1. Sinusoidal voltages and currents:

• Root Mean Square (RMS) Value:
  – The RMS value of a sinusoidal voltage or current is the most commonly used value in AC systems.
  – It’s the square root of the mean of the squared values of the waveform.
  – RMS value is denoted by [math] V_{\text{rms}} \, \text{ or } \, I_{\text{rms}} [/math].
  – For a sinusoidal waveform, [math] V_{\text{rms}} = V_0 / \sqrt{2} [/math] (approximately [math] 0.707 x V_0 [/math] ) ,.

  
  Figure 1 An alternating electrical waveform

• Peak Value:
  – The peak value of a sinusoidal voltage or current is the maximum value reached by the waveform.
  – Peak value is denoted by [math] V_0 \, \text{or} \, I_0 [/math].
• Peak-to-Peak Value:
  – The peak-to-peak value of a sinusoidal voltage or current is the difference between the positive and negative peaks.
  – Peak-to-peak value is denoted by [math] V_{p-p} \, \text{or} \, I_{p-p} [/math]
  – [math] V_{p-p} = 2 x V_0 [/math] (since the waveform is symmetric)
• The peak voltage, [math] V_0 [/math] , of the alternating waveform is half the peak-to-peak voltage, and is equivalent to the amplitude of the waveform.
• For a given component such as a resistor, the peak current [math]I_0 [/math] and peak voltage [math] V_0 [/math] are related to each other through the equation.
• [math] V_0 = I_0 R [/math]

• ⇒ Comparing alternating current (ac) and direct current (dc) equivalent:

• The average values cannot be used, because the average values are both zero – there is the same amount of signal above zero as there is below zero.
• The values chosen are the root mean square (r.m.s) voltage and current.
• When multiplied together, these quantities produce the same power in a resistor as would be produced by the same de values.
• [math] P = V_{\text{dc}} I_{\text{dc}} \\ P = V_{\text{rms}} I_{\text{rms}} [/math]
• A sinusoidal alternating voltage, V, varying with time, t, can be represented by the equation
• [math] V = V_0 \sin(2\pi f t) [/math]
• where [math]V_o [/math] is the peak voltage, and is the frequency of the supply. This is shown on the graph in Figure 2

  
  Figure 2 Alternating voltage

• If this voltage is applied across a fixed resistor, R, then the power dissipated by the resistor is equal to
• [math] P = \frac{V^2}{R} \\ P = \frac{V_0^2 \sin^2(2\pi f t)}{R} [/math]
• The average power is the power we need to compare to an equivalent constant dc value, but as [math] \frac {V_0^2}{R} [/math] is constant in this equation, the average value of [math] \sin^2(2\pi f t) [/math]. This can be done by analyzing the graph of the function in figure 3

  
  Figure 3 The [math] \sin^2 [/math] graph

• The average value of [math] \sin^2(2\pi f t) = 0.5 [/math], so
• [math] P = \frac{\frac{1}{2} V_0^2}{R} [/math]
• This will be the same power as that for an equivalent constant dc value of voltage, [math] V_{dc} [/math]
• [math] P = \frac{\frac{1}{2} V_0^2}{R} \\ P = \frac{V_{\text{dc}}^2}{R} [/math]
• Hence
• [math] V_{\text{dc}}^2 = \frac{1}{2} V_0^2 \\ \sqrt{V_{\text{dc}}^2} = \sqrt{\frac{1}{2} V_0^2} \\ V_{\text{dc}} = \frac{V_0}{\sqrt{2}} [/math]
• And
• [math] V_{\text{dc}} = V_{\text{rms}} = \frac{V_0}{\sqrt{2}} [/math]

• AS the alternating current varies in phase with the voltage

• [math] I_{\text{rms}} = \frac{I_0}{\sqrt{2}} [/math]

• The mean alternating power, [math] P_{\text{mean}} [/math], which is equivalent to the dc power is given by

• [math] P_{\text{mean}} = V_{\text{rms}} I_{\text{rms}} [/math]

• The peak alternating power [math] P_{\text{peak}} [/math], is given by

• [math]  P_{\text{peak}} = V_0 I_0 [/math]

• finally,

• [math] P_{\text{mean}} = \frac{V_0}{\sqrt{2}} * \frac{I_0}{\sqrt{2}} \\ P_{\text{mean}} = \frac{V_0 I_0}{2} \\ P_{\text{mean}} = \frac{P_{\text{peak}}}{2} [/math]

• In other words, the mean power dissipated through a fixed resistor by an alternating current and voltage is equal to half the peak power dissipated.

