How to Read a Graph Electric Potential Vs Resistance

eleven.2 Ohm's Law (ESBQ6)

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3 quantities which are fundamental to electric circuits are current, voltage (potential difference) and resistance. To recap:

1. Electrical current, \(I\), is defined as the rate of menses of charge through a circuit.

2. Potential divergence or voltage, \(Five\), is the amount of energy per unit charge needed to move that charge between two points in a circuit.

3. Resistance, \(R\), is a mensurate of how `hard' it is to push button current through a circuit element.

We will now await at how these 3 quantities are related to each other in electrical circuits.

An important relationship between the current, voltage and resistance in a circuit was discovered past Georg Simon Ohm and it is chosen Ohm's Law.

Ohm'due south Law

The amount of electric current through a metal usher, at a constant temperature, in a circuit is proportional to the voltage across the conductor and can exist described past

\(I = \frac{V}{R}\)

where \(I\) is the current through the conductor, \(V\) is the voltage across the conductor and \(R\) is the resistance of the conductor. In other words, at constant temperature, the resistance of the conductor is constant, contained of the voltage applied beyond it or current passed through it.

Ohm's Police force tells the states that if a conductor is at a abiding temperature, the current flowing through the conductor is directly proportional to the voltage across it. This means that if we plot voltage on the 10-axis of a graph and current on the y-axis of the graph, we will get a directly-line.

The gradient of the directly-line graph is related to the resistance of the conductor as \[\frac{I}{V} = \frac{one}{R}\] This can be rearranged in terms of the constant resistance as: \[R = \frac{5}{I}\]

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Ohm's Law

Aim

To decide the relationship between the current going through a resistor and the potential deviation (voltage) across the same resistor.

Appliance

4 cells, 4 resistors, an ammeter, a voltmeter, connecting wires

Method

This experiment has two parts. In the offset role we will vary the practical voltage across the resistor and measure the resulting current through the circuit. In the 2d role we volition vary the current in the excursion and measure the resulting voltage beyond the resistor. Afterward obtaining both sets of measurements, we will examine the human relationship betwixt the electric current and the voltage beyond the resistor.

1. Varying the voltage:

   1. Set upwardly the circuit according to circuit diagram i), starting with only ane cell.

   2. Depict the following table in your lab volume.

      Number of cells	Voltage, Five (\(\text{V}\))	Electric current, I (\(\text{A}\))
      \(\text{1}\)
      \(\text{2}\)
      \(\text{iii}\)
      \(\text{4}\)

   3. Become your teacher to bank check the circuit before turning the ability on.

   4. Measure out the voltage across the resistor using the voltmeter, and the current in the excursion using the ammeter.

   5. Add one more \(\text{one,5}\) \(\text{V}\) cell to the circuit and repeat your measurements.

   6. Repeat until you lot have four cells and you take completed your table.

2. Varying the current:

   1. Fix up the circuit according to circuit diagram ii), starting with only ane resistor in the excursion.

   2. Draw the following table in your lab book.

      Voltage, V (\(\text{V}\))	Current, I (\(\text{A}\))

   3. Get your teacher to check your circuit before turning the power on.

   4. Measure the current and mensurate the voltage across the unmarried resistor.

   5. Now add another resistor in series in the circuit and measure the electric current and the voltage across simply the original resistor again. Continue adding resistors until y'all take four in serial, but remember to but measure the voltage beyond the original resistor each time. Enter the values you measure into the table.

Analysis and results

1. Using the information you recorded in the beginning table, draw a graph of current versus voltage. Since the voltage is the variable which we are directly varying, it is the independent variable and will be plotted on the \(x\)-axis. The electric current is the dependent variable and must be plotted on the \(y\)-axis.

2. Using the data you recorded in the second table, draw a graph of voltage vs. current. In this case the independent variable is the current which must be plotted on the \(x\)-axis, and the voltage is the dependent variable and must be plotted on the \(y\)-axis.

Conclusions

1. Examine the graph you fabricated from the first table. What happens to the current through the resistor when the voltage across it is increased? i.due east. Does it increase or decrease?

2. Examine the graph yous made from the 2d table. What happens to the voltage beyond the resistor when the electric current increases through the resistor? i.e. Does information technology increase or decrease?

3. Do your experimental results verify Ohm's Law? Explain.

Questions and discussion

1. For each of your graphs, calculate the slope and from this decide the resistance of the original resistor. Do you get the same value when you summate information technology for each of your graphs?

2. How would you lot become about finding the resistance of an unknown resistor using only a power supply, a voltmeter and a known resistor \(R_0\)?

Ohm's Law

Textbook Exercise eleven.i

Plot a graph of voltage (on the 10-axis) and electric current (on the y-axis).

What type of graph do you obtain (straight-line, parabola, other bend)

directly-line

Summate the gradient of the graph.

The gradient of the graph (\(m\)) is the alter in the current divided by the change in the voltage:

\begin{align*} m & = \frac{\Delta I}{\Delta V} \\ & = \frac{(\text{i,6}) - (\text{0,four})}{(\text{12}) - (\text{3})} \\ & = \text{0,thirteen} \finish{align*}

Do your experimental results verify Ohm's Law? Explain.

Yes. A straight line graph is obtained when we plot a graph of voltage vs. current.

How would you get about finding the resistance of an unknown resistor using only a power supply, a voltmeter and a known resistor \(R_{0}\)?

You lot start by connecting the known resistor in a circuit with the power supply. Now you read the voltage of the ability supply and note this downward.

Next you connect the ii resistors in series. Yous tin now take the voltage measurements for each of resistors.

Then we can find the voltages for the two resistors. Now we note that:

\[Five = IR\]

So using this and the fact that for resistors in series, the current is the aforementioned everywhere in the circuit we can discover the unknown resistance.

