How To Find Unknown Resistor In Parallel Circuit

eleven.2 Ohm'south Law (ESBQ6)

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Iii quantities which are primal to electric circuits are current, voltage (potential departure) and resistance. To recap:

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

2. Potential deviation or voltage, \(5\), is the amount of energy per unit of measurement charge needed to move that accuse between ii points in a circuit.

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

We will now expect at how these three quantities are related to each other in electric circuits.

An important human relationship between the current, voltage and resistance in a excursion was discovered by Georg Simon Ohm and information technology is chosen Ohm's Law.

Ohm's Police force

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

\(I = \frac{5}{R}\)

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 abiding, independent of the voltage practical across information technology or current passed through information technology.

Ohm's Law tells us that if a conductor is at a constant temperature, the current flowing through the conductor is directly proportional to the voltage across it. This means that if nosotros plot voltage on the 10-centrality of a graph and current on the y-axis of the graph, we will get a straight-line.

The gradient of the straight-line graph is related to the resistance of the conductor as \[\frac{I}{V} = \frac{1}{R}\] This can exist rearranged in terms of the constant resistance as: \[R = \frac{Five}{I}\]

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Ohm'south Law

Aim

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

Apparatus

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

Method

This experiment has two parts. In the starting time part we volition vary the applied voltage beyond the resistor and measure the resulting current through the circuit. In the 2d part we will vary the electric current in the circuit and mensurate the resulting voltage across the resistor. Afterward obtaining both sets of measurements, we will examine the human relationship between the current and the voltage across the resistor.

1. Varying the voltage:

   1. Set upward the circuit according to circuit diagram 1), starting with just i cell.

   2. Describe the following table in your lab book.

      Number of cells	Voltage, V (\(\text{5}\))	Current, I (\(\text{A}\))
      \(\text{ane}\)
      \(\text{ii}\)
      \(\text{3}\)
      \(\text{4}\)

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

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

   5. Add one more \(\text{1,5}\) \(\text{Five}\) cell to the circuit and repeat your measurements.

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

2. Varying the electric current:

   1. Fix the circuit according to circuit diagram 2), starting with but i resistor in the circuit.

   2. Depict the following table in your lab book.

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

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

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

   5. Now add some other resistor in series in the circuit and measure out the current and the voltage across only the original resistor once again. Keep adding resistors until y'all accept 4 in series, but remember to simply measure the voltage beyond the original resistor each time. Enter the values you mensurate into the table.

Analysis and results

1. Using the information y'all recorded in the outset tabular array, draw a graph of electric 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 current is the dependent variable and must be plotted on the \(y\)-axis.

2. Using the information you lot recorded in the 2d 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 made from the first tabular array. What happens to the current through the resistor when the voltage across information technology is increased? i.e. Does it increment or decrease?

2. Examine the graph y'all made from the second tabular array. What happens to the voltage across the resistor when the electric current increases through the resistor? i.due east. Does information technology increment or decrease?

3. Do your experimental results verify Ohm'southward Police? Explain.

Questions and word

1. For each of your graphs, summate the gradient and from this determine the resistance of the original resistor. Do you lot get the same value when you calculate it for each of your graphs?

2. How would you lot go nigh finding the resistance of an unknown resistor using but a ability supply, a voltmeter and a known resistor \(R_0\)?

Ohm's Police force

Textbook Practice 11.1

Plot a graph of voltage (on the x-axis) and current (on the y-centrality).

What type of graph practise you lot obtain (direct-line, parabola, other curve)

straight-line

Calculate the gradient of the graph.

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

\begin{align*} thousand & = \frac{\Delta I}{\Delta V} \\ & = \frac{(\text{ane,half dozen}) - (\text{0,4})}{(\text{12}) - (\text{three})} \\ & = \text{0,13} \end{align*}

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

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

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

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

Next you connect the two resistors in series. You can now accept the voltage measurements for each of resistors.

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

\[5 = IR\]

And then using this and the fact that for resistors in serial, the current is the same everywhere in the circuit nosotros tin can notice 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{marshal*}

Ohmic and not-ohmic conductors (ESBQ7)

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

As y'all take seen, at that place is a mention of constant temperature when we talk nigh Ohm'southward Law. This is because the resistance of some conductors changes as their temperature changes. These types of conductors are called non-ohmic conductors, considering they do not obey Ohm's Police force. A light bulb is a common example of a non-ohmic conductor. Other examples of non-ohmic conductors are diodes and transistors.

