# Electric Circuit Analysis/Resistors in Series

## Lesson 2 : Review

What you need to remember from Simple resistive Circuits. If you ever feel lost, do not be shy to go back to the previous lesson & go through it again. You can learn by repetition.

• Voltage: (V or v - Volts)The electrical potential between two points in a circuit.
• Current: (I or i - Amperes)The amount of charge flowing through a part of a circuit.
• Power: (W - Watts)Simply P = IV. It is the current times the voltage.
• Source: A voltage or current source is the supplier for the circuit.
• Resistor: (R measured in Ω - Ohms)A circuit element that "constricts" current flow.

## Lesson 3: Preview

This Lesson is about Resistors in Series. The student/User is expected to understand the following at the end of the lesson:

• Total Series Resistance: ($R_{E}=R_{1}+R_{2}\!$ )
The total resistance of resistors in series is the sum of all resitors in series.

## Part 1

### Resistors in Series

A Series of resistors means resistors connected end to end in a line. This means that the resistance for the circuit is different from any one resistor. Take two resistors in series in a circuit with a voltage supply.

To find the overall resistance of the circuit, add up the individual resistances of the each of the resistors. i.e. ($R=R1+R2\!$ )

## Part 2

Equation 3.1

$R_{E}=R_{1}+R_{2}\!$ So what if there were 10 resistors in series? Just add up all the resistances and you have the equvalent overall resistance. In general this can be expressed as:

Equation 3.2

$R_{E}=\sum _{k=1}^{n}R_{k}\!$ Where R equivalent is the sum of all n resistors in series. So it really doesn't matter how many resistors there are. If resistors are set up in series they can be added up into an equivalent resistance.

## Part 3

### Voltage Divider

There comes a time when the boss or the project demands that you know what the voltage is between these millions of resistors in series. No need to panic though because it isn't too much harder.

Lets take the two-resistor problem first. There is a voltage source with two resistors in series. We know that the overall voltage drop across the two resistors is the same as the voltage the source is supplying in our example world. So the voltage drop across one resistor would be a portion of the overall drop. What proportion would we use to figure out the answer? One resistor over the two added together times the overall voltage drop:

Equation 3.3

$V_{1}={\frac {R_{1}}{R_{1}+R_{2}}}*V_{s}$ Remember, this is the voltage drop across the first resistor. If you want the actual voltage there you still need to do some adding or subtracting to get it. Say that you have a 12V source and a drop over the first resistor of 3V. Then you actually need to subtract 3V from 12V to get the actual voltage between the resistors.

At this point it seems that everything isn't quite as simple as it started. With our example and equation for two resistors in series something else can happen. What if the second resistor was set in the first resistor's place in the equation? Well, we would simply get the other side of the proportion:

## Part 4

Equation 3.4

$V_{1}={\frac {R_{2}}{R_{1}+R_{2}}}*V_{s}$ This is the drop over the second resistor. But if it is dropping to zero, ground, or the negative side of the source then adding it to zero would give us the same answer as above.

For more than two resistors in series it is just a matter of keeping track of which resistor is on which side and summing appropriately.

Equation 3.5

$V_{n}={\frac {\sum _{j=1}^{n}R_{j}}{\sum _{k=1}^{m}R_{k}}}*V_{s}$ Where $V_{n}$ is the voltage drop over n resistors out of a total of m resistors. Remember that the resistors where the voltage drop is being calculated should be continuous. If they aren't all that can be said about the answer derived from the equation is that it is part of the whole voltage drop and somewhat worthless otherwise.

If the resistors are in the middle of the series then it will be necessary to calculate the voltage drop on one of the sides in order to calculate the voltage.

## Part 5

It becomes clear, then, that two equal resistors will divide the source voltage into two equal voltages (half of the source's voltage is dropped across each resistor). If the ratio of the resistance values is 3 to 1, there will be 3/4 of the source voltage dropped across the higher resistance, and ${\frac {1}{4}}$ of the source voltage dropped across the lower resistance.

Three equal resistances in a series circuit with a single voltage source would drop 1/3 of the source voltage across each resistor. If the three had 1-2-3 proportionality (100,200,300 ohms for instance) they would drop ${\frac {1}{6}},{\frac {1}{3}}$ , and ${\frac {1}{2}}$ of the source voltage each. That is:

${\frac {1}{(1+2+3)}}$ × VTotal, ${\frac {2}{(1+2+3)}}$ × VTotal,and ${\frac {3}{(1+2+3)}}$ × VTotal.

## Part 6

### Current

Where does current come into any of this? Current, in this case, plays a similar role to that of the current in the Simple Resistive Circuits. Once the equivalent resistance of all the resistors in a series is found, effectively making a simple circuit again, then the current can be found with:

Equation 3.6

$V=I*R\!$ Just as a reminder. But the interesting thing is that the current through all resistors in series is the same. If the resistor is 30Ω it has the same current flow as the resistor with 500Ω, so long as it is in series. Thinking about everything above we are adding up all of the resistors to make a single equivalent resistor. So current isn't different from 30Ω to 500Ω because together the resistance is 530Ω. That resistance is used then to calculate the current.

## Part 7

### More Examples

Figure 3.1 shows a Series resistive circuit with the following parameters. Vs=100Volts ; R1=15; R2=30; Find V1 and V2.

Solution: from Equation 3.3 we see that.

${\begin{matrix}\ V1&=&{\frac {(Vs\times R1)}{(R1+R2)}}\\\ \\\ &=&{\frac {(100V\times 15\Omega )}{(15\Omega +30\Omega )}}\\\ \\\ &=&{\frac {1500}{45}}V\\\ \\\ &=&33.33V\\\ \\\ V1&=&{\frac {1}{3}}Vs\end{matrix}}$ .

Similarily:

${\begin{matrix}\ V2&=&{\frac {(Vs\times R2)}{(R1+R2)}}\\\ \\\ &=&{\frac {(100V\times 30\Omega )}{(15\Omega +30\Omega )}}\\\ \\\ &=&{\frac {3000}{45}}V\\\ \\\ &=&66.66V\\\ \\\ V2&=&{\frac {2}{3}}Vs\end{matrix}}$ .

Thus it can be said that The Supply Voltage has been divided between R1 and R2 by ${\frac {1}{3}}$ and ${\frac {2}{3}}$ respectively.

### Related Topic(s) in Wikiversity

Please visit the following page to supplement material covered in this lesson.

## Part 8: Exercise 3

Here are some questions to test yourself with.

1. Given 2 Resistors: R1 = R2 = 5$\Omega$ and Supply Volatage is 20 Volts find The V1; V2 and Current drawn by the Resistors.
2. Given 3 Resistors: R1 = 2$\Omega$ ; R2 = 3$\Omega$ and R3 = 7$\Omega$ and Supply Voltage is 15 Volts. Find V1; V2 & V3 and Current drawn by these Resistors.
3. Given 3 Resistors: R1 = 2$\Omega$ ; R2 = 3$\Omega$ and R3 = 7$\Omega$ and 3 Batteries with negligible internal resistances connected in series as follows: Vs1 = 3V ; Vs2 = 5V and Vs3 = 1.5V, Find V1; V2 & V3 dropped by individual Resistors.

## Completion list

Once you finish your Exercises you can post your score here! To post your score just e-mail your course co-ordinator, your name, and score Click Here .

1. Ozzimotosan -- 100%
2. Doldham -- 100% & Corrected
3. Sonu rockin -- 100% and corrected