\brainstorm{align*} V_{0} & = IR_{0} \\ I & = \frac{V_{0}}{R_{0}} \\ V_{U} & = IR_{U} \\ I & = \frac{V_{U}}{R_{U}} \\ \frac{V_{U}}{R_{U}} & = \frac{V_{0}}{R_{0}} \\ \therefore R_{U} & = \frac{V_{U}R_{0}}{V_{0}} \end{align*}

Ohmic and non-ohmic conductors (ESBQ7)

Conductors which obey Ohm's Law have a constant resistance when the voltage is varied across them or the current through them is increased. These conductors are called ohmic conductors. A graph of the current vs. the voltage across these conductors will exist a directly-line. Some examples of ohmic conductors are excursion resistors and nichrome wire.

Every bit you have seen, there is a mention of constant temperature when we talk nigh Ohm's Police. This is considering the resistance of some conductors changes equally their temperature changes. These types of conductors are called non-ohmic conductors, considering they exercise not obey Ohm's Law. A light seedling is a common example of a non-ohmic conductor. Other examples of non-ohmic conductors are diodes and transistors.

In a light seedling, the resistance of the filament wire will increase dramatically as it warms from room temperature to operating temperature. If we increase the supply voltage in a real lamp circuit, the resulting increase in current causes the filament to increment in temperature, which increases its resistance. This effectively limits the increase in current. In this case, voltage and current practice not obey Ohm'south Law.

The miracle of resistance changing with variations in temperature is 1 shared by almost all metals, of which most wires are fabricated. For most applications, these changes in resistance are small-scale enough to be ignored. In the application of metal lamp filaments, which increase a lot in temperature (up to almost \(\text{1 000}\) \(\text{℃}\), and starting from room temperature) the change is quite large.

In general, for not-ohmic conductors, a graph of voltage against current will not exist a straight-line, indicating that the resistance is not constant over all values of voltage and current.

A recommended experiment for informal cess is included. In this experiment learners will obtain current and voltage data for a resistor and lite seedling and determine which obeys Ohm's law. You volition need light bulbs, resistors, connecting wires, power source, ammeter and voltmeter. Learners should discover that the resistor obeys Ohm's law, while the light bulb does not.

Ohmic and non-ohmic conductors

Aim

To determine whether two circuit elements (a resistor and a lightbulb) obey Ohm's Law

Apparatus

4 cells, a resistor, a lightbulb, connecting wires, a voltmeter, an ammeter

Method

The two circuits shown in the diagrams to a higher place are the same, except in the first at that place is a resistor and in the second there is a lightbulb. Set up both the circuits higher up, starting with 1 cell. For each circuit:

1. Measure the voltage beyond the circuit chemical element (either the resistor or lightbulb) using the voltmeter.

2. Measure the current in the circuit using the ammeter.

3. Add another cell and echo your measurements until you take 4 cells in your circuit.

Results

Describe two tables which look similar the following in your book. You should accept one tabular array for the first circuit measurements with the resistor and another tabular array for the second circuit measurements with the lightbulb.

Number of cells	Voltage, V (\(\text{V}\))	Current, I (\(\text{A}\))
\(\text{i}\)
\(\text{2}\)
\(\text{3}\)
\(\text{4}\)

Analysis

Using the data in your tables, draw two graphs of \(I\) (\(y\)-axis) vs. \(V\) (\(x\)-axis), one for the resistor and one for the lightbulb.

Questions and Discussion

Examine your graphs closely and answer the following questions:

1. What should the graph of \(I\) vs. \(5\) look similar for a usher which obeys Ohm's Constabulary?

2. Do either or both your graphs wait like this?

3. What can you conclude most whether or non the resistor and/or the lightbulb obey Ohm's Police?

4. Is the lightbulb an ohmic or non-ohmic conductor?

Using Ohm's Police (ESBQ8)

Nosotros are now ready to see how Ohm'southward Law is used to analyse circuits.

Consider a circuit with a cell and an ohmic resistor, R. If the resistor has a resistance of \(\text{5}\) \(\text{Ω}\) and voltage across the resistor is \(\text{5}\) \(\text{Five}\), then we can employ Ohm's Law to calculate the current flowing through the resistor. Our beginning task is to describe the circuit diagram. When solving any problem with electrical circuits information technology is very important to make a diagram of the excursion earlier doing whatever calculations. The excursion diagram for this problem looks like the post-obit:

The equation for Ohm's Law is: \[R = \frac{V}{I}\]

which can be rearranged to: \[I = \frac{Five}{R}\]

The electric current flowing through the resistor is:

\brainstorm{align*} I &= \frac{5}{R} \\ &= \frac{\text{v}\text{ V}}{\text{five }\Omega} \\ &= \text{1}\text{ A} \end{align*}

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Worked example 1: Ohm's Police force

Written report the excursion diagram beneath:

The resistance of the resistor is \(\text{x}\) \(\text{Ω}\) and the current going through the resistor is \(\text{iv}\) \(\text{A}\). What is the potential divergence (voltage) across the resistor?

Determine how to arroyo the problem

We are given the resistance of the resistor and the current passing through it and are asked to summate the voltage across it. We can apply Ohm's Law to this trouble using: \[R = \frac{5}{I}.\]

Solve the problem

Rearrange the equation above and substitute the known values for \(R\) and \(I\) to solve for \(V\). \begin{align*} R &= \frac{V}{I} \\ R \times I&= \frac{V}{I} \times I\\ V &= I \times R \\ &= \text{10} \times \text{4} \\ &= \text{forty}\text{ 5} \end{align*}

Write the final answer

The voltage across the resistor is \(\text{forty}\) \(\text{Five}\).

Ohm's Police force

Textbook Exercise xi.2

Summate the resistance of a resistor that has a potential difference of \(\text{8}\) \(\text{5}\) across it when a current of \(\text{2}\) \(\text{A}\) flows through it. Draw the excursion diagram earlier doing the calculation.

The resistance of the unknown resistor is:

\brainstorm{align*} R & = \frac{Five}{I} \\ & = \frac{8}{2} \\ & = \text{4}\text{ Ω} \end{marshal*}

What current will menstruum through a resistor of \(\text{6}\) \(\text{Ω}\) when at that place is a potential departure of \(\text{18}\) \(\text{5}\) across its ends? Depict the circuit diagram earlier doing the calculation.