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

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

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

A recommended experiment for informal assessment is included. In this experiment learners will obtain current and voltage data for a resistor and light bulb and determine which obeys Ohm's law. Y'all volition need light bulbs, resistors, connecting wires, power source, ammeter and voltmeter. Learners should find that the resistor obeys Ohm's law, while the light bulb does non.

Ohmic and not-ohmic conductors

Aim

To determine whether 2 circuit elements (a resistor and a lightbulb) obey Ohm's Law

Appliance

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

Method

The ii circuits shown in the diagrams above are the same, except in the showtime at that place is a resistor and in the second there is a lightbulb. Gear up both the circuits above, starting with 1 cell. For each excursion:

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

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

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

Results

Draw two tables which look like the post-obit in your volume. You lot should take ane table for the first circuit measurements with the resistor and another table for the 2nd circuit measurements with the lightbulb.

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

Assay

Using the data in your tables, describe two graphs of \(I\) (\(y\)-axis) vs. \(Five\) (\(ten\)-axis), one for the resistor and ane for the lightbulb.

Questions and Give-and-take

Examine your graphs closely and answer the following questions:

1. What should the graph of \(I\) vs. \(5\) await like for a conductor which obeys Ohm'due south Police force?

2. Exercise either or both your graphs wait similar this?

3. What can you conclude about whether or not the resistor and/or the lightbulb obey Ohm'southward Police?

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

Using Ohm'southward Law (ESBQ8)

We are now fix to see how Ohm's Law is used to analyse circuits.

Consider a excursion with a cell and an ohmic resistor, R. If the resistor has a resistance of \(\text{5}\) \(\text{Ω}\) and voltage beyond the resistor is \(\text{five}\) \(\text{Five}\), so we can employ Ohm'southward Police to summate the electric current flowing through the resistor. Our offset task is to describe the circuit diagram. When solving any problem with electric circuits it is very of import to brand a diagram of the circuit before doing whatever calculations. The excursion diagram for this problem looks like the post-obit:

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

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

The current flowing through the resistor is:

\begin{align*} I &= \frac{5}{R} \\ &= \frac{\text{5}\text{ V}}{\text{five }\Omega} \\ &= \text{i}\text{ A} \end{marshal*}

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Worked example 1: Ohm's Constabulary

Study the excursion diagram below:

The resistance of the resistor is \(\text{10}\) \(\text{Ω}\) and the electric current going through the resistor is \(\text{4}\) \(\text{A}\). What is the potential difference (voltage) across the resistor?

Determine how to approach the problem

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

Solve the problem

Rearrange the equation above and substitute the known values for \(R\) and \(I\) to solve for \(V\). \begin{marshal*} R &= \frac{V}{I} \\ R \times I&= \frac{V}{I} \times I\\ V &= I \times R \\ &= \text{10} \times \text{four} \\ &= \text{40}\text{ V} \end{align*}

Write the final answer

The voltage across the resistor is \(\text{40}\) \(\text{V}\).

Ohm's Law

Textbook Exercise 11.2

Summate the resistance of a resistor that has a potential deviation of \(\text{8}\) \(\text{5}\) beyond it when a current of \(\text{2}\) \(\text{A}\) flows through information technology. Describe the excursion diagram before doing the adding.

The resistance of the unknown resistor is:

\begin{align*} R & = \frac{V}{I} \\ & = \frac{viii}{2} \\ & = \text{four}\text{ Ω} \end{align*}

What current will flow through a resistor of \(\text{6}\) \(\text{Ω}\) when there is a potential difference of \(\text{xviii}\) \(\text{5}\) across its ends? Describe the circuit diagram before doing the calculation.

The resistance of the unknown resistor is:

\begin{align*} I & = \frac{5}{R} \\ & = \frac{eighteen}{6} \\ & = \text{3}\text{ A} \cease{marshal*}

What is the voltage across a \(\text{x}\) \(\text{Ω}\) resistor when a electric current of \(\text{1,5}\) \(\text{A}\) flows though information technology? Draw the circuit diagram before doing the calculation.

The resistance of the unknown resistor is:

\begin{align*} V & = I \cdot R \\ & = (\text{one,v})(ten) \\ & = \text{15}\text{ V} \end{align*}

Epitomize of resistors in serial and parallel (ESBQ9)

In Grade ten, yous learnt most resistors and were introduced to circuits where resistors were continued in series and in parallel. In a series circuit there is one path along which current flows. In a parallel circuit there are multiple paths along which electric current flows.