• Example:
• (1)
• The given UK mains voltage is 230V ac – this is an r.m.s. value. The largest power that can usually be drawn from a local grid is 15kW – this is the mean value, equivalent to a dc supply.
  1- What is the r.m.s. current?
  2- What are the peak power, peak voltage and peak current?
• Given data:
  Main voltage = [math]V_{\text{rms}} = 230V [/math]
  Mean power = [math] P_{\text{peak}} = 15kW = 15 * 10^3 W [/math]
• Find data:
  1. Main current [math] = I_{\text{rms}} = ? [/math]
  2. Peak voltage [math] = V_0 = ? [/math]
  3. Peak current [math] = I_0 = ? [/math]
  4. Peak power [math] = P_{\text{peak}} = ? [/math]
• Formula:
• [math] P_{\text{mean}} = V_{\text{rms}} I_{\text{rms}} \\\text{(i)} \quad I_{\text{rms}} = \frac{P_{\text{mean}}}{V_{\text{rms}}} \\
  V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \\
  \text{(ii)} \quad V_0 = V_{\text{rms}} * \sqrt{2} \\
  I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \\
  \text{(iii)} \quad I_0 = I_{\text{rms}} * \sqrt{2} \\
  \text{(iv)} \quad P_{\text{peak}} = V_0 I_0 [/math]
• Solution:
• [math]\text{(i)} \quad I_{\text{rms}} = \frac{P_{\text{mean}}}{V_{\text{rms}}} \\
  I_{\text{rms}} = \frac{15 \times 10^3}{230} \\
  I_{\text{rms}} \approx 65.2 \, \text{A} \\
  \text{(ii)} \quad V_0 = V_{\text{rms}} \cdot \sqrt{2} \\
  V_0 = 230 *\sqrt{2} \\
  V_0 \approx 325.3 \, \text{V} \\
  \text{(iii)} \quad I_0 = I_{\text{rms}} * \sqrt{2} \\
  I_0 = 65.2 * \sqrt{2} \\
  I_0 \approx 92.2 \, \text{A} \\
  \text{(iv)} \quad P_{\text{peak}} = V_0 I_0 \\
  P_{\text{peak}} = (325.3)(92.2) \\
  P_{\text{peak}} \approx 29993 \, \text{W} [/math]
• We can then use these values to calculate the peak-to-peak values of p.d. and current:
• [math]V_{\text{peak-to-peak}} = 2V_0 \\
  V_{\text{peak-to-peak}} = 2(325.3) \\
  V_{\text{peak-to-peak}} = 650 \, \text{V} \\
  I_{\text{peak-to-peak}} = 2I_0 \\
  I_{\text{peak-to-peak}} = 2(92.2) \\ 
  I_{\text{peak-to-peak}} = 180 \, \text{A} [/math]

2. Use of an oscilloscope as a dc and ac voltmeter:

• ⇒ Oscilloscope with a dc signal:

• An oscilloscope is a fantastic tool for visualizing and analyzing electrical signals, including DC signals.

  
  Figure 4 Oscilloscope display DC signals

• When using an oscilloscope with DC signals, you’ll typically see a flat line on the screen, indicating the constant voltage level. Here are some key points to keep in mind:
  – DC signals have no frequency component, so the oscilloscope will display a static voltage level.
  – The oscilloscope’s time base and voltage scale controls won’t affect the appearance of the DC signal.
  – You can use the oscilloscope’s offset feature to shift the DC signal up or down on the screen for better visibility.
  – DC signals can be measured using the oscilloscope’s voltage measurement tools, such as the voltage cursors or the measurement markers.
• Some common applications of using an oscilloscope with DC signals include:
  – Troubleshooting electronic circuits
  – Measuring voltage levels in power supplies
  – Analyzing battery performance
  – Verifying reference voltage levels in analog circuits

• ⇒ Oscilloscope with an ac waveform:

• The oscilloscope in Figure 9.6 illustrates an ac waveform from a signal generator. Describe the waveform.

