Ohm's law: Formula, Graph and 3 Easy Examples
Important Formula
What do the formula symbols mean?
Voltage
$$ U $$ Unit $$ \mathrm{V} = \frac{ \mathrm J }{ \mathrm C } = \frac{ \mathrm{kg} \, \mathrm{m}^2 }{ \mathrm{A} \, \mathrm{s}^3 } $$In order for an electric current \(I\) to flow between two points on the conductor, positive and negative charges must be separated, that is there must be a voltage between these points. The mutual attraction of the opposite charges creates an electric current.
Electrical Resistance
$$ \class{brown}{R} $$ Unit $$ \mathrm{\Omega} = \frac{ \mathrm{kg} \, \mathrm{m}^2 }{ \mathrm{A}^2 \, \mathrm{s}^3 } $$The Ohm's law is characterized by the fact that the resistance \(R\) is constant! So it doesn't matter what voltage is applied, the current through the conductor will adjust so that the ratio \(U/I\), that is the resistance, always remains constant.
At a current of \(I = 0.1 \, \text{A} \) and a voltage of \( U = 10 \, \text{V}\), the resistance is \( R = 100 \, \Omega \).
Electric current
$$ \class{red}{\boldsymbol I} $$ Unit $$ \mathrm{A} = \frac{ \mathrm C }{ \mathrm s } $$At a current of \(1 \, \text{A}\) (1 amp), \(1 \, \text{C}\) (1 coulomb) of charge flows through the conductor per second.
With a resistance of \(R = 10 \, \Omega \) and a voltage of \( U = 1 \, \text{V}\), a current of \( I = 0.1 \, \text{A} \) flows.
Table of contents
 Important Formula
 The 1st ingredient: Voltage
 The 2nd ingredient: Electrical conductor
 The 3rd ingredient: Electric current
 Ohm's law as a graph Here you will learn how to identify Ohm's law in a currentvoltage diagram.
 Ohm's law as a formula Here you will learn about the famous URI formula, how you can remember it easily and how it comes about.
 Electrical resistance Here you will learn what the unit of electrical resistance is and what it has to do with conductors.
 3 Examples of how the URI formula is applied Here you will learn with three examples how to calculate voltage, current or resistance when two other quantities are given.
 Exercises with Solutions
Ohm's law, named after the German physicist Georg Simon Ohm, describes how electric current \(I\) and voltage \(U\) are related.
So if you want to understand and build up electric circuits, you won't get very far without Ohm's Law. So it's worth to understand this law. Especially because it is the simplest physical law you can imagine.
In previous lessons, you learned what electric current and voltage physically mean. Here is a quick recap!
The 1st ingredient: Voltage
If you separate positive and negative electric charges and put them in two boxes, the bunch of negative charges forms a minus pole and the bunch of positive charges forms a plus pole.
A voltage is created between the negative and positive poles. The voltage is abbreviated by the letter \(U\) and is sketched like this, for example. The voltage tells how much kinetic energy a positive charge would gain if it traveled from the positive to the negative pole. Voltage is measured in volts, abbreviated with a \(\text{V}\). Let's assume that there is a voltage of \( 10 \, \text{V} \) between the poles.
The 2nd ingredient: Electrical conductor
We can't do much with voltage alone. Next, we need an electrical conductor. This could be a wire made of copper or aluminum, for example. We need this conductor to connect the two poles with each other. Of course, it makes a difference whether we use copper or aluminum for the connection, because different materials conduct the charges differently. But more about that later.
The 3rd ingredient: Electric current
Since the two poles are now conductively connected, the positive charges can move to the negative pole along the conductor. They are eventually attracted by the negative charges. Thus, there is an electric current \(I\) through this conductor. The negative charges, of course, would also travel to the positive pole. However, we have fixed them in the box so that they cannot move, in order not to make our thought experiment unnecessarily complicated.
The resulting electric current \(I\) is measured in amperes (or short amps), abbreviated with the letter \(\text{A}\). Let us assume that a current of \( 1 \, \text{A}\) flows through the conductor.
