Ohm's law: Formula, Graph and 3 Easy Examples

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 \( \Omega \cdot \text{A} \) (ohm-ampere) 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.

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.

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 \( 250 \, \Omega \).

You want to connect a self-made 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 self-made device must have a resistance of at least \( 2300 \, \Omega \) to avoid being destroyed by excessive current.

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.

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.

1. What is the magnitude of the current \(I\) in this section of the wire?
2. 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?

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 cross-sectional 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: \( \rho = 1.68 \cdot 10^{-8} \, \Omega \mathrm{m} \).

A sodium chloride solution with a charge carrier density of \( n = 10^{25} \, \text{1}/\text{m} \) is placed in an electric time-dependent 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 \( \rho = 1 \, \Omega \text{m}\).

1. What is the maximum velocity \(v_0\) of displacement?
2. 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 time-dependent 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 time-independent 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: \( a_0 = 10 \, \mu\text{m} \).

Consider a wire of length \(1 \, \text{m}\). Apply a voltage of \(12 \, \text{V}\) between the ends of the wire. The cross-section of the wire is not constant along its entire length; rather, it decreases from one end to the other. The diameter of the largest cross-section is \(d_1 = 2 \, \text{mm}\), and the diameter of the smallest cross-section is \(d_2 = 1 \, \text{mm}\). The electrical resistivity of the wire is \( \rho = 8.7 \cdot 10^{-8} \, \Omega\text{m} \).

1. What is the total resistance \(R\) of the wire?
2. 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 cross-sectional 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 cross-sectional area. However, the cross-section is not constant; it depends on which point of the wire is being considered. For this, the x-axis is laid along the wire. Then, the cross-section 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 cross-sectional 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 cross-sectional 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: \(P = 261 \, \text{W}\).