   Alternating Current
Introduction

The electricity generated in power stations is different from the simple direct current produced by the chemical reactions in a battery.

Power stations generate alternating current (a.c.). This has its voltage stepped up by transformers before being distributed by a network of overhead and underground cables. The voltage of the supply is stepped down at local substations and sent to individual houses.

Generating alternating currents
The alternating
voltage
The voltage across a component is the electrical energy transferred by 1 coulomb of charge passing through the component.
voltage
produced by the generator shown in Fig.1 varies in a
sinusoidal
Sinusoidal variations are continuous changes over time which, when plotted on a graph, have the shape of a sine curve.
sinusoidal
manner.

 Figure 1. Simple illustration of how alternating current is generated. The alternating voltage produced in Fig.1 will cause an alternating
current
The rate of flow of charge past any specific point in a circuit. The base unit of current is the Ampere.
current
in any resistor connected across the terminals of the supply. Note that the brightness of the lamp in Fig.2 increases and decreases as the current rises and falls.

Click on the figure below to interact with the model.  Figure 2.  Alternating current used in a circuit.

As you look at the lamp and graph, can you see how the brightness of the lamp changes with changing p.d.?

Decide which one of the following statements is true.
• Consequently, the lamp reaches its maximum brightness twice during each cycle of the alternating voltage.

At any instant, the current in the lamp can be calculated using
Ohm's Law
Ohm's Law states that at constant temperature the current in a conductor is directly proportional to the d.p. across the conductor. The constant of proportionality is called the resistance of the sample.
Ohm's Law
. Since the generated p.d. is alternating sinusoidally, the current in the circuit also changes in the same way.

Comparing a.c. and d.c. voltages
The peak voltages applied across the bulbs in the circuits of Fig.3 have the same values.

Click on the figure below to interact with the model.  Figure 3.  Circuits with equal peak voltages.

The lamp in circuit 1 is being powered from an alternating voltage with a peak value of 5 V while the lamp in circuit 2 is connected to a 5 V d.c. battery.

Since the peak p.d.s across the lamp are the same, the maximum brightnesses will also be equal. However, since the lamp in circuit 2 is permanently on while the lamp in circuit 1 flashes, we can say that more
energy
A system has energy when it has the capacity to do work. The scientific unit of energy is the joule.
energy
is being taken from the
power
The power of system is a measurement of the rate at which energy is transferred from one form to another. The scientific unit of power is the watt.
power
supply in circuit 2 than in circuit 1.

Complete the following sentences to compare the energy taken from a.c. and d.c. power sources in Fig.3.

• Over a certain time interval the d.c. source supplies energy than the a.c. power supply. The a.c. supply with the peak of 5 V provides energy than the 5 V d.c. power supply.
• One way to equalize the energy supplied to the lamps by both the a.c. and d.c. supplies would be to decrease the value of the d.c. voltage.

Click on the figure below to interact with the model.  Figure 4.  Circuits with different peak voltages.

Both lamps in Fig.4 now radiate the same quantity of energy. When the lamp in circuit 3 is fully on, it is brighter than the lamp in circuit 4. However, it is off for some of the time, so the quantities of energy taken from the power sources are the same.

Root mean square values
The previous circuit is shown again in Fig.5, but this time the power is plotted on the graph as well as the voltage. You should see from the voltage graph that the d.c. voltage in circuit 4 is set at a steady 6 V while the a.c. supply in circuit 3 has a peak value of 8.5 V.

Click on the figure below to interact with the model.  Figure 5.  Circuits with equal energy outputs.

You should also notice that the power graphs show that on average, both lamps provide the same amount of energy (where the energy is equal to power multiplied by time). In other words, an a.c. supply with a peak value of 8.5 V is effectively equal to a d.c. supply with a value of 6 V. The name for this equivalent d.c. voltage is the root mean square (or r.m.s.) value of voltage. In this case, the 8.5 V a.c. supply has an
r.m.s.
Root mean square is often abbreviated to r.m.s. The r.m.s. value of an a.c. signal is the voltage of an equivalent d.c. signal that provides the same amount of energy.
r.m.s.
value of 6 V.

The peak and r.m.s. values for any voltage are linked in the following way:

This means that the r.m.s. value is approximately 70 per cent of the peak value.

A similar equation connecting the peak current and the r.m.s. current in a simple a.c. circuit can also be stated: You will be able to find out more details about the mathematics linking peak and r.m.s. values for sinusoidal variations in more advanced physics or electronics texts.

Power in a.c. circuits
The power delivered to a component in a simple d.c. circuit is calculated using the formula:

In a circuit with an a.c. supply, the effective current and p.d. are given by the respective r.m.s. values. We can therefore rewrite this equation as:

This equation allows us to calculate the average power supplied to a resistor connected to an alternating current supply. However, care needs to be taken when using this equation to design a.c. circuits as the actual power being delivered to a component at any instant will usually be higher or lower than the average value.

The maximum power delivered to a component can be calculated by multiplying the peak values for current and p.d. Good circuit designers will use components capable of operating at this maximum power even though it is only the actual power for a very small part of the time.

 Figure 6. High power appliance. Converting a.c. to d.c.
Power supplies that deliver d.c. are required by the electronic systems within many appliances. In most domestic appliances the power supplies are concealed, but d.c. power supplies such as that shown in Fig.7 are used to turn the a.c. provided by the mains into the d.c. needed by circuits.

 Figure 7. Simple d.c. power supply. These devices use circuits such as that shown in Fig.8 to convert a.c. to d.c. This process is called rectification.

Click on the figure below to interact with the model.  Figure 8.  Converting a.c. to d.c.

