In this explainer, we will learn how to determine the values of electrical quantities in circuits powered by alternating voltages.

The direction of a conventional current in a circuit is from the positive terminal to the negative terminal of a cell, as seen in the diagram below.

If we reverse the battery in this circuit, the direction of the current also reverses, like in the diagram below.

When the current is in one direction, this is called a direct current, or dc. Many electrical systems, including the electricity in homes, do not use dc, but rather ac.

AC is alternating current and differs in that, instead of being in one direction, it alternates its direction. Letβs look at dc for two different circuits in the graph below.

For both circuits, represented by the blue and green lines, the current is not constant, but it is always either positive or negative. This means its direction is always the same.

The same type of graph for an alternating current shows that the current fluctuates between positive and negative, as seen below.

Alternating currents are generated using ac generators that consist of loops of wire rotating at constant speeds through magnetic fields.

A typical ac generator design consists of a loop of wire free to rotate in one axis. A uniform magnetic field B is perpendicular to the wire like in the diagram below.

This rotation through the magnetic field induces an electromotive force, emf, in the loop. This emf creates an alternating current, first in one direction, then switching to the other direction as it rotates. If the loop were to stand still, no emf would be inducedβit must be moving.

To find this induced emf, letβs first look at the emf induced on a simple straight wire with length in a uniform magnetic field. In order to induce an emf in such a wire, this wire must be moving. The diagram below shows a wire of length , indicated by the blue line, moving with a velocity, , indicated by the yellow line. It moves with an angle to the direction of a uniform magnetic field with strength , indicated by the green lines.

The equation that describes the induced emf, , in this wire is where is the magnetic field strength, is the length of the wire, is the velocity of the wire as it travels through the magnetic field, and is the angle the wireβs velocity makes with the direction of the magnetic field.

A straight wire moving with constant velocity is not how ac generators use motion to induce emf though. They have the wire loops within them spin rapidly, meaning that instead of expressing the motion with a velocity , it is expressed as an angular frequency that is measured in radians per second. A conversion between the velocity and angular frequency can be performed using the equation where is the distance from the axis of rotation to the wire.

The diagram below shows a wire spinning in a circle a distance away from the axis of rotation.

This makes the equation for induced emf in a single straight spinning wire but now the angle between the wireβs velocity and the direction of the magnetic field is rapidly changing in the wire as it rotates.

Due to the equationβs relationship with the angle , the induced emf is at its maximum value when the wire is moving perpendicular to the magnetic field direction, and it is at its minimum of 0 when it is moving parallel to the magnetic field direction. The diagram below shows a wire rotating with constant angular speed through a uniform magnetic field at various points.

When the angle is 90 or 270 degrees ( or in radians), the induced emf is at its maximum. When the angle is 0 or 180 degrees (0 or in radians), the induced emf is 0.

When looking at the coils of wire found in an ac generator, only two of the wires will actually count toward the induced emf: the ones on the top and bottom, as the side wires in the loop will have no induced emf regardless of the angle. Note the location of the top and bottom wires of the loop in the diagram below that shows a complete rotation.

This means that, for a coil, double the emf is being induced compared to a straight wire, as there are two wires rather than just one. This creates a factor of 2 in the equation for induced emf when using a coil but this is just for a single loop. When there are multiple loops, the final induced emf is multiplied by , where is the number of loops, giving an equation of

This equation can be further simplified though, by noting that the area, , of a square loop is the product of two of its sides. Since the axis of rotation is through the center of the wires, one side length is , and the other is . This means that the area of the coil, , is which simplifies the equation to

One last step can be done to simplify this equation further, by relating to the angular frequency. The wire loop rotates with a constant angular frequency , which is in units of radians per second. The exact angle can be found then, by just multiplying the angular frequency by the time that has passed which can then be substituted in for in the sine term to give the final equation.

### Definition: Induced emf due to Rotation of a Conducting Coil in a Uniform Magnetic Field

The induced emf, , of a rotating circuit in a uniform magnetic field is where is the number of loops the coil has, is the area of the coil, is the strength of the magnetic field, is the angular frequency, and is time.

Letβs look at an example using this equation.