The resistance of the unknown resistor is:

\begin{marshal*} I & = \frac{V}{R} \\ & = \frac{eighteen}{6} \\ & = \text{3}\text{ A} \end{align*}

What is the voltage across a \(\text{ten}\) \(\text{Ω}\) resistor when a current of \(\text{one,5}\) \(\text{A}\) flows though information technology? Draw the circuit diagram before doing the calculation.

The resistance of the unknown resistor is:

\brainstorm{align*} V & = I \cdot R \\ & = (\text{1,5})(10) \\ & = \text{15}\text{ V} \cease{align*}

Recap of resistors in series and parallel (ESBQ9)

In Grade 10, you learnt virtually resistors and were introduced to circuits where resistors were connected in series and in parallel. In a serial circuit there is one path along which current flows. In a parallel excursion at that place are multiple paths along which current flows.

When there is more than one resistor in a excursion, we are usually able to summate the total combined resistance of all the resistors. This is known as the equivalent resistance.

Equivalent series resistance

In a excursion where the resistors are connected in series, the equivalent resistance is just the sum of the resistances of all the resistors.

Equivalent resistance in a series circuit,

For n resistors in serial the equivalent resistance is:

\[R_{s} = R_{1} + R_{2} + R_{iii} + \ldots + R_{due north}\]

Let the states apply this to the following excursion.

The resistors are in series, therefore:

\brainstorm{align*} R_{s} & = R_{1} + R_{two} + R_{3} \\ & = \text{3}\text{ Ω} + \text{ten}\text{ Ω} + \text{5}\text{ Ω} \\ & =\text{18}\text{ Ω} \cease{align*}

Equivalent parallel resistance

In a excursion where the resistors are connected in parallel, the equivalent resistance is given by the following definition.

Equivalent resistance in a parallel circuit

For \(n\) resistors in parallel, the equivalent resistance is:

\[\frac{1}{R_{p}} = \frac{one}{R_{1}} + \frac{1}{R_{2}} + \frac{ane}{R_{3}} + \ldots + \frac{1}{R_{north}}\]

Let us apply this formula to the following excursion.

What is the total (equivalent) resistance in the excursion?

\begin{align*} \frac{ane}{R_{p}} & = \left( \frac{1}{R_{ane}} + \frac{1}{R_{two}} + \frac{1}{R_{three}} \correct) \\ & = \left( \frac{1}{\text{x}\text{ Ω}} + \frac{1}{\text{2}\text{ Ω}} + \frac{1}{\text{1}\text{ Ω}} \right) \\ & = \left( \frac{\text{1}\text{ Ω} + \text{five}\text{ Ω} + \text{10}\text{ Ω}}{\text{10}\text{ Ω}} \right) \\ & = \left( \frac{\text{16}\text{ Ω}}{\text{10}\text{ Ω}} \right) \\ R_{p} & = \text{0,625}\text{ Ω} \cease{marshal*}

Series and parallel resistance

Textbook Practice 11.iii

Two \(\text{ten}\) \(\text{kΩ}\) resistors are connected in serial. Summate the equivalent resistance.

Since the resistors are in series we can utilise:

\[R_{south} = R_{1} + R_{two}\]

The equivalent resistance is:

\begin{align*} R_{south} & = R_{one} + R_{two} \\ & = \text{10}\text{ kΩ} + \text{10}\text{ kΩ} \\ & = \text{xx}\text{ kΩ} \terminate{align*}

Two resistors are connected in series. The equivalent resistance is \(\text{100}\) \(\text{Ω}\). If i resistor is \(\text{10}\) \(\text{Ω}\), calculate the value of the second resistor.

Since the resistors are in serial we can use:

\[R_{south} = R_{1} + R_{two}\]

The equivalent resistance is:

\begin{align*} R_{s} & = R_{1} + R_{two} \\ R_{2} & = R_{s} - R_{i} \\ & = \text{100}\text{ Ω} - \text{10}\text{ Ω} \\ & = \text{90}\text{ Ω} \finish{align*}

Two \(\text{10}\) \(\text{kΩ}\) resistors are connected in parallel. Calculate the equivalent resistance.

Since the resistors are in parallel we tin utilize:

\[\frac{i}{R_{p}} = \frac{1}{R_{i}} + \frac{1}{R_{2}}\]

The equivalent resistance is:

\begin{align*} \frac{ane}{R_{p}} & = \frac{ane}{R_{1}} + \frac{1}{R_{2}} \\ & = \frac{1}{\text{100}} + \frac{1}{\text{ten}} \\ & = \frac{1 + 10}{\text{100}} \\ & = \frac{11}{\text{100}} \\ R_{p} & = \text{9,09}\text{ kΩ} \finish{align*}

Ii resistors are continued in parallel. The equivalent resistance is \(\text{3,75}\) \(\text{Ω}\). If one resistor has a resistance of \(\text{x}\) \(\text{Ω}\), what is the resistance of the second resistor?

