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Electric Circuits: Resistance

Module by: Free High School Science Texts Project. E-mail the author

Resistance

The resistance of a circuit element can be thought of as how much it opposes the flow of electric current in the circuit.

Definition 1: Resistance

The resistance of a conductor is defined as the potential difference across it divided by the current flowing though it. We use the symbol R to show resistance and it is measured in units called Ohms with the symbol ΩΩ.

1 Ohm = 1 Volt Ampere . 1 Ohm = 1 Volt Ampere .
(1)

What causes resistance?

We have spoken about resistors that reduce the flow of charge in a conductor. On a microscopic level, electrons moving through the conductor collide with the particles of which the conductor (metal) is made. When they collide, they transfer kinetic energy. The electrons lose kinetic energy and slow down. This leads to resistance. The transferred energy causes the resistor to heat up. You can feel this directly if you touch a cellphone charger when you are charging a cell phone - the charger gets warm because its circuits have some resistors in them!

All conductors have some resistance. For example, a piece of wire has less resistance than a light bulb, but both have resistance. A lightbulb is a very thin wire surrounded by a glass housing The high resistance of the filament (small wire) in a lightbulb causes the electrons to transfer a lot of their kinetic energy in the form of heat1. The heat energy is enough to cause the filament to glow white-hot which produces light. The wires connecting the lamp to the cell or battery hardly even get warm while conducting the same amount of current. This is because of their much lower resistance due to their larger cross-section (they are thicker).

An important effect of a resistor is that it converts electrical energy into heat energy. Light is a by-product of the heat that is produced.

Note: Interesting Fact :

There is a special type of conductor, called a superconductor that has no resistance, but the materials that make up all known superconductors only start superconducting at very low temperatures (approximately -170C).

Why do batteries go flat?

A battery stores chemical potential energy. When it is connected in a circuit, a chemical reaction takes place inside the battery which converts chemical potential energy to electrical energy which powers the charges (electrons) to move through the circuit. All the circuit elements (such as the conducting leads, resistors and lightbulbs) have some resistance to the flow of charge and convert the electrical energy to heat and, in the case of the lightbulb, heat and light. Since energy is always conserved, the battery goes flat when all its chemical potential energy has been converted into other forms of energy.

Resistors in electric circuits

It is important to understand what effect adding resistors to a circuit has on the total resistance of a circuit and on the current that can flow in the circuit.

Resistors in series

When we add resistors in series to a circuit, we increase the resistance to the flow of current. There is only one path along which the current can flow and the current is the same at all places in the series circuit. Take a look at the diagram below: On the left there is a circuit with a single resistor and a battery. No matter where we measure the current, it is the same in a series circuit. On the right, we have added a second resistor in series to the circuit. The total resistance of the circuit has increased and you can see from the reading on the ammeter that the current in the circuit has decreased and is still the same everywhere in the circuit.

Figure 1
Figure 1 (PG10C9_032.png)

Potential difference and resistors in series

When resistors are in series, one after the other, there is a potential difference across each resistor. The total potential difference across a set of resistors in series is the sum of the potential differences across each of the resistors in the set. This is the same as falling a large distance under gravity or falling that same distance (difference) in many smaller steps. The total distance (difference) is the same.

Look at the circuits below. If we measured the potential difference between the black dots in all of these circuits it would be the same; it is just the potential difference across the battery which is the same as the potential difference across the rest of the circuit. So we now know the total potential difference is the same across one, two or three resistors. We also know that some work is required to make charge flow through each one. Each is a step down in potential energy. These steps add up to the total voltage drop which we know is the difference between the two dots. The sum of the potential differences across each individual resistor is equal to the potential difference measured across all of them together. For this reason, series circuits are sometimes called voltage dividers.

Figure 2
Figure 2 (PG10C9_018.png)

Let us look at this in a bit more detail. In the picture below you can see what the different measurements for 3 identical resistors in series could look like. The total voltage across all three resistors is the sum of the voltages across the individual resistors.

Figure 3
Figure 3 (PG10C9_019.png)

Equivalent Series Resistance

When there is more than one resistor in a circuit, we are usually able to calculate the total combined resitance of all the resistors. The resistance of the single resistor is known as equivalent resistance or total resistance. Consider a circuit consisting of three resistors and a single cell connected in series.