  
  Figure 5 Oscilloscope displaying an ac waveform

Solution:

• In this case the time base and y-sensitivities have not changed.
• There are five horizontal divisions between the two successive peaks or troughs, and this corresponds to a time period of 100 ms. The frequency of the signal is therefore.
• Given data:
• Time period [math] = t =100ms = 100 * 10^{-3} s [/math]
  Find data:
  Frequency = f =?
  Formula:
• [math] \text{frequency} = \frac{1}{\text{time period}} [/math]
• Solution:
• [math] \text{frequency} = \frac{1}{\text{time period}} \\ \text{frequency} = \frac{1}{100 * 10^{-3}s} \\ \text{frequency} = f =10Hz [/math]
• The peak-to-peak voltage measured from the bottom of a trough to the top of a peak on the screen is six divisions, corresponding to 6V. This corresponds to a peak voltage of 3V and
• [math] V_{\text{rms}} = \frac{3V}{\sqrt{2}} \\ V_{\text{rms}} = 2.1 \, \text{V} [/math]

The operation of a transformer

3. Transformer:

• A transformer is an electrical device that transfers electrical energy between two or more circuits through electromagnetic induction.
• It consists of two coils of wire, known as the primary and secondary coils, which are wrapped around a common core.
• Figure 6 working of transformer
• – Step-up transformer: Increases voltage from primary to secondary
  – Step-down transformer: Decreases voltage from primary to secondary
  – Isolation transformer: Provides electrical isolation between primary and secondary
  – Autotransformer: Single coil that acts as both primary and secondary
• Transformers are used in:
  – Power transmission and distribution
  – Electronic devices (e.g., power supplies, amplifiers)
  – Medical equipment (e.g., MRI machines)
  – Audio equipment (e.g., speakers, microphones)
• The benefits of transformers include:
  – Voltage transformation
  – Isolation
  – Current limiting
  – Efficient energy transfer
• For an ideal transformer, with no power losses, the ratio of the turns on each coil equals the ratio of the primary and secondary voltages. That is
• [math] \frac{V_s}{V_p} = \frac{N_s}{N_p} [/math]
• where [math] V_s = \text{secondary voltage,} V_p = \text{primary voltage} \, (V),  N_s = \text{turns on the secondary coil and} N_p \text{turns on the primary coil} [/math].
• A step-up transformer is a transformer that increases voltage, so [math] \frac{N_s}{N_p} [/math]  is more than 1. A step-down transformer is a transformer that decreases voltage, so [math] \frac{N_s}{N_p} [/math], is less than 1. Figure 7 shows a simple circuit diagram for a transformer, with the symbols for an ac supply, a step-up transformer and a bulb.
• 
  Figure 7 Transformer with attach a bulb

• Transformers cannot increase the power output of the supply. In an ideal transformer, with no power losses, the power input to the transformer must be equal to the power output. Therefore, we can write the following equation

• [math] V_p I_p = V_s I_s [/math]

• where [math] V_s [/math] = secondary voltage (V), [math] V_p [/math] =primary voltage (V), [math] I_s [/math]= current in the secondary coil (A) and [math] I_p [/math]  current in the primary coil (A).

  This means that a transformer that reduces the output voltage compared to the input voltage has a larger current in the secondary coil compared to the primary coil.