Since the resistors are in parallel nosotros can use:

\[\frac{one}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\]

The equivalent resistance is:

\brainstorm{align*} \frac{i}{R_{p}} & = \frac{1}{R_{i}} + \frac{1}{R_{2}} \\ \frac{one}{R_{2}} & = \frac{1}{R_{p}} - \frac{1}{R_{1}} \\ & = \frac{1}{\text{three,75}} - \frac{1}{\text{10}} \\ & = \frac{\text{10} - \text{3,75}}{\text{37,5}} \\ & = \frac{\text{6,25}}{\text{37,five}} \\ R_{ii} & = \text{6}\text{ Ω} \finish{align*}

Summate the equivalent resistance in each of the following circuits:

a) The resistors are in parallel so we use:

\[\frac{one}{R_{p}} = \frac{1}{R_{i}} + \frac{one}{R_{2}}\]

The equivalent resistance is:

\begin{align*} \frac{1}{R_{p}} & = \frac{1}{R_{ane}} + \frac{1}{R_{two}} \\ & = \frac{1}{\text{3}} + \frac{1}{\text{2}} \\ & = \frac{\text{ii} + \text{three}}{\text{6}} \\ & = \frac{\text{5}}{\text{6}} \\ R & = \text{i,two}\text{ Ω} \end{marshal*}

b) The resistors are in parallel and then nosotros employ:

\[\frac{1}{R_{p}} = \frac{1}{R_{one}} + \frac{i}{R_{2}} + \frac{1}{R_{three}} + \frac{1}{R_{4}}\]

The equivalent resistance is:

\begin{marshal*} \frac{i}{R_{p}} & = \frac{1}{R_{1}} + \frac{one}{R_{2}} + \frac{1}{R_{iii}} + \frac{1}{R_{four}} \\ & = \frac{1}{\text{ii}} + \frac{1}{\text{3}} + \frac{1}{\text{4}} + \frac{1}{\text{i}} \\ & = \frac{\text{6} + \text{four} + \text{iii} + \text{12}}{\text{12}} \\ & = \frac{\text{25}}{\text{12}} \\ R & = \text{0,48}\text{ Ω} \end{align*}

c) The resistors are in series and so we use:

\[R_{s} = R_{1} + R_{2}\]

The equivalent resistance is:

\begin{align*} R_{s} & = R_{1} + R_{two} \\ & = \text{two}\text{ Ω} + \text{3}\text{ Ω} \\ & = \text{five}\text{ Ω} \end{align*}

d) The resistors are in series and so we use:

\[R_{southward} = R_{1} + R_{2} + R_{3} + R_{four}\]

The equivalent resistance is:

\begin{align*} R_{south} & = R_{1} + R_{ii} + R_{3} + R_{4} \\ & = \text{2}\text{ Ω} + \text{iii}\text{ Ω} + \text{4}\text{ Ω} + \text{1}\text{ Ω} \\ & = \text{10}\text{ Ω} \terminate{align*}

Use of Ohm's Law in serial and parallel circuits (ESBQB)

Using the definitions for equivalent resistance for resistors in series or in parallel, nosotros can analyse some circuits with these setups.

Series circuits

Consider a circuit consisting of three resistors and a single cell connected in series.

The first principle to understand about serial circuits is that the corporeality of current is the aforementioned through any component in the excursion. This is because there is only one path for electrons to flow in a series circuit. From the mode that the battery is continued, we tin can tell in which direction the electric current will flow. We know that current flows from positive to negative by convention. Conventional current in this excursion will flow in a clockwise management, from point A to B to C to D and dorsum to A.

We know that in a series circuit the electric current has to be the same in all components. So we tin can write:

\[I = I_{1} = I_{ii} = I_{three}.\]

We also know that total voltage of the excursion has to be equal to the sum of the voltages over all three resistors. And then we tin can write:

\[V = V_{1} + V_{two} + V_{3}\]

Using this information and what nosotros know about computing the equivalent resistance of resistors in series, we tin approach some circuit problems.

Worked instance 2: Ohm'due south Law, series circuit

Summate the current (I) in this excursion if the resistors are both ohmic in nature.

Determine what is required

Nosotros are required to calculate the current flowing in the circuit.

Decide how to approach the trouble

Since the resistors are ohmic in nature, we tin can use Ohm's Police force. There are however two resistors in the circuit and nosotros need to find the total resistance.

Find total resistance in excursion

Since the resistors are continued in series, the full (equivalent) resistance R is:

\[R = R_{1} + R_{two}\]

Therefore,

\brainstorm{align*} R & = \text{two} + \text{4} \\ & = \text{half dozen}\text{ Ω} \terminate{align*}

Apply Ohm's Constabulary

\begin{align*} R & = \frac{5}{I} \\ R \times \frac{I}{R} & = \frac{V}{I} \times \frac{I}{R} \\ I & = \frac{V}{R} \\ & = \frac{12}{6} \\ & = \text{2}\text{ A} \end{marshal*}

Write the final answer

A current of \(\text{2}\) \(\text{A}\) is flowing in the excursion.

Worked example three: Ohm's Law, series circuit

2 ohmic resistors (\(R_{i}\) and \(R_{2}\)) are connected in series with a jail cell. Find the resistance of \(R_{ii}\), given that the electric current flowing through \(R_{ane}\) and \(R_{ii}\) is \(\text{0,25}\) \(\text{A}\) and that the voltage beyond the cell is \(\text{1,5}\) \(\text{V}\). \(R_{ane}\) =\(\text{1}\) \(\text{Ω}\).

Depict the excursion and fill in all known values.

Make up one's mind how to arroyo the problem.

We tin use Ohm's Law to find the total resistance R in the circuit, and so summate the unknown resistance using:

\[R = R_{1} + R_{2}\]

considering information technology is in a series circuit.

Find the full resistance

\begin{align*} R & = \frac{Five}{I} \\ & = \frac{\text{ane,5}}{\text{0,25}} \\ & = \text{6}\text{ Ω} \end{align*}

Find the unknown resistance

We know that:

\[R = \text{six}\text{ Ω}\]

and that

\[R_{ane} = \text{one}\text{ Ω}\]

Since

\[R = R_{1} + R_{2}\] \[R_{ii} = R - R_{1}\]

Therefore,

\[R_{1} = \text{5}\text{ Ω}\]

Worked instance 4: Ohm'due south Law, series circuit

For the following circuit, calculate:

1. the voltage drops \(V_1\), \(V_2\) and \(V_3\) across the resistors \(R_1\), \(R_2\), and \(R_3\)

2. the resistance of \(R_3\).