Figure 4
Figure 4 (PG11C9_007.png)

We can define the total resistance in a series circuit as:

Definition 2: Equivalent resistance in a series circuit, RsRs

For nn resistors in series the equivalent resistance is:

R s = R 1 + R 2 + R 3 + + R n R s = R 1 + R 2 + R 3 + + R n
(2)

The more resistors we add in series, the higher the equivalent resistance in the circuit. Since the resistors act as obstacles to the flow of charge through the circuit, the current in the circuit is reduced. Therefore, the higher the resistance in the circuit, the lower the current through the battery and the circuit. We say that the current in the battery is inversely proportional to the resistance in the circuit. Let us apply the rule of equivalent resistance in a series circuit to the following circuit.

Figure 5
Figure 5 (PG11C9_008.png)

The resistors are in series, therefore:

R s = R 1 + R 2 + R 3 = 3 Ω + 10 Ω + 5 Ω = 18 Ω R s = R 1 + R 2 + R 3 = 3 Ω + 10 Ω + 5 Ω = 18 Ω
(3)
Experiment : Current in Series Circuits
Aim:

To determine the effect of multiple resistors on current in a circuit

Apparatus:

  • Battery
  • Resistors
  • Light bulb
  • Wires

Method:

  1. Construct the following circuits
    Figure 6
    Figure 6 (PG10C9_027.png)
  2. Rank the three circuits in terms of the brightness of the bulb.

Conclusions:

The brightness of the bulb is an indicator of how much current is flowing. If the bulb gets brighter because of a change then more current is flowing. If the bulb gets dimmer less current is flowing. You will find that the more resistors you have the dimmer the bulb.

Exercise 1: Equivalent series resistance I

Two 10 kΩΩ resistors are connected in series. Calculate the equivalent resistance.

Solution
  1. Step 1. Determine how to approach the problem :

    Since the resistors are in series we can use:

    R s = R 1 + R 2 R s = R 1 + R 2
    (4)
  2. Step 2. Solve the problem :
    R s = R 1 + R 2 = 10 k Ω + 10 k Ω = 20 k Ω R s = R 1 + R 2 = 10 k Ω + 10 k Ω = 20 k Ω
    (5)
  3. Step 3. Write the final answer :

    The equivalent resistance of two 10 kΩΩ resistors connected in series is 20 kΩΩ.

Exercise 2: Equivalent series resistance II

Two resistors are connected in series. The equivalent resistance is 100 ΩΩ. If one resistor is 10 ΩΩ, calculate the value of the second resistor.

Solution
  1. Step 1. Determine how to approach the problem :

    Since the resistors are in series we can use:

    R s = R 1 + R 2 R s = R 1 + R 2
    (6)

    We are given the value of RsRs and R1R1.

  2. Step 2. Solve the problem :
    R s = R 1 + R 2 R 2 = R s - R 1 = 100 Ω - 10 Ω = 90 Ω R s = R 1 + R 2 R 2 = R s - R 1 = 100 Ω - 10 Ω = 90 Ω
    (7)
  3. Step 3. Write the final answer :

    The second resistor has a resistance of 90 ΩΩ.

Figure 7
Khan academy video on circuits - 2

Resistors in parallel

In contrast to the series case, when we add resistors in parallel, we create more paths along which current can flow. By doing this we decrease the total resistance of the circuit!

Take a look at the diagram below. On the left we have the same circuit as shown on the left in Figure 1 with a battery and a resistor. The ammeter shows a current of 1 ampere. On the right we have added a second resistor in parallel to the first resistor. This has increased the number of paths (branches) the charge can take through the circuit - the total resistance has decreased. You can see that the current in the circuit has increased. Also notice that the current in the different branches can be different (in this case 1 A and 2 A) but must add up to the current through the battery (3 A). Since the total current in the circuit is equal to the sum of the currents in the parallel branches, a parallel circuit is sometimes called a current divider.

Figure 8
Figure 8 (PG10C9_033.png)

Potential difference and parallel resistors

When resistors are connected in parallel the start and end points for all the resistors are the same. These points have the same potential energy and so the potential difference between them is the same no matter what is put in between them. You can have one, two or many resistors between the two points, the potential difference will not change. You can ignore whatever components are between two points in a circuit when calculating the difference between the two points.