When at that place is more than one resistor in a circuit, we are usually able to summate the full combined resistance of all the resistors. This is known every bit the equivalent resistance.

Equivalent series resistance

In a circuit 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 series the equivalent resistance is:

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

Let us apply this to the following circuit.

The resistors are in serial, therefore:

\brainstorm{marshal*} R_{s} & = R_{one} + R_{2} + R_{three} \\ & = \text{3}\text{ Ω} + \text{10}\text{ Ω} + \text{5}\text{ Ω} \\ & =\text{18}\text{ Ω} \stop{align*}

Equivalent parallel resistance

In a excursion where the resistors are continued in parallel, the equivalent resistance is given past the post-obit definition.

Equivalent resistance in a parallel excursion

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

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

Let us use this formula to the following circuit.

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

\brainstorm{align*} \frac{1}{R_{p}} & = \left( \frac{i}{R_{1}} + \frac{1}{R_{ii}} + \frac{1}{R_{3}} \right) \\ & = \left( \frac{i}{\text{10}\text{ Ω}} + \frac{one}{\text{ii}\text{ Ω}} + \frac{1}{\text{1}\text{ Ω}} \right) \\ & = \left( \frac{\text{1}\text{ Ω} + \text{5}\text{ Ω} + \text{x}\text{ Ω}}{\text{10}\text{ Ω}} \correct) \\ & = \left( \frac{\text{16}\text{ Ω}}{\text{10}\text{ Ω}} \right) \\ R_{p} & = \text{0,625}\text{ Ω} \end{align*}

Series and parallel resistance

Textbook Practice 11.3

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

Since the resistors are in series we can use:

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

The equivalent resistance is:

\begin{align*} R_{s} & = R_{1} + R_{2} \\ & = \text{10}\text{ kΩ} + \text{10}\text{ kΩ} \\ & = \text{20}\text{ kΩ} \end{marshal*}

Ii resistors are connected in serial. The equivalent resistance is \(\text{100}\) \(\text{Ω}\). If i resistor is \(\text{ten}\) \(\text{Ω}\), calculate the value of the 2d resistor.

Since the resistors are in series we can use:

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

The equivalent resistance is:

\brainstorm{align*} R_{s} & = R_{1} + R_{two} \\ R_{ii} & = R_{s} - R_{1} \\ & = \text{100}\text{ Ω} - \text{10}\text{ Ω} \\ & = \text{xc}\text{ Ω} \cease{align*}

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

Since the resistors are in parallel we tin can use:

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

The equivalent resistance is:

\begin{marshal*} \frac{one}{R_{p}} & = \frac{i}{R_{i}} + \frac{i}{R_{2}} \\ & = \frac{1}{\text{100}} + \frac{ane}{\text{10}} \\ & = \frac{ane + 10}{\text{100}} \\ & = \frac{11}{\text{100}} \\ R_{p} & = \text{9,09}\text{ kΩ} \end{align*}

Two resistors are continued in parallel. The equivalent resistance is \(\text{3,75}\) \(\text{Ω}\). If one resistor has a resistance of \(\text{10}\) \(\text{Ω}\), what is the resistance of the 2d resistor?

Since the resistors are in parallel we can use:

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

The equivalent resistance is:

\begin{marshal*} \frac{1}{R_{p}} & = \frac{1}{R_{one}} + \frac{1}{R_{2}} \\ \frac{one}{R_{2}} & = \frac{i}{R_{p}} - \frac{ane}{R_{i}} \\ & = \frac{1}{\text{three,75}} - \frac{1}{\text{10}} \\ & = \frac{\text{ten} - \text{3,75}}{\text{37,5}} \\ & = \frac{\text{vi,25}}{\text{37,5}} \\ R_{ii} & = \text{6}\text{ Ω} \stop{marshal*}

Summate the equivalent resistance in each of the following circuits:

a) The resistors are in parallel and so we employ:

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

The equivalent resistance is:

\begin{align*} \frac{ane}{R_{p}} & = \frac{1}{R_{i}} + \frac{1}{R_{2}} \\ & = \frac{1}{\text{3}} + \frac{ane}{\text{2}} \\ & = \frac{\text{ii} + \text{iii}}{\text{half-dozen}} \\ & = \frac{\text{5}}{\text{6}} \\ R & = \text{1,ii}\text{ Ω} \end{align*}