Over time, the number of positive charges at the positive pole will decrease because the charges are moving to the negative pole all the time. Thus, the voltage and current would also decrease over time because fewer charges are separated. To prevent this, we will constantly supply the charges to maintain the charge separation. This resupply of charges is the purpose of a voltage source.
With a voltage source, we make sure that the voltage and current stay constant and don't get lower.
Ohm's law as a graph
Our voltage between the poles is set to \( 10 \, \text{V}\) and a current of \( 1 \, \text{A}\) flows. Now let's change the voltage and see how this affects the electric current. We can change voltage by changing the number of separated charges.

If we increase the voltage from \( 10 \, \text{V}\) to \( 20 \, \text{V}\), then the current increases from \( 1 \, \text{A}\) to \( 2 \, \text{A}\). We have doubled the voltage \(U\) and thus the current \(I\) has also doubled.

If we increase the voltage from \( 20 \, \text{V}\) to \( 60 \, \text{V}\), then the current increases from \( 2 \, \text{A}\) to \( 6 \, \text{A}\). We have tripled the voltage \(U\) and thus the current \(I\) has also tripled.

If we decrease the voltage from \( 60 \, \text{V}\) to \( 40 \, \text{V}\), then the current decreases from \( 6 \, \text{A}\) to \( 4 \, \text{A}\). We have decreased the voltage \(U\) by the factor \(\frac{2}{3}\) and thus the current \(I\) has also decreased by the factor \(\frac{2}{3}\).
No matter by which factor we change the voltage, the current also changes by the same factor.
We can illustrate the measured values in a diagram. On the \(y\) axis, that is the vertical axis, we plot the voltage. And on the \(x\) axis, that is the horizontal axis, we plot the current. This graph is called voltagecurrent graph because it illustrates the relationship between voltage and current.
If we now connect the measured data points with each other, we get a straight line. Whenever a straight line comes out on a plot, then we say that the plotted quantities are related linearly to each other. So we have found a law that voltage and current are linearly related. And this is exactly the statement of Ohm's law!
What if we plot our voltage and current measurements and get such a graph like in illustration (4)?
Does this conductor, which connects the poles and through which a current flows, fulfill Ohm's law? No, it does not! Because it is not a straight line. A conductor fulfills Ohm's law only if the voltagecurrent graph results in a straight line. An electric current through such conductors, which give a straight line, are called ohmic conductors, because they fulfill exactly Ohm's law. For example, a conductor made of copper is an ohmic conductor. This is because if we apply a voltage to the copper conductor and a current flows as a result, then the current and voltage are linearly related.
The straight currentvoltage line is represented by its slope. We denote the slope as electrical resistance and abbreviate it with the letter \(R\):
We can therefore say:

A steep straight line has a large slope and thus represents a large electrical resistance.

A shallow straight line has a small slope and thus represents a small electrical resistance.
What effect does the slope of the straight line have on the voltage and current?

A shallow straight line, that is a small resistance, means: If you increase the voltage \(U\) only very slightly, then the current \(I\) increases very much.

On the other hand, a steep straight line, that is a large resistance, means: If you increase the voltage \(U\) just a little bit, then the current \(I\) also increases only very slightly.
Ohm's law as a formula
How can we now unite the three quantities, voltage \(U\), current \(I\) and resistance \(R\) in one formula? Let's use the language of physics, that is mathematics, to translate this straight line into a formula.
