩ potentiometer is set to its 90% position (m = 0.9), and the 100 kΩ potentiometer is set to its 40% position (m = 0.4).
A student builds the following voltage divider circuit so she can power a 6volt lamp from a 15volt power supply:

When built, the circuit works just as it should. However, after operating successfully for hours, the lamp suddenly goes dark. Identify all the possible faults you can think of in this circuit which could account for the lamp not glowing anymore.
Old vacuumtube 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, highvoltage 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 bleeder current, designed to discharge the high voltage output of the power supply quickly when it was turned off.
Design a highvoltage 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):

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