Make up one's mind how to approach the problem

We are given the voltage beyond the cell and the current in the circuit, as well as the resistances of two of the three resistors. Nosotros can utilize Ohm's Police force to calculate the voltage driblet across the known resistors. Since the resistors are in a series circuit the voltage is \(V = V_1 + V_2 + V_3\) and we can calculate \(V_3\). At present we can utilize this data to notice the voltage across the unknown resistor \(R_3\).

Calculate voltage drop across \(R_1\)

Using Ohm's Law: \begin{align*} R_1 &= \frac{V_1}{I} \\ I \cdot R_1 &= I \cdot \frac{V_1}{I} \\ V_1 &= {I}\cdot{R_1}\\ &= ii \cdot 1 \\ V_1 &= \text{ii}\text{ V} \cease{align*}

Calculate voltage drib beyond \(R_2\)

Once again using Ohm's Constabulary: \brainstorm{marshal*} R_2 &= \frac{V_2}{I} \\ I \cdot R_2 &= I \cdot \frac{V_2}{I} \\ V_2 &= {I}\cdot{R_2}\\ &= 2 \cdot 3 \\ V_2 &= \text{6}\text{ V} \end{align*}

Calculate voltage drop across \(R_3\)

Since the voltage drop across all the resistors combined must exist the same as the voltage drop across the jail cell in a serial circuit, we can observe \(V_3\) using: \begin{align*} V &= V_1 + V_2 + V_3\\ V_3 &= V - V_1 - V_2 \\ &= xviii - two - 6 \\ V_3&= \text{10}\text{ V} \finish{align*}

Find the resistance of \(R_3\)

We know the voltage across \(R_3\) and the current through information technology, and so we can apply Ohm's Law to calculate the value for the resistance: \brainstorm{align*} R_3 &= \frac{V_3}{I}\\ &= \frac{10}{2} \\ R_3&= \text{5 } \Omega \cease{align*}

Write the final answer

\(V_1 = \text{2}\text{ V}\)

\(V_2 = \text{half dozen}\text{ V}\)

\(V_3 = \text{10}\text{ 5}\)

\(R_1 = \text{5 } \Omega\)

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Parallel circuits

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

The first principle to sympathise about parallel circuits is that the voltage is equal beyond all components in the circuit. This is because there are only two sets of electrically common points in a parallel circuit, and voltage measured between sets of common points must always be the same at any given time. So, for the circuit shown, the following is truthful:

\[Five = V_{1} = V_{2} = V_{3}.\]

The second principle for a parallel circuit is that all the currents through each resistor must add together up to the total electric current in the circuit:

\[I = I_{1} + I_{2} + I_{iii}.\]

Using these principles and our cognition of how to calculate the equivalent resistance of parallel resistors, nosotros can now arroyo some excursion problems involving parallel resistors.

Worked example 5: Ohm's Constabulary, parallel circuit

Calculate the current (I) in this circuit if the resistors are both ohmic in nature.

Determine what is required

We are required to summate the current flowing in the excursion.

Decide how to approach the trouble

Since the resistors are ohmic in nature, we can use Ohm's Law. At that place are still two resistors in the excursion and nosotros need to find the total resistance.

Notice the equivalent resistance in excursion

Since the resistors are connected in parallel, the total (equivalent) resistance R is:

\[\frac{i}{R} = \frac{ane}{R_{1}} + \frac{one}{R_{2}}.\] \begin{marshal*} \frac{1}{R} &= \frac{1}{R_1} + \frac{1}{R_2} \\ &= \frac{i}{2} + \frac{1}{iv} \\ &= \frac{two+1}{4} \\ &= \frac{3}{four} \\ \text{Therefore, } R &= \text{one,33 } \Omega \end{align*}

Apply Ohm'south Police force

\brainstorm{align*} R&= \frac{V}{I} \\ R \cdot \frac{I}{R} &= \frac{V}{I} \cdot \frac{I}{R} \\ I &= \frac{V}{R}\\ I &= Five \cdot \frac{1}{R}\\ &= (12) \left(\frac{3}{iv}\right) \\ &= \text{9}\text{ A} \end{align*}

Write the final respond

The current flowing in the circuit is \(\text{9}\) \(\text{A}\).

Worked case 6: Ohm's Law, parallel circuit

Two ohmic resistors (\(R_1\) and \(R_2\)) are continued in parallel with a cell. Detect the resistance of \(R_2\), given that the current flowing through the cell is \(\text{4,8}\) \(\text{A}\) and that the voltage beyond the jail cell is \(\text{9}\) \(\text{V}\).

Determine what is required

We demand to summate the resistance \(R_2\).

Make up one's mind how to approach the problem

Since the resistors are ohmic and nosotros are given the voltage across the cell and the current through the cell, we can apply Ohm's Constabulary to find the equivalent resistance in the circuit. \brainstorm{marshal*} R & = \frac{V}{I} \\ & = \frac{9}{\text{4,8}} \\ & = \text{1,875} \ \Omega \end{align*}

Calculate the value for \(R_2\)

Since we know the equivalent resistance and the resistance of \(R_1\), we can use the formula for resistors in parallel to find the resistance of \(R_2\). \begin{align*} \frac{1}{R} & = \frac{1}{R_1} + \frac{1}{R_2} \end{align*} Rearranging to solve for \(R_2\): \begin{align*} \frac{1}{R_2} & = \frac{ane}{R} - \frac{1}{R_1} \\ & = \frac{1}{\text{1,875}} - \frac{1}{3}\\ & = \text{0,2} \\ R_2 & = \frac{1}{\text{0,2}} \\ & = \text{5} \ \Omega \end{align*}

Write the final answer

The resistance \(R_2\) is \(\text{v}\) \(\Omega\)

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Worked instance 7: Ohm's Law, parallel circuit

An 18 volt cell is continued to two parallel resistors of \(\text{4}\) \(\Omega\) and \(\text{12}\) \(\Omega\) respectively. Summate the current through the cell and through each of the resistors.

First depict the circuit before doing any calculations

Determine how to approach the problem

Nosotros need to decide the electric current through the cell and each of the parallel resistors. We have been given the potential difference beyond the cell and the resistances of the resistors, so nosotros can utilise Ohm's Law to calculate the current.