### Example 1: Induced emf in an Alternating Current Generator

An alternating current generator contains 5 rectangular loops of conducting wire with side lengths 15 cm and 25 cm, the ends of which form terminals. The sides of the loops with the same lengths as each other are parallel to each other. The loops rotate at 15 revolutions per second within a 620 mT uniform magnetic field. What is the peak potential difference across the terminals? Give your answer to two decimal places.

### Answer

Recall the equation for induced emf due to the rotation of a conducting coil:

We want to find the peak potential difference, where the emf is absolutely at its maximum. The maximum occurs when , so we can just substitute that whole term with 1 in the equation for emf:

Now, we have to find the other values. The number of loops is given, 5, as is the magnetic field, 620 mT, but we want it in regular teslas.

There are 1βββ000 milliteslas in 1 tesla: and we can convert 620 mT to teslas by multiplying it by this relation:

So the value of is 0.62 T.

Next, we need to calculate the area of the loops, . We are told that these loops are rectangular, with side lengths 15 cm and 25 cm, and the sides of the same length are parallel to each other. This means that the loop looks like this.

Before going any further, letβs convert these side lengths into metres from centimetres. There are 100 centimetres in 1 metre: so, obtaining the side lengths in metres means multiplying by this relation:

In terms of metres, 15 cm is 0.15 m, and 25 cm is 0.25 m.

The area of a rectangle is the product of two of its nonparallel sides. So we multiply these side lengths together to get the area:

The area of this rectangle is 0.0375 m^{2}.

Finally, we need the angular frequency, . This must be in units of radians per second, so we have to convert from the given value of revolutions per second.

A complete revolution, going all the way around a circle, is . So to put the angular frequency in terms of radians, we have to multiply the revolutions per second by :

So, is moving through Β per second. We now have all the terms we need to complete the equation

We can now substitute in the values. is 5 loops, is
0.0375 m^{2}, is
0.62 T, and is :

We can expand out the unit of teslas. 1 tesla is 1 volt-second per square metre: so using this in the equation gives

Multiplying the angular frequency and magnetic field strength together cancels the units of seconds: and multiplying the last three terms together cancels the units of square metre, leaving behind volts:

Rounded to two decimal places, the peak potential difference across the terminals is 10.96 volts.

The graph below shows the change in induced emf with time.

We see that it is both positive and negative at various points in time, with the highest and lowest points having magnitudes of .

As this graph continues forever in time, the portions when it is positive are canceled completely by the portions where it is negative. This means that the average emf will simply be zero:

This tells us nothing about a circuit though, since it means every single ac circuit will have an average of 0 emf, regardless of the number of loops, area, or magnetic field strength. The way we can instead look at it is by using a root-mean-square. A root-mean-square (rms) is found by taking the square of every possible number at every point on the graph, then finding the mean of all these numbers, then taking the square root of it.

When a negative value is squared, it becomes positive, so the squared values are all positive. This changes the graph to look like the following diagram.

Then all these values are added up and divided by the total number of values to find the mean:

The square root of the mean is then taken:

When this square root is taken of any sinusoidal graph, the final rms value is always of the peak value:

The root-mean-square of the emf is times the highest possible value of emf, called the peak emf. This means that the rms of emf of an alternating current is

Letβs look at an example.

### Example 2: Root-Mean-Square Value of emf in an Alternating Current Generator

An alternating current generator contains 50 rectangular loops of conducting wire with side lengths of 55 cm and 35 cm, the ends of which form terminals. The sides of the loops with the same lengths as each other are parallel to each other. The loops rotate within a uniform magnetic field at 18 revolutions per second within a 360 mT uniform magnetic field. What is the root-mean-square voltage across the terminals? Give your answer to the nearest volt.

### Answer

Recall the equation for the induced emf due to the rotation of a conducting coil:

To find the rms voltage, we first need to find the peak potential difference, where the emf is at its maximum. This maximum occurs when , so we can just substitute that whole term with 1 in the equation for emf:

Now we have to find the other terms. Letβs convert the magnetic field of 360 mT to tesla. There are 1βββ000 milliteslas in 1 tesla: and multiplying 360 mT by this relation will give an answer in teslas:

So, the value of is 0.36 T.

Next we need to calculate the area of the loops, . We are told that these loops are rectangular, with side lengths 55 cm and 35 cm. Letβs convert these side lengths into metres from centimetres. There are 100 centimetres in 1 metre: and multiplying the side lengths by this relation gives

So 55 cm is 0.55 m, and 35 cm is 0.35 m.