From mathematics we know that a straight line passing through the origin of the coordinate system is described by the equation of a straight line:
Here \(m\) is the slope of the straight line. In our case the slope of the resistance is \(R\). So let's replace \(m\) with \(R\): \( y = R \, x \). The \(y\) is at the \(y\) axis and is the voltage \(U\). So we replace the \(y\) with \(U\): \( U = R \, x \). The \(x\) is on the \(x\) axis and in our case represents the current \(I\). Let's replace the \(x\) with \(I\). And we have a straight line in the diagram translated into a formula. So Ohm's law as a formula is:
Electrical resistance
You can find out the unit of resistance from the formula for Ohm's law. You only have to solve the formula for the resistance \(R\). Bring \(I\) to the other side and you get:
The voltage has the unit Volt and the current has the unit Ampere. The resistance must therefore have the unit Volt per Ampere:
&~=~ \frac{\mathrm{V}}{\mathrm{A}} \end{align} $$
We abbreviate Volts per Ampere briefly with the unit Ohm:
By the way, this character \(\Omega\) is a Greek letter 'Omega'. The value of this electrical resistance \(R\) depends on the conductor used to connect the positive and negative poles.

A conductor made of copper has a lower resistance \(R\) than a conductor made of aluminum. For the copper conductor we expect a shallower straight line than for the aluminum conductor.

And a conductor made of aluminum has a smaller resistance \(R\) than a conductor made of iron. So an iron conductor has a steeper straight line than an aluminum conductor.
3 Examples of how the URI formula is applied
Let's take a look at some concrete examples of how you can apply Ohm's Law formula.
So, that's it! You learned that a conductor obeys Ohm's law if you get a straight line in the voltagecurrent diagram. In addition, Ohm's law is described by the \( U ~=~ R\,I \) formula. Where the resistance \(R\) is the slope of the straight line.
In the next lesson, you'll learn how to apply Ohm's Law to simple electric circuits.
Exercises with Solutions
Use this formula eBook if you have problems with physics problems.Exercise #1: Voltage across the Conductor
What voltage must you apply to a conductor so that a current of \(0.3\, \text{A}\) flows through this conductor? (The resistance of the conductor is \(200\,\Omega\)).
Solution to Exercise #1
The resistance \( R \) of the conductor is: \(R = 200 \, \Omega\). The desired current \(I\) through the conductor is: \( I = 0.3 \, \text{A} \). To find the voltage \(U\) that must be applied between the ends of the conductor, use Ohm's law: 1 \[ U ~=~ R \, I \]
You don't need to rearrange anything here. Simply plug in the resistance \(R\) and the current \(I\) to calculate \(U\): 1.1 \[ U ~=~ 200 \, \Omega ~\cdot~ 0.3 \, \text{A} ~=~ 60 \, \text{V} \] Here we have used the fact that the unit \( \Omega \cdot \text{A} \) (ohmampere) is equivalent to the unit \(\text{V}\) (volt).
You need to apply \(60 \, \text{V}\) across the two ends of the conductor to allow \(0.3 \, \text{A}\) to flow through it.
Exercise #2: Current through the Toaster
The datasheet of the toaster states that it has a resistance of \(50 \, \Omega\). What is the current flowing through the toaster when you plug it into the power outlet? (Voltage at the power outlet is \(230 \, \text{V}\)).
Solution to Exercise #2
Your toaster has a resistance: \( R = 50 \, \Omega\). You connect it to the mains voltage, so the voltage is \( U = 230 \, \text{V}\). To find the electric current \(I\) flowing through the toaster, you need to rearrange Ohm's law for current \(I\): 2 \[ I ~=~ \frac{U}{R} \]
Substitute the given resistance \(R\) and voltage \(U\): 2.1 \[ I ~=~ \frac{ 230 \, \text{V} }{ 50 \, \Omega } ~=~ 4.6 \, \text{A} \] Here we have used the fact that the unit \(\frac{ \text{V} }{ \Omega }\) (volt per ohm) is equivalent to the unit \(\text{A}\) (ampere).
So, a current of \(4.6 \, \text{A}\) flows through the toaster.
Exercise #3: Resistance of a Bulb
A flashlight is powered by a \(1.5\,\text{V}\) AA battery. As soon as the flashlight is turned on, a current of \(0.006\, \text{A}\) (6 milliamperes) flows through the bulb. What is the resistance of the bulb?