Calculate the current through the jail cell

To summate the current through the cell we first need to make up one's mind the equivalent resistance of the remainder of the excursion. The resistors are in parallel and therefore: \begin{align*} \frac{i}{R} &= \frac{1}{R_1} + \frac{one}{R_2} \\ &= \frac{ane}{4} + \frac{i}{12} \\ &= \frac{3+1}{12} \\ &= \frac{4}{12} \\ R &= \frac{12}{4} = \text{iii} \ \Omega \end{align*} Now using Ohm's Law to find the current through the cell: \begin{align*} R &= \frac{V}{I} \\ I &= \frac{5}{R} \\ &= \frac{xviii}{iii} \\ I &= \text{6}\text{ A} \end{marshal*}

Now determine the current through ane of the parallel resistors

We know that for a purely parallel circuit, the voltage across the cell is the same equally the voltage across each of the parallel resistors. For this circuit: \brainstorm{align*} Five &= V_1 = V_2 = \text{xviii}\text{ V} \end{align*} Let'southward start with calculating the current through \(R_1\) using Ohm's Law: \begin{align*} R_1 &= \frac{V_1}{I_1} \\ I_1 &= \frac{V_1}{R_1} \\ &= \frac{eighteen}{4} \\ I_1 &= \text{4,v}\text{ A} \cease{align*}

Summate the electric current through the other parallel resistor

We can use Ohm's Constabulary again to find the current in \(R_2\): \begin{marshal*} R_2 &= \frac{V_2}{I_2} \\ I_2 &= \frac{V_2}{R_2} \\ &= \frac{18}{12} \\ I_2 &= \text{1,v}\text{ A} \end{align*} An alternative method of calculating \(I_2\) would have been to employ the fact that the currents through each of the parallel resistors must add together up to the total current through the prison cell: \begin{marshal*} I &= I_1 + I_2 \\ I_2 &= I - I_1 \\ &= 6 - 4.5 \\ I_2 &= \text{1,5}\text{ A} \stop{marshal*}

Write the last respond

The current through the jail cell is \(\text{six}\) \(\text{A}\).

The electric current through the \(\text{iv}\) \(\Omega\) resistor is \(\text{4,5}\) \(\text{A}\).

The current through the \(\text{12}\) \(\Omega\) resistor is \(\text{1,5}\) \(\text{A}\).

Ohm's Law in serial and parallel circuits

Textbook Do xi.iv

Calculate the value of the unknown resistor in the excursion:

We commencement use Ohm's police force to summate the total series resistance:

\begin{align*} R & = \frac{Five}{I} \\ & = \frac{9}{i} \\ & = \text{9}\text{ Ω} \terminate{align*}

Now we can find the unknown resistance:

\brainstorm{align*} R_{south} & = R_{1} + R_{2} + R_{3} + R_{4} \\ R_{4} & = R_{s} - R_{one} - R_{two} - R_{iii} \\ & = ix - iii - 3 - 1 \\ & = \text{2}\text{ Ω} \cease{align*}

Summate the value of the electric current in the following circuit:

Nosotros beginning notice the total resistance:

\begin{align*} R_{southward} & = R_{i} + R_{2} + R_{3} \\ & = \text{i} + \text{2,5} + \text{ane,five} \\ & = \text{five}\text{ Ω} \end{marshal*}

Now we tin summate the current:

\begin{align*} I & = \frac{Five}{R} \\ & = \frac{9}{5} \\ & = \text{1,8}\text{ A} \end{align*}

Three resistors with resistance \(\text{1}\) \(\text{Ω}\), \(\text{5}\) \(\text{Ω}\) and \(\text{ten}\) \(\text{Ω}\) respectively, are continued in serial with a \(\text{12}\) \(\text{5}\) battery. Summate the value of the current in the excursion.

We depict the circuit diagram:

We now notice the total resistance:

\begin{marshal*} R_{s} & = R_{1} + R_{2} + R_{3} \\ & = \text{1} + \text{5} + \text{ten} \\ & = \text{16}\text{ Ω} \end{marshal*}

Now we can calculate the current:

\begin{align*} I & = \frac{V}{R} \\ & = \frac{12}{xvi} \\ & = \text{0,75}\text{ A} \end{align*}

Calculate the current through the cell if the resistors are both ohmic in nature.

We kickoff discover the total resistance:

\begin{align*} \frac{1}{R_{p}} & = \frac{1}{R_{1}} + \frac{1}{R_{ii}} \\ & = \frac{1}{\text{1}} + \frac{ane}{\text{3}} \\ & = \frac{iii + 1}{\text{iii}} \\ & = \frac{4}{\text{3}} \\ & = \text{0,75}\text{ Ω} \end{align*}

Now we can calculate the electric current:

\begin{marshal*} I & = \frac{V}{R} \\ & = \frac{9}{\text{0,75}} \\ & = \text{12}\text{ A} \end{align*}

Calculate the value of the unknown resistor \(R_{four}\) in the circuit:

We outset find the full resistance:

\begin{align*} R & = \frac{5}{I} \\ & = \frac{24}{\text{2}} \\ & = \text{12}\text{ Ω} \end{align*}

At present nosotros can calculate the unknown resistance:

\begin{align*} \frac{1}{R_{p}} & = \frac{i}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{iii}} + \frac{one}{R_{4}}\\ \frac{1}{R_{4}} & = \frac{1}{R_{p}} - \frac{1}{R_{1}} - \frac{1}{R_{2}} - \frac{1}{R_{3}}\\ & = \frac{one}{\text{12}} - \frac{1}{\text{120}} - \frac{1}{\text{40}} - \frac{1}{\text{lx}} \\ & = \frac{10 - 1 - three - 2}{\text{120}} \\ & = \frac{four}{\text{120}} \\ & = \text{30}\text{ Ω} \end{marshal*}

the value of the current through the battery

We draw a circuit diagram:

To calculate the value of the current through the bombardment we commencement need to calculate the equivalent resistance:

\begin{align*} \frac{1}{R_{p}} & = \frac{1}{R_{i}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}\\ & = \frac{1}{\text{1}} + \frac{one}{\text{5}} + \frac{ane}{\text{10}} \\ & = \frac{x + 2 + i}{\text{10}} \\ & = \frac{13}{\text{ten}} \\ & = \text{0,77}\text{ Ω} \cease{align*}

Now we can calculate the current through the bombardment:

\begin{align*} I & = \frac{V}{R} \\ & = \frac{xx}{\text{0,77}} \\ & = \text{26}\text{ A} \terminate{marshal*}

the value of the current in each of the three resistors.