Move the red probe in Fig.8 and place it on the circuit at the point marked A. You should see that the four diodes convert the a.c. signal from the supply into a voltage with only positive values. The output from the diodes is a d.c. signal whose value rises and falls. This variation or ripple on the output voltage is a remnant of the a.c. variation and is often called the ripple voltage. Reposition the probe at point B and note that the same rectified p.d. also appears across the load resistors.

Close the switch to connect the
capacitor
A capacitor is a system of closely spaced conductors designed to store charge. The larger the p.d. across the capacitor plates the more charge stored
capacitor
into the circuit and note that this makes the output voltage vary less. The capacitor has reduced the size of the ripple voltage.

Alter the value of the variable resistor and note that the size of the ripple voltage depends on the magnitude of the current drawn from the supply.

Summarize the behaviour of Fig.8 by completing the following sentences:

• When the variable resistor is at it maximum value the current supplied to the load resistors is at its and the ripple voltage is . Reducing the value of the variable resistor allows currents to pass through the lamp. Increasing the current drawn from the power supply the size of the ripple voltage.
• Click on the figure below to interact with the model.  Figure 9.  A stabilized power supply.

In the circuit shown in Fig.9, some components have been added to improve the performance of the power supply. Use the probe to confirm that the output voltage of the power supply is stable.

The addition of the Zener diode and transistor has stabilized the voltage being supplied to the lamp. Place the cursor over the line connecting the
transistor
A transistor is a semiconductor device that is used extensively in amplifier and switching circuits.
transistor
to the lamp and note that the current supplied is virtually constant.

Summary

The root mean square value (r.m.s. value) of an a.c. signal is the voltage of an equivalent d.c. signal that provides the same amount of energy.

The r.m.s. value is approximately 70 per cent of the peak value.

The average power delivered to components connected to an a.c. supply can be calculated using the r.m.s. current and p.d.s. Well-designed circuits take account of both the average power and the peak power delivered to components.

Exercises
1. An alternating voltage with a peak of 7 V is connected in series with a 100 Ω resistor in an ammeter.

• Calculate the r.m.s. current in the circuit.
mA   (to 1 d.p.)

Calculate the power dissipated in the resistor.
mW   (to the nearest whole number)

Another 100 Ω resistor is now connected in parallel with the first one. Calculate the current in the circuit.
mA   (to the nearest whole number)

• 2. An a.c. signal with a
frequency
The wave frequency f is the number of complete waves passing any point each second. Frequency is measured in hertz, Hz.
frequency
of 80 Hz and a peak voltage of 10 V is connected in a circuit with a 100 Ω resistor.

• What d.c. voltage would produce the same heating effect in the 100 Ω resistor?
V   (to 2 d.p.)

The a.c. frequency is now doubled. What d.c. voltage is now required to produce the same heating effect in the 100 Ω resistor?
V   (to 2 d.p.)

• 3. An alternating voltage with a peak
amplitude
The amplitude of a wave is the distance between the peak of the crest and the undisturbed position.
amplitude
of 7 V is connected in series with an ammeter and a 100 Ω resistor. The r.m.s. current recorded by the ammeter is:
• 4. An a.c. signal with a frequency of 80 Hz and a peak amplitude of 10 V is connected to a circuit with a 100 Ω
resistance
The opposition to the flow of current provided by a circuit is called resistance. Resistance is measured in units called Ohms.
resistance
. What d.c. voltage would produce the same heating effect in the 100 Ω resistor?
• Figure 10. 5. An alternating voltage with a peak value of 9 V is connected in series with a 1 kΩ resistor and an ammeter, as shown in Fig.10.

• Calculate the r.m.s. current in the circuit.
mA   (to 1 d.p.)

Calculate the power dissipated in the resistor.
mA   (to the nearest whole number)
• Figure 11. 6. Another 1 kΩ resistor is now connected in parallel with the first one, as shown in Fig.11.

• Calculate the effective resistance of the resistors in parallel.
Ω   (to the nearest whole number)

Calculate the new current in the circuit.
mA   (to 1 d.p.)

• 7. The a.c. supply, as shown in Fig.10, is replaced with a battery and the ammeter is switched to measure direct current. Calculate the value of the voltage required for the d.c. supply to produce the same power dissipation in the parallel resistors.
•  V   (to 2 d.p.)
• 8. A 6 V, 0.15 A lamp is connected to a 6 V battery of negligible
internal resistance
All batteries or power supplies have internal resistance. This resistance has the effect of reducing the output p.d. as the current supplied increases.
internal resistance
, as shown in Fig.12. The ammeter in the circuit indicates that the current in the lamp is 0.15 A. The p.d. of 6 V and the current of 0.15 A are described as the optimum operating conditions of the lamp.

Calculate the resistance of the lamp filament.
•  V   (to the nearest whole number)
• Figure 12. 9. What is the power dissipated by the lamp in Fig.12 when lit at normal brightness?
•  mW   (to the nearest whole number)
• 10. The 6 V battery in Fig.12 is now replaced with an a.c. supply providing an r.m.s. output voltage of 6 V.

Calculate the peak output p.d. of the a.c. supply.
•  V   (to 2 d.p.)
• 11. What is the peak current in the filament of the lamp shown in Fig.12?
•  A   (to 2 d.p.)
• 12. Calculate the peak power dissipated by the filament of the lamp in Fig.12 when connected to the a.c. supply.
•  W   (to 2 d.p.)
• 13. What is the average power dissipated by the filament of the lamp in Fig.12?
•  W   (to 2 d.p.)
• Well done!
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