# Power in an Alternating Circuit. Transformers

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16.17Power in an Alternating Circuit. Transformers

In this Physics tutorial, you will learn:

• How to calculate the electric power in alternating circuits
• What is the difference between average and total power in AC circuits?
• How is the rms current and voltage related to electric power in AC circuits
• What is power factor? How it is related to reactances and resistance in the circuit?
• What can we do to maximize the power in an AC circuit?
• What are electric transformers? What are they used for?
• How many types of transformers are there?
• What is the difference between ideal and real transformers?
• How the current and emf are related to each other in transformers?

## Introduction

Do you remember what does power represent? How do we calculate the mechanical power?

How the equation of mechanical power is transformed when dealing with electricity?

Do you think a correct value of power is obtained when considering the amplitudes of current and voltage in an AC circuit? What do you suggest in this case?

How can we change the voltage in a circuit? Discuss.

In this tutorial, we will consider again a series RLC circuit to provide answer to the above questions. This is because RLC circuits include all possible power consuming circuit elements. In this way, the explanation is more comprehensive.

## Electric Power as a Rate of Electric Energy Transfer

As explained in tutorial 5.5 "Power and Efficiency", power is the work done by system in the unit time. In other words, power as a generalized concept, is the rate of useful energy transfer - a process that brings a change in the energy of the system, i.e.

P = W/t = Euseful/t = ∆E/t

In addition, from the definition of work (the energy used by a system to displace an object or to transfer heat to it), we can include the thermal energy transfer (heat energy therefore) in the definition of power. The advantage of this approach is observed when dealing with electric power. We all know that electricity (electrical energy) is used for several purposes where one of the most important applications is to dissipate heat energy in the environment through a resistor. When considering also the time interval during which this heat energy is dissipated in the environment, we obtain the concept of useful electric power (explained in tutorial 15.6 "Electric Power and Efficiency"), which is found by applying the Joule's Law, i.e.

Po = ∆E/t = I ∙ ∆V ∙ t/t = I ∙ ∆V

The above equation is applied when calculating the power delivered by an energy-consuming device, such as a resistor. When calculating the power delivered by the source, we must replace potential difference ΔV with the electromotive force ε in the equation of electrical power, i.e.

Pi = I ∙ ε

In DC circuits we don't have any difficulty in calculating the power delivered by the source (input or total power) and that of any circuit component (output or useful power) because the values of emf, current and voltage are constant. In this way, the efficiency of circuit (the part of the total energy converted in the desired form, i.e. as useful energy) is calculated by

e = W/Einput × 100% = Po/Pi × 100%

In AC circuits however, the discussion about electrical power is more complicated as both current and voltage in circuit are not constant; they both vary in a sinusoidal fashion. We can transform the equation

P(t) = i(t) ∙ ε(t)

in such a way that one of the above quantities is expressed in terms of the other or through a constant. Hence, giving that in DC circuits

P = I ∙ ε
= I ∙ I ∙ R
= I2 ∙ R

we use the same approach in AC circuits as well, to eliminate the induced emf from the formula. Thus, we write

P(t) = i(t) ∙ ε(t)
= i(t) ∙ [i(t) ∙ R]
= i2 (t) ∙ R

where R is the resistance of the AC circuit.

In this way, since current in an AC circuit as a function of time, is

i(t) = imax ∙ sin⁡(ω ∙ t + φ)

we obtain for the power dissipated in such a circuit:

P(t) = [imax ∙ sin⁡(ω ∙ t + φ) ]2 ∙ R

or

P(t) = imax2 ∙ R ∙ sin2 (ω ∙ t + φ)

It is not necessary to calculate the power through this method however; we can use the concept of average power to simplify this process. For this, we must write the rms current instead of the maximum current in the above formula. In this way, we obtain for the average power in an AC circuit

= irms2 ∙ R
= (imax/2)2 ∙ R
= imax2 ∙ R/2
= Pmax/2

The above result is also proven by considering the fact that the average value of y = sin2 x function during a complete cycle is 1/2 despite the average value of y = sin x is zero. This is clarified through the two graphs below.  The electrical measuring devices such as ammeter and voltmeter are calibrated in such a way that they measure the average (rms) voltage and current instead of their maximum values. Thus, for example if a multimeter (a device that includes both an ammeter and a voltmeter) shows the values 4A and 220V, these are the average values of the corresponding quantities. If we want to known the amplitudes of current and voltage in the circuit, we write

imax = irms ∙ √2
= (4A) ∙ √2
= 5.66 A

and

∆Vmax = ∆Vrms ∙ √2
= (220 V) ∙ √2
= 311.13 V

If we want to calculate the average power in the circuit, we obtain

< P > = irms ∙ ∆Vrms
= (4A) ∙ (220V)
= 880 W

The maximum power delivered by the circuit is obviously twice this value, i.e.