For a parallel circuit the voltage across jail cell is the same as the voltage across each of the resistors. For this circuit:

\[V = V_{ane} = V_{two} = V_{3} = \text{twenty}\text{ V}\]

Now we can calculate the current through each resistor. We will start with \(R_{1}\):

\begin{marshal*} I & = \frac{Five}{R} \\ & = \frac{xx}{\text{ane}} \\ & = \text{20}\text{ A} \end{marshal*}

Next nosotros calculate the electric current through \(R_{two}\):

\begin{marshal*} I & = \frac{Five}{R} \\ & = \frac{20}{\text{5}} \\ & = \text{4}\text{ A} \end{align*}

And finally nosotros calculate the current through \(R_{3}\):

\begin{align*} I & = \frac{V}{R} \\ & = \frac{xx}{\text{x}} \\ & = \text{2}\text{ A} \terminate{align*}

You can check that these add up to the total electric current.

Series and parallel networks of resistors (ESBQC)

Now that you know how to handle simple serial and parallel circuits, y'all are fix to tackle circuits which combine these ii setups such equally the following circuit:

Figure eleven.1: An example of a series-parallel network. The dashed boxes indicate parallel sections of the circuit.

Information technology is relatively like shooting fish in a barrel to work out these kind of circuits considering yous use everything you have already learnt nearly series and parallel circuits. The only deviation is that you lot do information technology in stages. In Figure 11.one, the circuit consists of 2 parallel portions that are then in series with a prison cell. To piece of work out the equivalent resistance for the circuit, you start by computing the full resistance of each of the parallel portions and and then add up these resistances in serial. If all the resistors in Figure 11.1 had resistances of \(\text{ten}\) \(\text{Ω}\), nosotros can summate the equivalent resistance of the unabridged circuit.

Nosotros start by computing the total resistance of Parallel Excursion 1.

The value of \(R_{p1}\) is: \begin{align*} \frac{1}{R_{p1}} &= \frac{1}{R_1} + \frac{1}{R_2} \\ R_{p1}&= \left(\frac{1}{10} + \frac{1}{ten} \correct)^{-1} \\ &= \left(\frac{1+1}{x} \correct)^{-1} \\ &= \left(\frac{2}{10} \correct)^{-ane} \\ &= \text{v} \, \Omega \cease{align*}

We tin similarly calculate the total resistance of Parallel Circuit 2: \brainstorm{align*} \frac{1}{R_{p2}} &= \frac{one}{R_3} + \frac{1}{R_4} \\ R_{p2}&= \left(\frac{1}{10} + \frac{1}{10} \right)^{-ane} \\ &= \left(\frac{1+one}{10} \right)^{-1} \\ &= \left(\frac{two}{x} \right)^{-one} \\ &= \text{v} \, \Omega \stop{marshal*}

You tin now care for the excursion like a elementary serial excursion as follows:

Therefore the equivalent resistance is: \begin{align*} R &= R_{p1} + R_{p2} \\ &= five + 5 \\ &= ten \, \Omega \cease{align*}

The equivalent resistance of the circuit in Figure xi.ane is \(\text{10}\) \(\text{Ω}\).

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Series and parallel networks

Textbook Exercise 11.five

We showtime past determining the equivalent resistance of the parallel combination:

\begin{align*} \frac{1}{R_{p}} & = \frac{1}{R_{one}} + \frac{1}{R_{2}} \\ & = \frac{1}{4} + \frac{1}{2} \\ & = \frac{3}{four}\\ R_{p} & = \text{one,33}\text{ Ω} \terminate{align*}

Now we have a circuit with 2 resistors in series so we can calculate the equivalent resistance:

\brainstorm{align*} R_{south} & = R_{iii} + R_{p} \\ & = \text{2} + \text{one,33} \\ & = \text{3,33}\text{ Ω} \end{align*}

We start by determining the equivalent resistance of the parallel combination:

\brainstorm{align*} \frac{1}{R_{p}} & = \frac{1}{R_{1}} + \frac{ane}{R_{2}} \\ & = \frac{1}{1} + \frac{1}{2} \\ & = \frac{3}{ii}\\ R_{p} & = \text{0,67}\text{ Ω} \terminate{marshal*}

Now we accept a circuit with three resistors in series and so we can calculate the equivalent resistance:

\brainstorm{align*} R_{due south} & = R_{3} + R_{iv} + R_{p} \\ & = \text{4} + \text{6} + \text{0,67} \\ & = \text{10,67}\text{ Ω} \end{align*}

Nosotros start by determining the equivalent resistance of the parallel combination:

\begin{align*} \frac{one}{R_{p}} & = \frac{1}{R_{1}} + \frac{one}{R_{2}} + \frac{ane}{R_{three}} \\ & = \frac{1}{iii} + \frac{1}{5} + \frac{1}{1} \\ & = \frac{23}{15}\\ R_{p} & = \text{0,652}\text{ Ω} \end{align*}

At present we have a circuit with ii resistors in serial and then we tin calculate the equivalent resistance:

\brainstorm{align*} R_{southward} & = R_{iv} + R_{p} \\ & = \text{2} + \text{0,652} \\ & = \text{ii,652}\text{ Ω} \terminate{marshal*}

the current \(I\) through the jail cell.