Look at the following circuit diagrams. The battery is the same in all cases. All that changes is that more resistors are added between the points marked by the black dots. If we were to measure the potential difference between the two dots in these circuits we would get the same answer for all three cases.

Figure 9
Figure 9 (PG10C9_016.png)

Let's look at two resistors in parallel more closely. When you construct a circuit you use wires and you might think that measuring the voltage in different places on the wires will make a difference. This is not true. The potential difference or voltage measurement will only be different if you measure a different set of components. All points on the wires that have no circuit components between them will give you the same measurements.

All three of the measurements shown in the picture below (i.e. A–B, C–D and E–F) will give you the same voltage. The different measurement points on the left (i.e. A, E, C) have no components between them so there is no change in potential energy. Exactly the same applies to the different points on the right (i.e. B, F, D). When you measure the potential difference between the points on the left and right you will get the same answer.

Figure 10
Figure 10 (PG10C9_017.png)

Definition 3: Equivalent resistance of two parallel resistor, RpRp

For 22 resistors in parallel with resistances R1R1 and R2R2, the equivalent resistance is:

Rp=R1R2R1+R2RpR1R2R1R2
(8)

Equivalent parallel resistance

Consider a circuit consisting of a single cell and three resistors that are connected in parallel.

Figure 11
Figure 11 (PG11C9_009.png)

Using what we know about voltage and current in parallel circuits we can define the equivalent resistance of several resistors in parallel as:

Definition 4: Equivalent resistance in a parallel circuit, RpRp

For nn resistors in parallel, the equivalent resistance is:

1 R p = 1 R 1 + 1 R 2 + 1 R 3 + + 1 R n 1 R p = 1 R 1 + 1 R 2 + 1 R 3 + + 1 R n
(9)

Let us apply this formula to the following circuit.

Figure 12
Figure 12 (PG11C9_010.png)

What is the total resistance in the circuit?

1 R p = 1 R 1 + 1 R 2 + 1 R 3 = 1 10 Ω + 1 2 Ω + 1 1 Ω = 1 + 5 + 10 10 = 16 10 R p = 0 , 625 Ω 1 R p = 1 R 1 + 1 R 2 + 1 R 3 = 1 10 Ω + 1 2 Ω + 1 1 Ω = 1 + 5 + 10 10 = 16 10 R p = 0 , 625 Ω
(10)
Experiment : Current in Parallel Circuits
Aim:

To determine the effect of multiple resistors on current in a circuit

Apparatus:

  • Battery
  • Resistors
  • Light bulb
  • Wires

Method:

  1. Construct the following circuits
    Figure 13
    Figure 13 (PG10C9_030.png)
  2. Rank the three circuits in terms of the brightness of the bulb.

Conclusions:

The brightness of the bulb is an indicator of how much current is flowing. If the bulb gets brighter because of a change then more current is flowing. If the bulb gets dimmer less current is flowing. You will find that the more resistors you have the brighter the bulb.

Why is this the case? Why do more resistors make it easier for charge to flow in the circuit? It is because they are in parallel so there are more paths for charge to take to move. You can think of it like a highway with more lanes, or the tube of marbles splitting into multiple parallel tubes. The more branches there are, the easier it is for charge to flow. You will learn more about the total resistance of parallel resistors later but always remember that more resistors in parallel mean more pathways. In series the pathways come one after the other so it does not make it easier for charge to flow.

Exercise 3

Two 8kΩ8kΩ resistors are connected in parallel. Calculate the equivalent resistance.

Solution
  1. Step 1. Determine how to approach the problem :

    Since the resistors are in parallel we can use:

    1 R p =1 R 1 + 1 R 2 1 R p =1 R 1 + 1 R 2
    (11)
  2. Step 2. Solve the problem :
    1 R p = 1 R 1 + 1 R 2 = 1 8 k Ω +1 10 k Ω Rp = 28 = 4 k Ω 1 R p = 1 R 1 + 1 R 2 = 1 8 k Ω +1 10 k Ω Rp = 28 = 4 k Ω
    (12)
  3. Step 3. Write the final answer :

    The equivalent resistance of two 8kΩ8kΩ resistors connected in parallel is 4kΩ4kΩ.