Solution to Exercise #3
A voltage of \(U = 1.5 \, \text{V} \) is applied across the bulb of the flashlight and a current of \( I = 0.006\, \text{A} \) flows. To find the resistance \(R\) of the bulb, you need to rearrange Ohm's law for resistance: 3 \[ R ~=~ \frac{U}{I} \]
Substitute the given voltage \(U\) and current \(I\): 3.1 \[ R ~=~ \frac{ 1.5 \, \text{V} }{ 0.006\, \text{A} } ~=~ 250 \, \Omega \] Here we have used the fact that the unit \(\frac{ \text{V} }{ \text{A} }\) (volt per ampere) is equivalent to the unit \(\Omega\) (ohm).
So, the bulb has a resistance of \( 250 \, \Omega \).
Exercise #4: Resistance of a Device
You want to connect a selfmade electronic device to the power outlet (\(230 \, \text{V}\) mains voltage). However, your device can only withstand a maximum current of \(0.1\, \text{A}\). You definitely don't want to exceed this current to prevent the device from being destroyed. What must be the minimum resistance of the device?
Solution to Exercise #4
The voltage \( U = 230 \, \text{V}\) is given. Also, the current \(I\) is given. It should be at most \( I = 0.1 \, \text{A} \). To determine the minimum resistance \(R\) that the device must have to avoid being destroyed by excessive current, use Ohm's law. Rearrange it for resistance \(R\): 4 \[ R ~=~ \frac{U}{I} \]
Substitute the given voltage \(U\) and maximum current \(I\): 4.1 \[ R ~=~ \frac{ 230 \, \text{V} }{ 0.1\, \text{A} } ~=~ 2300 \, \Omega \] Here we have used the fact that the unit \(\frac{ \text{V} }{ \text{A} }\) (volt per ampere) is equivalent to the unit \(\Omega\) (ohm).
So, your selfmade device must have a resistance of at least \( 2300 \, \Omega \) to avoid being destroyed by excessive current.
Exercise #5: Current through the Human Body
From about \(0.025\, \text{A}\) (25 milliamperes) onwards, the current becomes dangerous for humans. You are standing on an insulating mat and accidentally touch both hands to the power outlet. What current flows through your arms if the resistance of each arm is approximately \(1000\, \Omega\)?
Solution to Exercise #5
You have touched the two poles of the mains voltage. So, the voltage between your hands is \( U = 230 \, \text{V} \). Since you are standing on an insulating mat, the current \(I\) does not flow through your body to the ground but rather from one hand through both arms to the other hand touching the opposite pole of the power outlet. The resistance of your arm is approximately \( 500 \, \Omega\). Since the current flows through both arms, the total resistance of both arms is \( 1000 \, \Omega\) (twice the resistance of one arm).
To find the current flowing through both arms, use Ohm's law. Rearrange it for current \(I\): 5 \[ I ~=~ \frac{U}{R} \]
Substitute the given mains voltage \(U\) and resistance \(R\) of the arms: 5.1 \[ I ~=~ \frac{ 230 \, \text{V} }{ 1000 \, \Omega } ~=~ 0.23 \, \text{A} \] Here we have used the fact that the unit \(\frac{ \text{V} }{ \Omega }\) (volt per ohm) is equivalent to the unit \(\text{A}\) (ampere).
So, a current of \(0.23 \, \text{A}\) flows through your arms. This is \(0.23 \cdot 1000 \, \mathrm{mA} = 230\, \mathrm{mA} \) (milliamperes). The current dangerous for humans starts from 25 milliamperes onwards.
Exercise #6: Current and Resistivity of a Cylindrical Wire
A cylindrical conductor with a radius \(r\), length \(L\), and electrical conductivity \(\sigma\) has a constant potential difference \(U\) between its ends, which causes the electric field inside the conductor to be constant.
 What is the magnitude of the current \(I\) in this section of the wire?
 Find out the material of the conductor. To do this, measure the voltage \(U = 0.055 \, \text{V} \) and current \(I = 10.28 \, \text{A} \) across a length of the conductor of \(L = 1 \, \text{m} \). You have also determined the radius: \( r = 1 \, \text{mm} \). What material is the conductor made of?