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

\[\frac{1}{R_{p}} = \frac{1}{R_{i}} + \frac{ane}{R_{two}} + \frac{1}{R_{3}} + \frac{i}{R_{4}}\]

The equivalent resistance is:

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

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

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

The equivalent resistance is:

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

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

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

The equivalent resistance is:

\begin{align*} R_{s} & = R_{ane} + R_{2} + R_{3} + R_{4} \\ & = \text{2}\text{ Ω} + \text{3}\text{ Ω} + \text{iv}\text{ Ω} + \text{ane}\text{ Ω} \\ & = \text{10}\text{ Ω} \finish{align*}

Apply of Ohm'south Law in series and parallel circuits (ESBQB)

Using the definitions for equivalent resistance for resistors in series or in parallel, we 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 near series circuits is that the amount of current is the same through any component in the circuit. This is because there is only one path for electrons to flow in a serial circuit. From the way that the bombardment is connected, we can tell in which management the electric current will period. Nosotros know that current flows from positive to negative past convention. Conventional current in this circuit will menstruation in a clockwise direction, from bespeak A to B to C to D and dorsum to A.

We know that in a serial circuit the current has to exist the aforementioned in all components. And so we can write:

\[I = I_{i} = I_{ii} = I_{3}.\]

We likewise know that total voltage of the circuit has to be equal to the sum of the voltages over all three resistors. So we tin can write:

\[V = V_{one} + V_{two} + V_{three}\]

Using this information and what we know most computing the equivalent resistance of resistors in series, we can approach some circuit issues.

Worked example 2: Ohm's Law, series excursion

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

Determine what is required

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

Decide how to approach the problem

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

Detect full resistance in circuit

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

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

Therefore,

\begin{align*} R & = \text{two} + \text{4} \\ & = \text{6}\text{ Ω} \stop{align*}

Apply Ohm's Police

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

Write the terminal answer

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

Worked example 3: Ohm's Constabulary, serial excursion

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

Draw the circuit and fill in all known values.

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

We can use Ohm'southward Law to find the full resistance R in the excursion, and then calculate the unknown resistance using:

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

because it is in a series circuit.

Find the total resistance

\begin{marshal*} R & = \frac{V}{I} \\ & = \frac{\text{1,5}}{\text{0,25}} \\ & = \text{half dozen}\text{ Ω} \stop{align*}

Observe the unknown resistance

We know that:

\[R = \text{half-dozen}\text{ Ω}\]

and that

\[R_{ane} = \text{i}\text{ Ω}\]

Since

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

Therefore,

\[R_{1} = \text{5}\text{ Ω}\]

Worked example 4: Ohm's Law, series excursion

For the following circuit, summate:

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

2. the resistance of \(R_3\).

Determine how to approach the problem

We are given the voltage across the cell and the current in the circuit, as well as the resistances of 2 of the three resistors. We tin can employ Ohm's Law to calculate the voltage drop beyond the known resistors. Since the resistors are in a series excursion the voltage is \(V = V_1 + V_2 + V_3\) and we can calculate \(V_3\). Now we can apply this information to observe the voltage across the unknown resistor \(R_3\).

Calculate voltage drop across \(R_1\)

Using Ohm's Law: \brainstorm{marshal*} 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{two}\text{ 5} \end{marshal*}

Calculate voltage drop beyond \(R_2\)

Again using Ohm's Police force: \brainstorm{align*} 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{ Five} \end{marshal*}

Calculate voltage driblet across \(R_3\)

Since the voltage drop across all the resistors combined must exist the aforementioned as the voltage drop across the cell in a series circuit, nosotros tin can find \(V_3\) using: \begin{align*} V &= V_1 + V_2 + V_3\\ V_3 &= V - V_1 - V_2 \\ &= 18 - ii - 6 \\ V_3&= \text{x}\text{ V} \end{marshal*}

Find the resistance of \(R_3\)

We know the voltage beyond \(R_3\) and the current through it, so we can use Ohm's Police to calculate the value for the resistance: \brainstorm{marshal*} R_3 &= \frac{V_3}{I}\\ &= \frac{10}{2} \\ R_3&= \text{5 } \Omega \terminate{align*}