Exercise 4

Two resistors are connected in parallel. The equivalent resistance is 100 Ω100Ω. If one resistor is 150Ω150Ω, calculate the value of the second resistor.

Solution
  1. Step 1. Determine how to approach the problem :

    Since the resistors are in parallel we can use:

    1 R p = 1 R 1 + 1 R 2 1 R p = 1 R 1 + 1 R 2
    (13)

    We are given the value of RpRp and R1R1.

  2. Step 2. Solve the problem :
    1 R p = 1 R 1 + 1 R 2 1 R 2 = 1 R p - 1 R 1 = 1 100 Ω - 1 150 Ω = 3-2 300 = 1 300 R2 = 300 Ω 1 R p = 1 R 1 + 1 R 2 1 R 2 = 1 R p - 1 R 1 = 1 100 Ω - 1 150 Ω = 3-2 300 = 1 300 R2 = 300 Ω
    (14)
  3. Step 3. Write the final answer :

    The second resistor has a resistance of 300Ω300Ω.

Figure 14
Khan academy video on circuits - 3
Resistance
  1. What is the unit of resistance called and what is its symbol? Click here for the solution
  2. Explain what happens to the total resistance of a circuit when resistors are added in series? Click here for the solution
  3. Explain what happens to the total resistance of a circuit when resistors are added in parallel? Click here for the solution
  4. Why do batteries go flat? Click here for the solution

Figure 15
Khan academy video on circuits - 4

The following presentation summarizes the concepts covered in this chapter.

Figure 16

Exercises - Electric circuits

  1. Write definitions for each of the following:
    1. resistor
    2. coulomb
    3. voltmeter
    Click here for the solution
  2. Draw a circuit diagram which consists of the following components:
    1. 2 batteries in parallel
    2. an open switch
    3. 2 resistors in parallel
    4. an ammeter measuring total current
    5. a voltmeter measuring potential difference across one of the parallel resistors
    Click here for the solution
  3. Complete the table below:
    Table 1
    QuantitySymbolUnit of meaurementSymbol of unit
    e.g. Distancee.g. de.g. kilometere.g. km
    Resistance   
    Current   
    Potential difference   
    Click here for the solution
  4. Draw a diagram of a circuit which contains a battery connected to a lightbulb and a resistor all in series.
    1. Also include in the diagram where you would place an ammeter if you wanted to measure the current through the lightbulb.
    2. Draw where and how you would place a voltmeter in the circuit to measure the potential difference across the resistor.
    Click here for the solution
  5. Thandi wants to measure the current through the resistor in the circuit shown below and sets up the circuit as shown below. What is wrong with her circuit setup?
    Figure 17
    Figure 17 (circuit1.png)
    Click here for the solution
  6. [SC 2003/11] The emf of a battery can best be explained as the
    1. rate of energy delivered per unit current
    2. rate at which charge is delivered
    3. rate at which energy is delivered
    4. charge per unit of energy delivered by the battery
    Click here for the solution
  7. [IEB 2002/11 HG1] Which of the following is the correct definition of the emf of a battery?
    1. It is the product of current and the external resistance of the circuit.
    2. It is a measure of the cell's ability to conduct an electric current.
    3. It is equal to the “lost volts” in the internal resistance of the circuit.
    4. It is the power supplied by the battery per unit current passing through the battery.
    Click here for the solution
  8. [IEB 2005/11 HG] Three identical light bulbs A, B and C are connected in an electric circuit as shown in the diagram below.
    Figure 18
    Figure 18 (PG10C9_037.png)
    1. How bright is bulb A compared to B and C?
    2. How bright are the bulbs after switch S has been opened?
    3. How do the currents in bulbs A and B change when switch S is opened?
      Table 2
       Current in ACurrent in B
      (a)decreasesincreases
      (b)decreasesdecreases
      (c)increasesincreases
      (d)increasesdecreases
    Click here for the solution
  9. [IEB 2004/11 HG1] When a current II is maintained in a conductor for a time of tt, how many electrons with charge e pass any cross-section of the conductor per second?
    1. It
    2. It/e
    3. Ite
    4. e/It
    Click here for the solution

Footnotes

  1. Flourescent lightbulbs do not use thin wires; they use the fact that certain gases glow when a current flows through them. They are much more efficient (much less resistance) than lightbulbs.

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