To find the current \(I\) we first need to observe the equivalent resistance. We get-go by calculating the equivalent resistance of the parallel combination:

\begin{align*} \frac{one}{R_{p}} & = \frac{ane}{R_{1}} + \frac{one}{R_{2}} + \frac{1}{R_{three}} \\ & = \frac{1}{three} + \frac{one}{5} + \frac{1}{1} \\ & = \frac{23}{15}\\ R_{p} & = \text{0,652}\text{ Ω} \end{align*}

Now nosotros have a circuit with 2 resistors in series so we can calculate the equivalent resistance:

\begin{align*} R_{southward} & = R_{4} + R_{p} \\ & = \text{2} + \text{0,652} \\ & = \text{2,652}\text{ Ω} \end{marshal*}

So the current through the cell is:

\begin{align*} I & = \frac{V}{R} \\ & = \frac{\text{12}}{\text{2,652}} \\ & = \text{four,52}\text{ A} \end{align*}

the current through the \(\text{5}\) \(\text{Ω}\) resistor.

The electric current through the parallel combination of resistors is \(\text{4,52}\) \(\text{A}\). (The current is the same through series combinations of resistors and we can consider the entire parallel prepare of resistors equally i serial resistor.)

Using this nosotros can find the voltage through the parallel combination of resistors (remember to use the equivalent parallel resistance and not the equivalent resistance of the circuit):

\begin{marshal*} V & = I \cdot R \\ & = (\text{4,52})(\text{0,652}) \\ & = \text{2,95}\text{ V} \cease{align*}

Since the voltage across each resistor in the parallel combination is the same, this is as well the voltage across the \(\text{5}\) \(\text{Ω}\) resistor.

So now nosotros can calculate the current through the resistor:

\brainstorm{align*} I & = \frac{5}{R} \\ & = \frac{\text{ii,95}}{\text{v}} \\ & = \text{0,59}\text{ A} \cease{align*}

If current flowing through the prison cell is \(\text{ii}\) \(\text{A}\), and all the resistors are ohmic, calculate the voltage across the jail cell and each of the resistors, \(R_1\), \(R_2\), and \(R_3\) respectively.

To find the voltage we first need to notice the equivalent resistance. Nosotros start by calculating the equivalent resistance of the parallel combination:

\brainstorm{align*} \frac{1}{R_{p}} & = \frac{i}{R_{two}} + \frac{1}{R_{3}} \\ & = \frac{1}{2} + \frac{1}{four} \\ & = \frac{3}{4}\\ R_{p} & = \text{1,33}\text{ Ω} \end{align*}

Now we have a excursion with 2 resistors in serial so we tin can calculate the equivalent resistance:

\begin{marshal*} R_{s} & = R_{1} + R_{p} \\ & = \text{4,66} + \text{1,33} \\ & = \text{5,99}\text{ Ω} \cease{align*}

So the voltage across the cell is:

\brainstorm{marshal*} 5 & = I \cdot R \\ & = (\text{2})(\text{five,99}) \\ & = \text{12}\text{ V} \finish{align*}

The current through the parallel combination of resistors is \(\text{2}\) \(\text{A}\). (The current is the same through serial combinations of resistors and nosotros can consider the entire parallel gear up of resistors every bit one serial resistor.)

Using this we tin find the voltage through the each of the resistors. We start by finding the voltage across \(R_{1}\):

\brainstorm{align*} 5 & = I \cdot R \\ & = (\text{two})(\text{four,66}) \\ & = \text{9,32}\text{ V} \end{align*}

Now we discover the voltage across the parallel combination:

\begin{align*} V & = I \cdot R \\ & = (\text{2})(\text{one,33}) \\ & = \text{2,66}\text{ Five} \end{marshal*}

Since the voltage across each resistor in the parallel combination is the same, this is likewise the voltage beyond resistors \(R_{two}\) and \(R_{3}\).

the current through the prison cell

To find the current nosotros first demand to discover the equivalent resistance. Nosotros commencement by computing the equivalent resistance of the parallel combination:

\begin{marshal*} \frac{1}{R_{p}} & = \frac{1}{R_{2}} + \frac{1}{R_{3}} \\ & = \frac{one}{i} + \frac{1}{1} \\ & = ii\\ R_{p} & = \text{0,five}\text{ Ω} \end{align*}

Now nosotros have a circuit with two resistors in serial so we can summate the equivalent resistance:

\brainstorm{marshal*} R_{s} & = R_{1} + R_{4} + R_{p} \\ & = \text{2} + \text{1,5} + \text{0,5} \\ & = \text{iv}\text{ Ω} \cease{align*}

So the current through the jail cell is:

\begin{align*} I & = \frac{5}{R} \\ & = \frac{\text{10}}{\text{4}} \\ & = \text{two,5}\text{ A} \stop{align*}

the voltage drop across \(R_4\)

The current through all the resistors is \(\text{two,5}\) \(\text{A}\). (The current is the same through series combinations of resistors and we tin can consider the entire parallel set up of resistors as one series resistor.)

Using this we can find the voltage through \(R_{4}\):

\brainstorm{align*} Five & = I \cdot R \\ & = (\text{2,5})(\text{1,v}) \\ & = \text{iii,75}\text{ V} \end{align*}

the current through \(R_2\)

The current through all the resistors is \(\text{ii,5}\) \(\text{A}\). (The current is the same through series combinations of resistors and we can consider the entire parallel set of resistors as one serial resistor.)

Using this nosotros can find the current through \(R_{2}\).

We first need to find the voltage across the parallel combination:

\brainstorm{align*} V & = I \cdot R \\ & = (\text{two,v})(\text{0,5}) \\ & = \text{1,25}\text{ 5} \end{align*}

Now nosotros can observe the current through \(R_{2}\) using the fact that the voltage is the same across each resistor in the parallel combination:

\begin{align*} I & = \frac{V}{R} \\ & = \frac{\text{1,25}}{\text{1}} \\ & = \text{1,25}\text{ A} \end{align*}

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