The role of resistors in electrical circuits

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The role of the resistor in an electrical circuit: NOTE: Diagrams, links and practice questions in this document are still to be added: In direct current (DC) or alternating current (AC) electrical circuits a resistor, as its name implies, resists the flow of electrons. It is one of the most basic of electrical components. It can be used to reduce the available voltage or current present in a circuit. While there are differences in how a resistor affects the two different types of current sources (DC or AC), depending on how the resistor is constructed and the AC frequency involved, it can be assumed that the following applies equally to both. For AC circuits it may be necessary to indicate how the voltage is represented, the average value, the peak value or the root mean square (RMS). If no indication is made of the type of AC voltage it is generally assumed to be the (RMS) value

As discussed in the < ohm's law section link>, in an electrical circuit, voltage (measured in volts and denoted by the letter V) equals current (measured in amps and denoted by the letter I) multiplied by the resistance (measured in ohms and denoted by the letter R) present in the circuit. This is represented by the following formula.

  V = IR  or  E = IR

(Voltage is sometimes denoted by the letter “E”, which stands for electro-motive force)

An electrical circuit may include many resistors. The way these resistors impact the circuit depends on the way they’re arranged in the circuit. Resistors may be arranged in series or in parallel with the voltage supply source. Refer to the example below.

File:Wiki circuit diagram.jpeg

Figure 1 represents an electrical circuit with two resistors in a series arrangement. For current to complete the electrical circuit it must flow from the voltage source (B1) and pass through both resistor 1 (R1) and resistor 2 (R2) and then back to B1.

The total resistance in the circuit is simply the sum of the two resistor values (measured in ohms, denoted by the Greek letter Ω). Therefore, in figure 1, the total circuit resistance (RT) is R1 + R2, which equals 100Ω.

Figure 2 represents an electrical circuit with two resistors in a parallel arrangement. For current to complete the electrical circuit it must flow from the voltage source (B1) and then the current has two available paths to get back to B1. Some of the current will pass through resistor 1 (R1) back to B1 and some will pass through resistor 2 (R2) and then back to B1.

The total resistance in a parallel circuit is not as simple as in a series circuit. The total resistance in the circuit for figure 2 is the reciprocal of the sum of the reciprocal sum of the two resistor values (measured in ohms, denoted by the Greek letter Ω). Therefore, in figure 2, the total circuit resistance (RT) is 1/(1/ R1 + 1/ R2), which equals 25Ω.

It is important to note the impact to the circuit that the arrangement of the resistors have. Using Ohm’s law we can determine the total current flowing in each of the two circuits is considerably different even though the same components were used for both.

By applying a little algebra to the Ohm’s law equation we can determine the total current for each circuit.

For the circuit in figure 1, the total circuit current flow is expressed by the equation: I=V/R. By substituting in the numbers we know, where V = 10 volts and R=100Ω, we get a total current flow in the circuit equal to 10/100, which equals 0.1 amps.

For the circuit in figure 2, the total circuit current flow is again expressed by the equation: I=V/R. By substituting in the numbers we know, where V = 10 volts and R=25Ω, we get a total current flow in the circuit equal to 10/25 which equals 0.4 amps


Resistors in series:

When used in series, resistors can be said to be a “voltage dividing network.” This is because in a series circuit, current flowing through each resistor is the same value but the voltage present across each resistor is only part of the total circuit voltage value. Looking again at the circuit from figure 1 we can derive the voltage present across each resistor.

<Diagram>

Based on the fact that in a series circuit the current flowing through each resistor is the same, we can again use Ohm’s law to predict how much voltage will be present across each resistor. Since we already know that the total circuit current equals 0.1 amps and R1 equals 50Ω, the total voltage present across R1 equals 0.1A X 50Ω = 5 volts. Since R2 is the same value as R1, 5 volts will also be present across R2.

We can double check our math by adding together all the voltages present across all the resistors. In this case 5V + 5V = 10V, which is in agreement with the total voltage present.


Resistors in parallel:

When used in parallel, resistors can be said to be a “current dividing network.” This is because in a parallel circuit, voltage across each resistor is the same value but the current flowing through each resistor only part of the total circuit current value. Looking again at the circuit from figure 2 we can derive the current flowing through each resistor.

Based on the fact that in a parallel circuit the voltage present across each resistor is the same, we can again use Ohm’s law to predict how much current will flow through each resistor. Since we already know that the total circuit voltage equals 10 volts and R1 equals 50Ω, the total current flowing through R1 equals 10V / 50Ω = 0.2 amps. Since R2 is the same value as R1, 0.2 amps will also be present across R2.

We can double check our math by adding together all the current flow present through all the resistors. In this case 0.2A + 0.2A = 0.4A, which is in agreement with the total current flow we previously determined for figure 2.


Complex circuit resistance:

In some circuits you will find both series and parallel resistors. The same rules apply in these more complex circuits, where both circuit types are present, as in simpler circuits where only one is present. With complex series/parallel resistive circuits it is best to redefine the parallel parts of the circuit into a series equivalent circuit and then use Ohm’s law to define the total current and resistance present. You can then use the total current and voltage values to determine the voltages and currents present at each of the resistors in the circuit.

<DIAGRAM>

Start by determining the total resistance of the parallel combination of R2 and R3, which equals:

  [R2&3 = 1/(1/R2 + 1/R3)] → [R2&3 = 1/(1/100 + 1/400)] → [R2&3 = 1/(0.01 + 0.0025)] → [R2&3 = 1/(0.0125)] →   R2&3 = 80Ω

You can then redraw the circuit in figure 3 to a series equivalent circuit that looks like figure 4.

<DIAGRAM>

We can now determine the total circuit resistance by simply adding all the resistors in the series equivalent circuit:

  [RT = R1 + R2&3 + R4 + R5] →   [RT = 50 + 80 + 100 + 20] →  RT = 250Ω

Armed with the total resistance for the circuit and the total circuit voltage we can now figure out the total circuit current using Ohm’s law:

  [VT = ITRT] → [IT = VT/RT] →  [IT = 10V/250Ω] →  IT = 0.04A


Now we can figure out the voltage and current present at each of the resistors by using Ohm’s law and the two rules for resistance circuits:

1) In a series circuit, the current is the same through all resistors – voltage divider circuit. 2) In a parallel circuit, the voltage present is the same for all resistors – current divider circuit.

For R1:

  [VR1 = IT X R1] → [VR1 = 0.04A X 50Ω] →  VR1 = 2V

For R2&3:

  [VR2&3 = IT X R2&3] → [VR2&3 = 0.04A X 80Ω] →  VR2&3 = 3.2V

For R2:

  [IR2 = VR2&3 / R2] → [IR2 = 3.2V / 100] →  I R2 = .032A

For R3:

  [IR3 = VR2&3 / R3] → [IR2 = 3.2V / 400] →  I R2 = .008A

For R4:

  [VR4 = IT X R4] → [VR4 = 0.04A X 100Ω] →  VR4 = 4V

For R5:

  [VR5 = IT X R5] → [VR5 = 0.04A X 20Ω] →  VR5 = 0.8V

A double check to verify the accuracy of our circuit analysis confirms that all individual voltages present across each resistor in the series equivalent circuit adds up to the total 10 volts available from the source and all currents in the parallel portion of the circuit adds up to the total current for the circuit of 0.04A.