Material  Resistivity \( \rho = 1/\sigma \) 

Silicium (Silicon)  2.5·10^{3}Ωm 
Cuprium (Copper)  1.68·10^{8}Ωm 
Pure water  2.5·10^{5}Ωm 
Solution to the Exercise #6.1
Since the electric field in this wire is constant, the current density is also constant: 1 $$ \class{red}{j} ~=~ \sigma \, \class{purple}{E} ~=~ \text{const.}$$
Current is given by : \( \class{red}{I} = \class{red}{j} \, A \). Insert current density: 2 $$\class{red}{I} ~=~ \sigma \, \class{purple}{E} \, A$$
The electric field \(\class{purple}{E}\) can be expressed as potential \(U\) per length \(L\). The crosssectional area through which the current flows is equivalent to the area of the circle: \(A = \pi \, r^2 \). Substituting \(\class{purple}{E}\) and \(A\), we obtain: 3 $$\class{red}{I} ~=~ \frac{\pi \, \sigma \, r^2}{L} \, U$$
The total current \(\class{red}{I}\) flowing from one end to the other end of the wire is therefore proportional to the potential difference \(U\) between its ends. The proportionality factor between current and voltage of a cylindrical wire is \(\frac{\pi\;\sigma\,r^2}{L}\) and is called the electrical conductance \(G\). Its reciprocal \(\frac{1}{G}\) is called resistance \(R\) and depends on the geometry of the conductor (area, length, etc.) and the conductivity of the medium between the ends of the conductor.
Solution to the Exercise #6.2
To determine the material of the cylindrical conductor, we use the derived equation for the current from Exercise #6.1 of the problem. Rearrange equation 3
to solve for the conductivity \(\sigma\):
$$\sigma ~=~ \frac{I \, L}{\pi \, r^2 \, U}$$
The problem statement provides a table of resistivities \(\rho \). You can obtain the conductivity by taking the reciprocal of the resistivity, that is, \( 1/\sigma = \rho \), to obtain the resistivity: $$\rho ~=~ \frac{\pi \, r^2 \, U}{I \, L}$$
Insert all given values and you will get a concrete value for the resistivity: $$\begin{align}\rho ~&=~ \frac{ \pi \cdot 0.001\,\mathrm{m}^2 \cdot 0.055\,\mathrm{V} }{ 10.28\,\mathrm{A} \cdot 1\,\mathrm{m}} \\\\ ~&=~ 1.68 \cdot 10^{8}\,\frac{\mathrm{Vm}}{\mathrm A}\end{align}$$
According to the table, the conductor is called Cuprium (commonly known as copper), with a resistivity of \( \rho = 1.68 \cdot 10^{8} \, \Omega \mathrm{m} \).
It is a conductor called cuprium ( commonly known as copper), with resistivity: \( \rho = 1.68 \cdot 10^{8} \, \Omega \mathrm{m} \).
Exercise #7: Amplitude and Velocity of NaCl in TimeDependent Electric Field
A sodium chloride solution with a charge carrier density of \( n = 10^{25} \, \text{1}/\text{m} \) is placed in an electric timedependent field with amplitude \( \class{purple}{E_0} = 4000 \, \text{V}/\text{m} \) and frequency \( f = 40 \, \text{Hz} \). The electrical resistivity of the NaCl solution is \( \rho = 1 \, \Omega \text{m}\).
 What is the maximum velocity \(v_0\) of displacement?
 What is the maximum displacement \(a_0\) between Na and Cl?
Hint: Use Ohm's law in differential form: $$ j ~=~ \sigma \, \class{purple}{E} $$ And the relationship between current density \(j\) and velocity \(v\).