Write the terminal answer

\(V_1 = \text{two}\text{ Five}\)

\(V_2 = \text{vi}\text{ V}\)

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

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

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Parallel circuits

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

The first principle to understand about parallel circuits is that the voltage is equal across all components in the circuit. This is considering there are only two sets of electrically common points in a parallel excursion, and voltage measured betwixt sets of common points must e'er be the same at whatsoever given time. And then, for the circuit shown, the post-obit is true:

\[V = V_{1} = V_{2} = V_{iii}.\]

The 2d principle for a parallel circuit is that all the currents through each resistor must add up to the total current in the circuit:

\[I = I_{1} + I_{2} + I_{3}.\]

Using these principles and our knowledge of how to calculate the equivalent resistance of parallel resistors, we can at present approach some circuit problems involving parallel resistors.

Worked example five: Ohm'south Law, parallel circuit

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

Determine what is required

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

Make up one's mind how to approach the problem

Since the resistors are ohmic in nature, we tin can employ Ohm's Police. There are however 2 resistors in the excursion and we need to find the full resistance.

Find the equivalent resistance in circuit

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

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

Apply Ohm'south Law

\brainstorm{marshal*} R&= \frac{Five}{I} \\ R \cdot \frac{I}{R} &= \frac{V}{I} \cdot \frac{I}{R} \\ I &= \frac{V}{R}\\ I &= V \cdot \frac{i}{R}\\ &= (12) \left(\frac{3}{4}\right) \\ &= \text{9}\text{ A} \end{marshal*}

Write the final answer

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

Worked example 6: Ohm's Police, parallel circuit

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

Determine what is required

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

Decide how to approach the problem

Since the resistors are ohmic and we are given the voltage beyond the cell and the current through the cell, we tin employ Ohm's Law to find the equivalent resistance in the excursion. \begin{align*} R & = \frac{V}{I} \\ & = \frac{9}{\text{four,8}} \\ & = \text{i,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 notice the resistance of \(R_2\). \begin{align*} \frac{1}{R} & = \frac{one}{R_1} + \frac{1}{R_2} \end{align*} Rearranging to solve for \(R_2\): \brainstorm{align*} \frac{one}{R_2} & = \frac{1}{R} - \frac{1}{R_1} \\ & = \frac{ane}{\text{1,875}} - \frac{1}{three}\\ & = \text{0,2} \\ R_2 & = \frac{ane}{\text{0,2}} \\ & = \text{v} \ \Omega \end{marshal*}

Write the final answer

The resistance \(R_2\) is \(\text{5}\) \(\Omega\)

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Worked example seven: Ohm'south Constabulary, parallel excursion

An 18 volt prison cell is connected to two parallel resistors of \(\text{iv}\) \(\Omega\) and \(\text{12}\) \(\Omega\) respectively. Calculate the current through the cell and through each of the resistors.

First describe the circuit before doing whatever calculations

Determine how to approach the problem

We need to determine the current through the cell and each of the parallel resistors. We have been given the potential difference across the cell and the resistances of the resistors, and so nosotros can use Ohm's Law to summate the current.

Calculate the current through the jail cell

To summate the current through the cell nosotros starting time demand to decide the equivalent resistance of the rest of the excursion. The resistors are in parallel and therefore: \begin{align*} \frac{1}{R} &= \frac{1}{R_1} + \frac{1}{R_2} \\ &= \frac{ane}{4} + \frac{1}{12} \\ &= \frac{3+1}{12} \\ &= \frac{four}{12} \\ R &= \frac{12}{iv} = \text{3} \ \Omega \end{marshal*} Now using Ohm's Police to find the electric current through the cell: \begin{align*} R &= \frac{5}{I} \\ I &= \frac{V}{R} \\ &= \frac{18}{three} \\ I &= \text{6}\text{ A} \end{align*}

Now determine the current through one of the parallel resistors

We know that for a purely parallel circuit, the voltage across the jail cell is the same equally the voltage across each of the parallel resistors. For this circuit: \brainstorm{align*} V &= V_1 = V_2 = \text{18}\text{ 5} \end{align*} Let'south showtime 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{18}{4} \\ I_1 &= \text{4,5}\text{ A} \cease{marshal*}

Summate the electric current through the other parallel resistor

We can apply Ohm's Law again to discover the current in \(R_2\): \begin{align*} R_2 &= \frac{V_2}{I_2} \\ I_2 &= \frac{V_2}{R_2} \\ &= \frac{xviii}{12} \\ I_2 &= \text{i,5}\text{ A} \stop{marshal*} An alternative method of calculating \(I_2\) would have been to use the fact that the currents through each of the parallel resistors must add upwards to the total electric current through the jail cell: \begin{align*} I &= I_1 + I_2 \\ I_2 &= I - I_1 \\ &= 6 - 4.5 \\ I_2 &= \text{1,5}\text{ A} \end{align*}

Write the concluding answer

The current through the cell is \(\text{half dozen}\) \(\text{A}\).