Pmax = < P > ∙ 2
= (880 W) ∙ 2
= 1760 W

Indeed,

Pmax = imax ∙ ∆Vmax
= (5.66 A) ∙ (311.13 V)
= 1760 W

If we consider all three quantities responsible for the impedance in the circuit, we obtain for the rms current

irms = εrms/Z = εrms/R2 + (XL-Xc )2

We can express the average power delivered in the circuit as

= εrms/Z ∙ irms ∙ R
= εrms ∙ irmsR/Z

The quantity R/Z is the cosine of the phase constant φ discussed in the previous tutorial, i.e.

cos φ = R/Z = R/R2 + (XL-Xc )2

Hence, combining the last two equations, yields

= εrms ∙ irms ∙ cos⁡φ

The 'cos φ' term is called "power factor." It is independent from the sign of the phase constant φ because cos φ = cos (-φ) for every value of φ.

The value of power factor is very important in maximizing the rate of energy supplied to a resistive load. To achieve this, the value of power factor must be as close to 1 as possible. This means the phase constant must be as close as possible to zero, i.e. the values of inductive and capacitive load must be very close to each other. Thus, for example if a circuit is more inductive than capacitive, we increase the capacitance in the circuit to balance the reactances. On the other hand, if the circuit is more capacitive than inductive, we decrease the capacitance by connecting an extra capacitor in the circuit (it is known that the total capacitance of a series combination of capacitors decreases the total capacitance in the system). After all, the main goal is to increase the power delivered in the circuit; no matter the method used.

### Example 1

A series RLC circuit operating at fd = 50 Hz is shown in the figure below. #### Calculate:

1. The power factor and phase constant of the circuit
2. The average rate at which the energy is dissipated in the circuit
3. The extra capacitance needed to maximize the average power if the other parameters in the circuit are kept unchanged

### Solution 1

1. Let's calculate the impedance Z first. Thus,
Z = √R2 + (XL-Xc )2
= √(80 Ω)2 + (40 Ω-60 Ω)2
= 82.5 Ω
The power factor therefore is
cos⁡φ = R/Z
= 80 Ω/82.5 Ω
= 0.97
Hence, the phase constant is
φ = cos-1 0.97
Since XC > XL we must consider the negative value only. Thus, φ = - 0.246 rad.
2. First, we have to find the rms current flowing through the circuit. Thus,
irms = εrms/Z
= 200 V/82.5 Ω
= 2.424 A
The average rate at which the energy is dissipated in the circuit means the average power delivered by the circuit. Thus, we have

= εrms ∙ irms ∙ cos φ
= (200 V) ∙ (2.424 A) ∙ (0.97)
= 470.3 W

3. The average power is maximized if the circuit is in resonance, i.e. if XL = XC. This is achieved by decreasing the value of capacitance, i.e. by connecting a new capacitor in series to the given one. Since
Xc = 1/ωd ∙ C = 1/(2π ∙ f ∙ C)
we obtain for the actual capacitance
C = 1/2π ∙ f ∙ Xc
= 1/2 ∙ 3.14 ∙ (50 Hz) ∙ (60 Ω)
= 5.31 × 10-5 F
The capacitance in the new conditions (when the circuit is in resonance) is obtained for
XL = Xc
Thus,
XL = 1/2π ∙ f ∙ Cnew
Cnew = 1/2π ∙ f ∙ XL
= 1/2 ∙ 3.14 ∙ (50 Hz) ∙ (40 Ω)
= 7.96 × 10-5 F
This means the extra capacitance added in the system is
Cextra = Cnew - C
= 7.96 × 10-5 F - 5.31 × 10-5 F
= 2.65 × 10-5 F

## Transformers

It is already known that electric power is the product of current and voltage. As explained earlier, electric power P, is the amount of electrical energy W produced or delivered in the unit time. Thus, we have

W = P ∙ t = I ∙ ∆V ∙ t

Since power is related to energy, we often use terms like "power station" or "power source" instead of "energy station" or "energy source". Therefore, we will henceforth use the term power but keeping in mind the energy.

Electricity is produced in power stations through generators. It must be transported from power stations to the end user (at our homes for example). Mains electricity transportation system is used for this purpose. It includes cables, poles, cabins and especially transformers, which are passive electric devices used to change the value of voltage in a circuit. Transformers use electromagnetic induction to transport electricity. We will explain below why transformers are so important in electricity transportation system.