The area of these rectangular loops is the product of these side lengths:
so the area of this rectangle is 0.1925 m^{2}.

Now, we have to find the angular frequency, , in radians per second. Each revolution is :

The angular frequency is moving through Β per second. We now have all the terms we need to complete the equation

We can now substitute in the values. is 50 loops, is
0.1925 m^{2}, is
0.36 T, and is
:

Letβs expand out the unit of tesla in the equation. 1 tesla is 1 volt-second per square metre:

Multiplying the magnetic field strength and angular frequency together cancels the units of seconds: and multiplying the last three terms together cancels the units of square metres, leaving volts as the only unit:

So, the peak value of potential difference is 391.88 V. We then take this value and multiply it by :

Rounded to the nearest volt, the rms voltage is thus 277 volts.

This relationship holds for any root-mean-square based on a sinusoidal function. If the emf is sinusoidal, then so is the subsequent current of the circuit too, meaning that it follows the same relationship for root-mean-square.

An alternating current, , can be represented sinusoidally. This means that the average current is 0, since the positives cancel with the negatives at every point. This is demonstrated in the graph below.

We can find the root-mean-square using the same relationship as emf, since they are both sinusoidal: meaning that the rms of current is the of the peak value:

Letβs look at some examples.

### Example 3: Root-Mean-Square Value of a Current

An alternating current has a peak value of 1.35 A. What is the root-mean-square value of the current? Give your answer to three decimal places.

### Answer

Recall the equation for finding rms current:

We are given the peak value of the current as 1.35 A. Substituting this value into the equation, we get and subsequent multiplication of the terms gives

To three decimal places, the root-mean-square value of this current is 0.955 amperes.

### Example 4: Graphical Representation of the Root-Mean-Square Value of a Current

The red line represents the change in the instantaneous value of the alternating current carried by a conductor. Which of the lines correctly represents the rootβmeanβsquare value of the current?

- The black line
- The green line
- The purple line
- The orange line
- The blue line

### Answer

The line that best represents the root-mean-square value of a current is the one that is closest to the value of of the peak. The red line represents the alternating current, so its peaks will be the absolute maximum value.

Letβs put in decimal terms:

So, the root-mean-square of the current will be the maximum times 0.707, or about of the maximum value.

So we are looking for the line about of the way up the graph. The black line is just barely above , so the lines below it, green and yellow, cannot be it.

The purple line is near about , so the best line is likely the one just below it, the blue line, which is closest to .

So the answer is E: the blue line.

The power in a circuit is related to the current through the equation where is power of the circuit, is the current in the circuit, and is the resistance of the circuit.

Since this equation contains the current , and current is sinusoidal in an ac circuit, the power is also sinusoidal. The maximum power in a circuit is when is at its peak: but the average power is not 0 like in the cases of current or emf. Power cannot be negative, as you cannot have negative energy. The average actually looks like

To put this in terms of rms values, we can bring the into the square value of current to get and since , we can substitute it into the equation to get

The square roots of 2 then cancel when multiplied together, leaving behind just :

Letβs look at an example.

### Example 5: Energy Dissipation in Alternating Current

An alternating current has a peak value of 1.75 A through a 148 Ξ© resistor. What is the energy dissipated by the current in a time of 365 s? Give your answer in kilojoules to one decimal place.

### Answer

Recall the general equation for power, : where is work and is the change in time.

To find the energy dissipated over a period of time for an alternating current, we want to use the average power instead, :

We want to find the work done by the circuit to find its energy, since work is the energy dissipated by the circuit.

To isolate work in the equation, we will multiply both sides by : causing the terms to cancel on the right side, leaving work:

The change in time, , is 365 seconds, but we do not know the average power. To find it, recall the equation that relates average power to rms current and resistance:

We are given the value of resistance in the circuit, but only the peak value of current, not the root-mean-square.

The rms current is the peak value of the current multiplied by : and the peak value of the current is given as 1.75 A, so substituting this into the equation yields

Now we have all the necessary values we need to substitute into the average power equation. The rms current is 1.237 A and resistance is 148 Ξ©: and the units of amperes multiplied by ohms become watts, giving an average power of

Now letβs put these values into the equation that finds work. Substituting in the values of 226.625 watts for average power and 365 seconds for change in time, and watts multiplied by seconds become joules, the SI unit of energy. This makes the energy dissipated by the circuit:

We now need to put this answer in kilojoules. There are 1βββ000 joules in 1 kilojoule: so to convert 82βββ718 joules to kilojoules, we need to multiply it by this proportion: the joules cancel, leaving behind kilojoules to give

This makes the energy dissipated by this circuit in 365 seconds about 82.7 kilojoules.