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

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

Ohm'southward Police force in series and parallel circuits

Textbook Exercise 11.4

Summate the value of the unknown resistor in the excursion:

Nosotros first use Ohm's law to summate the total series resistance:

\begin{align*} R & = \frac{V}{I} \\ & = \frac{nine}{i} \\ & = \text{nine}\text{ Ω} \end{align*}

Now we can discover the unknown resistance:

\brainstorm{marshal*} R_{south} & = R_{1} + R_{2} + R_{3} + R_{4} \\ R_{4} & = R_{due south} - R_{one} - R_{ii} - R_{3} \\ & = ix - 3 - three - one \\ & = \text{2}\text{ Ω} \end{align*}

Calculate the value of the electric current in the following excursion:

We showtime find the total resistance:

\begin{align*} R_{s} & = R_{1} + R_{2} + R_{3} \\ & = \text{ane} + \text{2,5} + \text{one,v} \\ & = \text{5}\text{ Ω} \terminate{marshal*}

Now we can calculate the current:

\begin{marshal*} I & = \frac{V}{R} \\ & = \frac{9}{5} \\ & = \text{ane,8}\text{ A} \end{align*}

Three resistors with resistance \(\text{1}\) \(\text{Ω}\), \(\text{v}\) \(\text{Ω}\) and \(\text{ten}\) \(\text{Ω}\) respectively, are connected in series with a \(\text{12}\) \(\text{5}\) bombardment. Calculate the value of the current in the circuit.

We draw the circuit diagram:

We now find the total resistance:

\begin{marshal*} R_{s} & = R_{1} + R_{ii} + R_{iii} \\ & = \text{1} + \text{v} + \text{10} \\ & = \text{16}\text{ Ω} \end{align*}

Now we tin calculate the current:

\begin{align*} I & = \frac{5}{R} \\ & = \frac{12}{16} \\ & = \text{0,75}\text{ A} \terminate{align*}

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

We first discover the total resistance:

\brainstorm{align*} \frac{1}{R_{p}} & = \frac{1}{R_{1}} + \frac{one}{R_{ii}} \\ & = \frac{1}{\text{1}} + \frac{1}{\text{three}} \\ & = \frac{3 + 1}{\text{3}} \\ & = \frac{4}{\text{3}} \\ & = \text{0,75}\text{ Ω} \end{align*}

Now we can summate the electric current:

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

Calculate the value of the unknown resistor \(R_{4}\) in the circuit:

We first notice the total resistance:

\brainstorm{marshal*} R & = \frac{Five}{I} \\ & = \frac{24}{\text{2}} \\ & = \text{12}\text{ Ω} \end{align*}

Now nosotros can calculate the unknown resistance:

\brainstorm{align*} \frac{1}{R_{p}} & = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{ane}{R_{3}} + \frac{ane}{R_{4}}\\ \frac{ane}{R_{4}} & = \frac{1}{R_{p}} - \frac{1}{R_{i}} - \frac{1}{R_{2}} - \frac{1}{R_{three}}\\ & = \frac{1}{\text{12}} - \frac{i}{\text{120}} - \frac{1}{\text{forty}} - \frac{1}{\text{lx}} \\ & = \frac{10 - 1 - iii - 2}{\text{120}} \\ & = \frac{iv}{\text{120}} \\ & = \text{30}\text{ Ω} \end{marshal*}

the value of the electric current through the battery

We draw a circuit diagram:

To summate the value of the current through the battery nosotros get-go need to summate the equivalent resistance:

\begin{align*} \frac{1}{R_{p}} & = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{ane}{R_{iii}}\\ & = \frac{one}{\text{1}} + \frac{1}{\text{five}} + \frac{i}{\text{10}} \\ & = \frac{10 + 2 + ane}{\text{x}} \\ & = \frac{xiii}{\text{10}} \\ & = \text{0,77}\text{ Ω} \cease{align*}