Solution to Exercise #7.1
To find out the velocity at which the Na and Cl ions are displaced, the relationship between velocity and the alternating electric field must be established. For this purpose, Ohm's law in differential form, which relates the electric current density \(j\) and the electric field \(\class{purple}{E}\) through the specific conductivity \(\sigma\), is used: 1 \[ j = \sigma \, \class{purple}{E} \]
Now, the current density \(j\) is expressed in terms of velocity \(v\) and charge density \( n \, e \), and the conductivity \(\sigma\) is the reciprocal of electrical resistivity \( \rho \): 2 \[ n\,e\,v ~=~ \frac{1}{\rho} \, \class{purple}{E} \]
Since it is an alternating electric field, \(\class{purple}{E}\) is not constant but timedependent and periodic with angular frequency \(\omega\): 3 \[ n\,e\,v ~=~ \frac{1}{\rho} \, \class{purple}{E_0} \, \cos(\omega \, t) \] Here, it doesn't matter whether the alternating field is described using cosine or sine.
Now, Eq. 3
can be rearranged for velocity:
4
\[ v(t) ~=~ \frac{\class{purple}{E_0}}{\rho \, n \, e} \, \cos(\omega \, t) \]
As seen in 4
, velocity oscillates. The maximum velocity \(v_0\) is the amplitude of the oscillation, that is, the maximum value of Eq. 4
. And that is precisely the value when the cosine is maximum, \( \cos(\omega t) = 1 \):
5
\[ v_0 = \frac{\class{purple}{E_0}}{\rho \, n \, e} \]
Substituting the concrete values yields: 6 \[ v_0 = 2.5 \, \frac{\text{mm}}{\text s} \]
Solution to Exercise #7.2
To determine the maximum displacement between the Na and Cl ions, the velocity function determined in Exercise #7.1 (see Eq. 4
) needs to be integrated:
7
\[ a(t) = \int v(t) \text{d}t \]
Substituting 4
into 7
and pulling out the timeindependent factor \(v_0\):
8
\[ a(t) = \frac{\class{purple}{E_0}}{\rho \, n \, e} \, \int \cos(\omega \, t) \text{d}t \]
The integration results in: 9 \[ a(t) = \frac{\class{purple}{E_0}}{\rho \, n \, e} \, \frac{1}{\omega} \, \sin(\omega \, t) \]
The amplitude, that is, the maximum displacement, occurs when the sine is equal to one: 10 \[ a_0 = \frac{\class{purple}{E_0}}{\rho \, n \, e} \, \frac{1}{\omega} \]
Now, only the unknown angular frequency \( \omega \) needs to be replaced with the known frequency \(f\): 11 \[ a_0 = \frac{\class{purple}{E_0}}{\rho \, n \, e} \, \frac{1}{2\pi \, f} \]
Substituting the concrete values yields: \( a_0 = 10 \, \mu\text{m} \).
Exercise #8: Resistance of a Wire with a Variable CrossSection
Consider a wire of length \(1 \, \text{m}\). Apply a voltage of \(12 \, \text{V}\) between the ends of the wire. The crosssection of the wire is not constant along its entire length; rather, it decreases from one end to the other. The diameter of the largest crosssection is \(d_1 = 2 \, \text{mm}\), and the diameter of the smallest crosssection is \(d_2 = 1 \, \text{mm}\). The electrical resistivity of the wire is \( \rho = 8.7 \cdot 10^{8} \, \Omega\text{m} \).
 What is the total resistance \(R\) of the wire?
 What is the total power \(P\) of the wire?
Tips: For Exercise #8.1: Use the relationship between resistance \(R\) and electrical resistivity \(\rho\): $$ R ~=~ \rho \, \frac{L}{A(x)} $$
Here, \(A(x)\) represents the variable crosssectional area of the wire, depending on the distance \(x\). Then, consider an infinitesimal length element of the wire in the above equation. Determine \(A(x)\).
For Exercise #8.2: Calculate the current from the resistance found in subtask #8.1, and then calculate the total electrical power dissipated in the wire.