Now that we know the relationships between rms value of emf, current, and power, letβs look at inductive, capacitive, and resistive circuits.

A resistive circuit is a circuit that contains a resistor, like in the diagram below.

The current in such a circuit is completely in phase with the emf, as they are proportional to each other, as shown by the equation . The diagram below shows the change in potential difference over time, represented by the yellow line, compared to the change in current over time, represented in blue.

The symbol at the left of the circuit with the wave in it represent an ac generator, which generates the alternating current in the circuit.

An alternating current circuit can also contain a capacitor, represented by two close lines in circuit diagrams. Such a circuit is called a capacitive circuit and is shown in the diagram below.

The relationship between the current and potential difference is different in a capacitive ac circuit compared to a resistive one. Charge cannot pass through a capacitor. Instead, charge accumulates on the sides of the charge plates. The current that causes this charge accumulation is proportional to the change in potential difference across the plates with time:

This means that the current will be at its maximum when the change in potential difference is at its maximum. The change in potential difference is maximum (it has the steepest slope) when the potential difference itself is 0. When potential difference is at its maximum, its slope is zero, which means its current is zero, as the potential difference is not changing at that point. The figure below shows the relationship between the current and potential difference in a capacitive ac circuit.

As the potential difference from the ac generator increases and decreases, it similarly changes the potential difference across the plates of the capacitor. The potential difference on these plates gradually switches sides as the potential difference in the ac generator similarly switches direction.

Instead of the current perfectly following the potential difference like in a resistive circuit, the current leads the potential difference. This is due to the constant charging and discharging of the capacitorβs plates.

In capacitive ac circuits, the potential difference across the capacitor and the charge-accumulating current are not in phase with each other. The change in potential difference leads the change in the current by 90 degrees, or by rad, as shown in the figure below.

An ac circuit can also contain an inductor, represented with a curled line in circuit diagrams. The figure below shows an inductor in an ac circuit.

When an inductor has a changing current within its coils, it induces a similarly changing magnetic field. This magnetic field then creates a potential difference across the inductor that creates a current that opposes the original direction of current. The change of the magnetic field, as well as the resultant potential difference, is proportional to the change in current with time:

The potential difference will be at its maximum when the change in current is maximum. The change in current is maximum (it has the steepest slope) when the current itself is 0. When the current is at its maximum, its slope is zero, which means its potential difference is zero, as the current is not changing at that point. The figure below shows the relationship between the current and potential difference in a inductive ac circuit.

Inductive ac circuits do not have the same phase difference that capacitive ac circuits have: the change in potential difference leads the change in current, rather than the current leading.

This is because the potential difference induced in the inductorβs coils opposes the change in current from the switching, causing the potential difference to lead the current. This is shown in the diagram below.

In inductive ac circuits, the change in potential difference is 90 degrees, or rad, behind the change in current.

Letβs now look at a graph that represents the currents of a resistive, capacitive, and inductive circuit with an ac source. The three colored lines represent the changes of the current with time in the circuit, depending on the properties of the circuit.

The line that corresponds to a circuit that is only resistive would be the line that matches the curves of the emf, as they are directly proportional to each other. This is only the blue line.

We expect the line representing the capacitive circuit to be 90 degrees, or ahead of the emf. This description most closely matches the orange line, as its peaks are behind the emf.

The emf will lead the line representing the inductive circuitβs current by 90 degrees, or . This means that the peak of the line of the inductive current will be after the emf peaks, which appears to be the red line.

Letβs summarize what we have learned in this explainer.

### Key Points

- An alternating current (ac) is a current that switches directions periodically.
- The induced emf, , due to the rotation of a conducting coil in a uniform magnetic field is where is the number of loops in the coil, is the area of the loops, is the strength of the magnetic field, is the angular frequency, and is the time.
- The root-mean-square values of emf, current, and power in ac are
- Resistive circuits have the current in phase with the emf, capacitive circuits have it leading by 90 degrees, and inductive circuits have it trailing by 90 degrees.