Now we can summate the current through the battery:

\begin{align*} I & = \frac{5}{R} \\ & = \frac{20}{\text{0,77}} \\ & = \text{26}\text{ A} \end{marshal*}

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

For a parallel circuit the voltage beyond cell is the same every bit the voltage across each of the resistors. For this circuit:

\[V = V_{1} = V_{two} = V_{3} = \text{20}\text{ Five}\]

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

\begin{align*} I & = \frac{V}{R} \\ & = \frac{xx}{\text{i}} \\ & = \text{twenty}\text{ A} \cease{align*}

Adjacent we summate the electric current through \(R_{2}\):

\brainstorm{marshal*} I & = \frac{V}{R} \\ & = \frac{twenty}{\text{five}} \\ & = \text{4}\text{ A} \stop{align*}

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

\begin{marshal*} I & = \frac{V}{R} \\ & = \frac{20}{\text{ten}} \\ & = \text{ii}\text{ A} \cease{align*}

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

Series and parallel networks of resistors (ESBQC)

At present that you know how to handle simple series and parallel circuits, you lot are ready to tackle circuits which combine these ii setups such equally the following circuit:

Figure 11.1: An example of a serial-parallel network. The dashed boxes indicate parallel sections of the circuit.

Information technology is relatively easy to work out these kind of circuits because you use everything y'all have already learnt virtually serial and parallel circuits. The only deviation is that you do it in stages. In Figure eleven.one, the circuit consists of 2 parallel portions that are then in serial with a cell. To work out the equivalent resistance for the excursion, you get-go by calculating the total resistance of each of the parallel portions and so add upwards these resistances in serial. If all the resistors in Figure 11.1 had resistances of \(\text{x}\) \(\text{Ω}\), nosotros can calculate the equivalent resistance of the unabridged circuit.

We start by calculating the total resistance of Parallel Circuit 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{ane}{ten} + \frac{1}{10} \correct)^{-1} \\ &= \left(\frac{1+one}{ten} \right)^{-1} \\ &= \left(\frac{2}{10} \right)^{-1} \\ &= \text{v} \, \Omega \end{marshal*}

We tin similarly summate the total resistance of Parallel Circuit 2: \begin{align*} \frac{i}{R_{p2}} &= \frac{1}{R_3} + \frac{i}{R_4} \\ R_{p2}&= \left(\frac{1}{10} + \frac{i}{10} \right)^{-1} \\ &= \left(\frac{1+one}{10} \right)^{-one} \\ &= \left(\frac{2}{10} \right)^{-1} \\ &= \text{5} \, \Omega \end{align*}

You can at present care for the excursion similar a simple series circuit equally follows:

Therefore the equivalent resistance is: \begin{align*} R &= R_{p1} + R_{p2} \\ &= 5 + 5 \\ &= 10 \, \Omega \end{align*}

The equivalent resistance of the circuit in Figure 11.1 is \(\text{10}\) \(\text{Ω}\).

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Serial and parallel networks

Textbook Exercise eleven.5

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

\begin{marshal*} \frac{1}{R_{p}} & = \frac{i}{R_{1}} + \frac{one}{R_{two}} \\ & = \frac{1}{4} + \frac{ane}{2} \\ & = \frac{iii}{4}\\ R_{p} & = \text{ane,33}\text{ Ω} \end{align*}

At present we have a circuit with two resistors in series so nosotros tin calculate the equivalent resistance:

\begin{marshal*} R_{s} & = R_{3} + R_{p} \\ & = \text{two} + \text{1,33} \\ & = \text{3,33}\text{ Ω} \end{marshal*}

We starting time past determining the equivalent resistance of the parallel combination:

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

Now we have a circuit with 3 resistors in series then we tin can calculate the equivalent resistance:

\brainstorm{align*} R_{s} & = R_{three} + R_{4} + R_{p} \\ & = \text{4} + \text{6} + \text{0,67} \\ & = \text{10,67}\text{ Ω} \end{align*}

We get-go by determining the equivalent resistance of the parallel combination:

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

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

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

the current \(I\) through the cell.