Solution to Exercise #8.1
The resistance \(R\) is related to the (given) electrical resistivity \(\rho\) through the geometry factor \(L/A\): 1 \[ R = \rho \, \frac{L}{A} \]
Here, \(L\) is the length of the wire and \(A\) is its crosssectional area. However, the crosssection is not constant; it depends on which point of the wire is being considered. For this, the xaxis is laid along the wire. Then, the crosssection depends on \(x\) and thus \(R(x)\) also depends on \(x\): 2 \[ R(x) = \rho \, \frac{L}{A(x)} \]
However, we're not looking for the resistance at a specific point \(x\), but rather the total resistance \(R\) of the wire. For this, consider an infinitesimal wire segment \( \text{d}x \), contributing resistance \( \text{d}R \) to the total resistance: 3 \[ \text{d}R = \rho \, \frac{ \text{d}x }{A(x)} \]
The goal now is to determine the function \(A(x)\) because it's still unknown. The crosssectional area is the area of a circle, but with a variable radius \(r(x)\): 4 \[ A(x) = \pi \, r(x)^2 \]
However, the given value is the diameter, not the radius. Therefore, Eq 4
is rewritten using the diameter function:
5
\[ A(x) = \frac{\pi}{4} \, d(x)^2 \]
The diameter must be a linear function because the wire diameter decreases uniformly. At \(d(0)\), the diameter is \(d_1\), and at \(d(L)\), the end of the wire has diameter \(d_2\). Between these points, the diameter function is a straight line! It has a negative slope \( (d_1  d_2) / L \): 6 \[ d(x) = d_1  \frac{d_1  d_2}{L} \, x \] \[ d(x) = d_1 + \frac{d_2  d_1}{L} \, x \]
Thus, \(d(x)\) has been determined and can now be substituted into 5
:
7
\[ A(x) = \frac{\pi}{4} \, \left( d_1 + \frac{d_2  d_1}{L} \, x \right)^2 \]
With the determined crosssectional area 7
, equation 3
is ready for integration:
8
\[ R = \rho \, \int_{0}^{L} \frac{1}{A(x)} \ \text{d}x \]
9
\[ R = \frac{4\rho}{\pi} \, \int_{0}^{L} \left( d_1 + \frac{d_2  d_1}{L} \, x \right)^{2} \ \text{d}x \]
To keep equation 9
compact, set \( a:= (d_2  d_1) / L \):
10
\[ R = \frac{4\rho}{\pi} \, \int_{0}^{L} \left( d_1 + a \, x \right)^{2} \, \text{d}x \]
The integration yields: 11 \[ R = \frac{4\rho}{\pi} \, \left[ \frac{1}{a} \, \left( d_1 + a \, x \right)^{1} \right]^{L}_{0} \]
Substituting the integration limits yields: 12 \[ R = \frac{4\rho}{\pi \, a} \, \left( \frac{1}{d_1 + a \, L}  \frac{1}{d_1} \right) \] \[ R = \frac{4\rho}{\pi \, a} \, \left( \frac{1}{d_1}  \frac{1}{d_1 + a \, L} \right) \]
Combine the fractions inside the brackets into the same denominator: 13 \[ R = \frac{4\rho}{\pi \, a} \, \left( \frac{d_1 + a \, L}{d_1 (d_1 + a \, L)}  \frac{d_1}{d_1 (d_1 + a \, L)} \right) \] \[ R = \frac{4\rho}{\pi \, a} \, \left( \frac{a \, L}{d_1 (d_1 + a \, L)} \right) \] \[ R = \frac{4\rho}{\pi} \, \frac{L}{d_1^2 + d_1 \, a \, L} \] \[ R = \frac{4\rho \, L}{\pi} \, \left( d_1^2 + d_1 \, \frac{d_2  d_1}{L} \, L \right)^{1} \] \[ R = \frac{4\rho \, L}{\pi} \, \left( d_1 \, d_2 \right)^{1} \] 14 \[ R = \frac{4\rho \, L}{\pi} \, \frac{1}{d_1 \, d_2} \]
Substituting specific values yields: \( R = 0.55 \, \Omega \).
Solution to Exercise #8.2
Using the calculated total resistance 15
, the total power dissipated in the wire can be calculated with:
$$ P ~=~ \frac{U^2}{R} $$
Substituting specific values yields: \(P = 261 \, \text{W}\).