4. Transformer efficiency:

• Transformer efficiency is a measure of how effectively a transformer converts electrical energy from the primary circuit to the secondary circuit. It’s defined as the ratio of the output power to the input power, usually expressed as a percentage.
• [math] \text{Efficiency } (\eta) = \frac{\text{Output power}}{\text{Input power}} \times 100\% \\  \text{Efficiency } (\eta) = \frac{V_s I_s}{V_p I_p} [/math]
• A higher efficiency value indicates that the transformer is converting most of the input energy into useful output energy, with minimal losses.
• Transformer efficiency is affected by factors such as:
  1. Core material and design
  2. Winding resistance and eddy currents
  3. Magnetic flux leakage
  4. Load conditions
• Typical efficiency values for transformers range from 80% to 99%, depending on the design and application.
• Some common efficiency classes for transformers include:
  1. Low-efficiency (80-89%)
  2. Medium-efficiency (90-94%)
  3. High-efficiency (95-98%)
  4. Very high-efficiency (99% and above)

5. Production of eddy currents:

• Eddy currents are induced electrical currents that flow in a conductor when it is exposed to a changing magnetic field. They are produced in a transformer due to the following reasons:
  1. Changing magnetic flux: The primary coil’s alternating current produces a changing magnetic flux in the core.
  2. Electromagnetic induction: This changing flux induces an electromotive force (EMF) in the secondary coil.
  3. Closed paths: The secondary coil provides a closed path for the induced currents to flow.
• Eddy currents flow in the core and other conductive parts of the transformer, causing:

• ⇒ Energy losses in transformers:

• Energy losses in transformers occur because of the following effects:
• Heat is produced in the copper wires of the primary coil and secondary coil when a current flows. Using low-resistance wire reduces these losses. This is particularly important for the secondary coil of a step-down transformer, because the current is larger in the secondary coil compared to the primary coil. A thicker wire is often used in the secondary coil of a step-down transformer.
• Some magnetic flux produced by the primary coil does not pass through the iron core, which means the flux linkage to the secondary coil is not 100%. This can be reduced by designing the transformer with coils close to each other or wound on top of each other, which improves the flux linkage.
• There is an effect called hysteresis. Some energy is lost as heat every time the direction of the magnetic field changes because energy is needed to realign the magnetic domains in the core. This is reduced by using a soft magnetic material such as iron, rather than steel which needs more energy to demagnetize and magnetize.
• Eddy currents form in the iron core due to the continuously changing flux. These currents heat the core up, increasing energy losses. Eddy currents are reduced by making the core using laminated sheets separated by thin layers of insulation. Eddy currents are discussed in more detail below.

• ⇒ Eddy current in transformer:

• Eddy currents are created in metal sheets when there is a change in magnetic flux.
• In the core of a transformer, the alternating supply creates alternating magnetic flux changes, and these create eddy currents.
• Eddy currents flow in loops, in a direction that opposes the magnetic flux changes that cause them.
• The result is that eddy currents in the iron core will reduce the e.m.f. induced in the secondary coil.
• In a core made from solid iron, eddy currents could become large enough to melt the core, because the resistance of the iron core is very low.
• To prevent these problems, the core is built from very thin laminations, or layers, of metal (Figure 8).
• Figure 8 Eddy currents are reduced by building the core from thin, insulated layers of iron.
• The eddy currents are smaller when there are thin laminations, because the induced voltage drives the current around longer paths so the resistance to flow increases.
• The laminations are insulated from each other, for example using layers of insulating varnish.

6. Causes of inefficiencies in a transformer:

• Inefficiencies in a transformer can be caused by:
  1.  Eddy currents: Induced currents in the core and other conductive parts, leading to energy loss and heat generation.
  2. Hysteresis loss: Energy lost due to the core material’s magnetic properties, causing a “memory” effect.
  3. Copper loss: Resistance in the primary and secondary coils, leading to energy dissipation.
  4. Flux leakage: Magnetic flux that escapes the core, reducing efficiency.
  5. Winding resistance: Electrical resistance in the coils, causing energy loss.
  6. Core saturation: Magnetic core material reaching its maximum flux density, leading to reduced efficiency.
  7. Overheating: Excessive temperature increases, affecting performance and lifespan.
  8. Poor design or construction: Suboptimal transformer design or manufacturing defects.
  9. Ageing and deterioration: Over time, insulation and materials can degrade, reducing efficiency.
  10. Overloading: Exceeding the transformer’s rated capacity, leading to increased losses and reduced efficiency.
• By understanding these causes, transformer designers and engineers can optimize performance, improve efficiency, and ensure reliable operation.