To find the current \(I\) nosotros start need to detect the equivalent resistance. Nosotros start past calculating the equivalent resistance of the parallel combination:

\brainstorm{align*} \frac{one}{R_{p}} & = \frac{one}{R_{1}} + \frac{1}{R_{2}} + \frac{ane}{R_{3}} \\ & = \frac{ane}{3} + \frac{1}{5} + \frac{1}{i} \\ & = \frac{23}{xv}\\ R_{p} & = \text{0,652}\text{ Ω} \end{marshal*}

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

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

So the current through the cell is:

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

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

The current through the parallel combination of resistors is \(\text{4,52}\) \(\text{A}\). (The electric current is the aforementioned through series combinations of resistors and we can consider the unabridged parallel fix of resistors as i serial resistor.)

Using this we tin find the voltage through the parallel combination of resistors (retrieve to employ the equivalent parallel resistance and non the equivalent resistance of the excursion):

\begin{align*} 5 & = I \cdot R \\ & = (\text{4,52})(\text{0,652}) \\ & = \text{2,95}\text{ 5} \end{align*}

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

And then at present we can summate the current through the resistor:

\begin{marshal*} I & = \frac{Five}{R} \\ & = \frac{\text{2,95}}{\text{five}} \\ & = \text{0,59}\text{ A} \cease{align*}

If current flowing through the cell is \(\text{2}\) \(\text{A}\), and all the resistors are ohmic, calculate the voltage beyond the prison cell and each of the resistors, \(R_1\), \(R_2\), and \(R_3\) respectively.

To find the voltage we outset need to observe the equivalent resistance. We start by calculating the equivalent resistance of the parallel combination:

\begin{marshal*} \frac{1}{R_{p}} & = \frac{i}{R_{ii}} + \frac{1}{R_{3}} \\ & = \frac{1}{2} + \frac{1}{4} \\ & = \frac{three}{4}\\ R_{p} & = \text{1,33}\text{ Ω} \end{marshal*}

Now we take a circuit with 2 resistors in serial so nosotros tin calculate the equivalent resistance:

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

So the voltage across the prison cell is:

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

The current through the parallel combination of resistors is \(\text{2}\) \(\text{A}\). (The electric current is the same through serial combinations of resistors and we can consider the unabridged parallel set of resistors every bit one series resistor.)

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

\begin{align*} V & = I \cdot R \\ & = (\text{2})(\text{4,66}) \\ & = \text{ix,32}\text{ Five} \end{align*}

Now we find the voltage across the parallel combination:

\begin{align*} Five & = I \cdot R \\ & = (\text{2})(\text{1,33}) \\ & = \text{two,66}\text{ V} \end{align*}

Since the voltage beyond each resistor in the parallel combination is the same, this is too the voltage across resistors \(R_{ii}\) and \(R_{iii}\).

the electric current through the prison cell

To notice the current we first need to find the equivalent resistance. We outset past calculating the equivalent resistance of the parallel combination:

\begin{align*} \frac{1}{R_{p}} & = \frac{ane}{R_{2}} + \frac{1}{R_{three}} \\ & = \frac{1}{i} + \frac{i}{one} \\ & = 2\\ R_{p} & = \text{0,5}\text{ Ω} \stop{align*}

Now we accept a circuit with 2 resistors in series and then we can calculate the equivalent resistance:

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

So the current through the cell is:

\begin{align*} I & = \frac{V}{R} \\ & = \frac{\text{10}}{\text{4}} \\ & = \text{two,v}\text{ A} \terminate{align*}

the voltage drib beyond \(R_4\)

The current through all the resistors is \(\text{2,5}\) \(\text{A}\). (The current is the same through series combinations of resistors and we tin consider the entire parallel ready of resistors as i series resistor.)

Using this we can observe the voltage through \(R_{four}\):

\begin{marshal*} 5 & = I \cdot R \\ & = (\text{2,5})(\text{1,5}) \\ & = \text{iii,75}\text{ Five} \end{marshal*}

the current through \(R_2\)

The current through all the resistors is \(\text{2,v}\) \(\text{A}\). (The current is the same through serial combinations of resistors and we tin can consider the entire parallel set of resistors every bit one serial resistor.)

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

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

\begin{align*} V & = I \cdot R \\ & = (\text{2,5})(\text{0,5}) \\ & = \text{one,25}\text{ V} \finish{align*}

Now we can find the electric 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{ane,25}}{\text{1}} \\ & = \text{ane,25}\text{ A} \finish{align*}

Source: https://ng.siyavula.com/read/science/grade-11/electric-circuits/11-electric-circuits-02

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