Power stations produce large amounts of energy in the unit time. Typical values range from hundreds of megawatt hour (MWh) to several gigawatt hour (GWh) of electricity. To carry this huge amount of energy throughout the transportation system, we can use two methods:

1. At high current and low voltage
2. At high voltage and low current

The first method is not suitable, as high current increases the temperature of cables and a lot of electricity is lost during the transportation because it is dissipated into the environment as heat energy. Therefore, we must increase the value of voltage during the transportation and then decrease it near the end user premises to make the electricity useable. Hence, during this process, we use two types of electric transformers: step-up transformers which are voltage increasers and step-down transformers, which are voltage reducers. Let's have a closer look at both of them.

Electricity is usually generated at 11kV in power stations although the generation voltage may vary in the range between 11kV and 33kV. Then the voltage increases during transportation to 220kV, 400kV or 765kV depending on the country. Step-up transformers are installed at high voltage poles to achieve this. Then, the voltage decreases to 20kV near residential areas (in electric substations) using step-down transformers and then, the voltage decreases further to 110V or 220V in electric cabins (using again step-down transformers) depending on the country. This is the value of voltage used in most part of daily activities.

Some electrical appliances require even lower voltages. For example, a TV needs 12V to operate. Hence, step-down transformers are also installed in specific appliances that require low operating voltages.

## How does a Transformer Work?

An ideal transformer contains a rectangular core (empty at middle) with two coils wound around the opposing arms. The core is made of insulating material to prevent the flow of undesired currents (Foucault currents). The coil connected with the input is called the primary coil while the one connected with the output is called secondary coil. As seen from the figures, step-up transformers have more turn in the secondary coil while step-down transformers have more turns in the primary coil. The alternating current flowing through the primary coil produces a variable magnetic flux in the core. Since the number of turns is different, the flux produced in the secondary coil changes, generating an induced emf in the secondary coil (in the primary coil the emf is due to the input source). However, the frequency of current (and voltage) doesn't change. This is an indicator of the relationship between currents and voltages in each coil of transformer.

The symbol of transformers in electric circuits is An ideal transformer has no any power loss during the induction from primary to secondary coil. However, this is not possible. Some of the energy (power) is lost during this process despite the coils are made of soft iron to reduce this loss. Hence, we can write

P1 ≥ P2

where 1 stands for input (primary coil) and 2 for output (secondary coil). The equality sign stands for ideal transformers. Thus, we obtain for the efficiency of transformer:

e = P2/P1 × 100%

Obviously, an ideal transformer is 100% efficient as no energy is lost during the induction.

We will explain the rest of transformers theory considering only ideal transformers. Thus, based on the power formula shown above, the relationship between currents and voltages in an ideal transformer is

i1 ∙ ∆V1 = i2 ∙ ∆V2

The relationship between the emf's and the number of turns in each coil is

ε1/ε2 = N1/N2

where N1 and N2 are the numbers of turns in the primary and secondary coil respectively. The ratio N1/N2 is called the turns factor.

For ideal transformers, the emf's in each coil are equal to the corresponding voltages at their ends. Hence, we can write

i1/i2 = ∆V2/∆V1

Combining all the above equations, we obtain for an ideal transformer

i1/i2 = ε2/ε1 = ∆V2/∆V1 = N2/N1

This means the current is inversely proportional to electromotive force, voltage and number of turns in the transformer.

The equation

∆V2/∆V1 = N2/N1

is known as the equation of transformer.

### Example 2

The voltage in the primary coil of a transformer is 220 V and the number of turns in this coil is 1000. If the number of turns in the secondary coil is 50, find:

1. What type of transformer is this?
2. The voltage in the secondary coil is the transformer is ideal
3. The voltage in the secondary coil if the transformer is 80% efficient
4. The current in the secondary coil for both situations above if the input current is 4.4 A.

### Solution 2

1. This is a step-down transformer as the number of turns in the secondary coil is smaller than in the primary coil (N1 = 1000, N2 = 50).
2. If the transformer is ideal, we have
∆V2/∆V1 = N2/N1
Substituting the known values, we obtain
∆V2/220 V = 50/1000
Thus,
∆V2 = 50 ∙ 220 V/1000
= 11 V
3. If the transformer is 80% efficient, we obtain for the output voltage
∆V1 = 80% ∙ ∆V2(ideal) = 0.80 ∙ 11V = 8.8 V
4. For ideal transformer we have
i1 ∙ ∆V1 = i2 ∙ ∆V2
Since i1 = 4.4 A, we have
(4.4 A) ∙ (220 V) = i2 ∙ (11 V)
i2 = (4.4 A) ∙ (220 V)/(11 V)
= 88 A
Since we have considered the efficiency when finding the output voltage, it is obvious that the current remains the same even in the real transformer. Hence, we have
i2 = 88 A

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