Alternating currents Flashcards
A.C in a wire is an example of
In a wire with a.c., the free electrons within the wire move back and forth with s.h.m. The variation of the current with time is a sine curve, so it is described as sinusoidal (having a magnitude that varies as a sine curve).
An equation for a.c.
The equation gives us the value of the current I at any time t:
I = I○ sinωt
where I is the current at time t, I○ is the peak value of the alternating current and ω is the angular frequency (ω=2πf) of the supply, measured in rad s−1 (radians per second).
[Remember that your calculator must be in the radian mode when using this equation]
Alternating Voltages
A generator consists of a coil rotating in a magnetic field. An e.m.f. is induced in the coil according to Faraday’s and Lenz’s laws of electromagnetic induction.
This e.m.f. V varies sinusoidally, and so we can write an equation to represent it that has the same form as the equation for alternating current:
V = V○ sinωt
where V○ is the peak value of the voltage.
How to measure frequency and voltage of an alternating current? (Practical Activity 27.1 - Pg. 547, 548)
An oscilloscope can be used to measure the frequency and voltage of an alternating current.
Practical Activity 27.1 - Pg. 547, 548
A CRO is a modified form of ‘Electron beam tube / Cathode ray tube’, as shown in Figure 25.4 (Ch 25 Oscillations), as it contains an extra set of ‘parallel’ deflective plates to produce a horizontal electric field at right angles to the beam. CRO is a high speed graph plotting device.
ELECTRON GUN: The filament attached to the cathode heats the cathode and due to the thermionic emission, electrons start to accumulate on the surface of the cathode. The anode (or set of anodes) accelerate the electrons in the form of a beam that is focused on the fluorescent screen).
DEFLECTIVE PLATES: Two sets of deflective plates are present outside the electron gun that deflect the electron beam based on the direction of their respective electric field. The x-plates are aligned vertically across the beam and the y-plates are aligned horizontally or parallel to the beam.
The ‘signal’ (which is to be observed/measured i.e. heart rate) into the CRO is a repetitively varying voltage. This is fed into the y-input, which deflects the beam up and down using the parallel plates Y1 and Y2 controlled by y-gain shown in Figure 27.5. The time-base produces a p.d. across the other set of parallel plates X1 and X2 to move the beam from left to right across the screen. (Voltage across the x-plates is provided by a built-in circuit known as ‘time-base’ generator.)
When the beam hits the screen of the CRO, it produces a small spot of light. If you look at the screen and slow the movement down, you can see the spot move from left to right (built-in circuit voltage through time-base), while the ‘applied’ signal moves the spot up and down (y-gain). When the spot reaches the right side of the screen, it flies back very quickly and waits for the next cycle of the signal to start before moving to the right once again. In this way, the ‘signal’ is displayed as a ‘stationary trace’ (waveform) on the screen.
Voltage applied across y-plates will appear on fluorescent screen if, simultaneously, time-base voltage is also applied on x-plates.
(The two controls that you must know about are the time-base and the Y-gain, or Y-sensitivity.)
You can see in Figure 27.6 that the time-base control has units marked alongside. Let us suppose that this reads 5 ms/cm, although it might be 5 ms/division. This shows that 1 cm (or 1 division) on the x-axis represents 5 ms. Varying the time-base control alters the speed with which the spot moves across the screen. If the time-base is changed to 1 ms/cm, then the spot moves faster and each centimetre represents a smaller time.
The Y-gain control has a unit marked in volts/cm, or sometimes volts/division. If the actual marking is 5 V/cm, then each centimetre on the y-axis represents 5 V in the applied signal.
Determining frequency and amplitude (peak value of voltage)
If you look at the CRO trace shown in Figure 27.7, you can see that the amplitude of the waveform, or the peak value of the voltage, is equivalent to 2 cm and the period of the trace is equivalent to 4 cm.
If the Y-gain or Y-sensitivity setting is 2 V/cm, then the peak voltage is 2 × 2 = 4 V. If the time-base setting is 5 ms/cm, then the period is 4 × 5 = 20 ms.
In the example:
frequency = 1 / time period = 1 / 0.02 = 50 Hz.
If the current and voltage are varying all the time, does this mean that the power is varying all the time too? Give examples.
The answer to this is yes.
Example 1: Some fluorescent lamps (tube-lights) flicker continuously, especially if you observe them out of the corner of your eye or when you move your head quickly from one side to the other.
Example 2: A tungsten filament lamp would flicker too, but the frequency of the mains has been chosen so that the filament does not have time to cool down noticeably between peaks in the supply. When the A.C voltage reaches zero, the filament is still hot enough to emit some light, resulting in the appearance of the bulb glowing even when the voltage is zero.
Define Root-mean-square (r.m.s) value
Root-mean-square current: The r.m.s value of an alternating current is that steady direct current that deliver the same average power (or energy) as the alternating current to a resistive load (i.e. lamp or resistor).
Root-mean-square voltage: The r.m.s value of an alternating voltage is that steady direct voltage that deliver the same average power (or energy) as the alternating voltage to a resistive load (i.e. lamp or resistor).
[Note that it is the r.m.s. value that is generally quoted, not the peak value, i.e. The mains supply to domestic consumers in many European countries has an r.m.s. value of 230 V for the alternating voltage.]
Root-mean-square (r.m.s.) values
There is a mathematical relationship between the peak value V○ of the alternating voltage and a direct voltage that delivers the same average electrical power. The direct voltage is about 70% of V○. (You might have expected it to be about half, but it is more than this, because of the shape of the sine graph.) This steady direct voltage is known as the root-mean-square (r.m.s.) value of the alternating voltage. In the same way, we can think of the root-mean-square value of an alternating current, I r.m.s.
Ｉr.m.s. = Ｉ○ / √ 2 = 0.707 × Ｉ○
where Ｉ○ is the peak (maximum) current.
Vr.m.s. = V○ / √ 2 = 0.707 × V○
where V○ is the peak (maximum) voltage.
[This is where the factor of 70% comes from. Note that this factor only applies to sinusoidal alternating currents.]
Explaining root-mean-square and how it’s derived
The equation P = Ｉ²R shows us that the power P is directly proportional to the square of the current Ｉ. Figure 27.10 shows how we can calculate Ｉ² for an alternating current. The current I varies sinusoidally, and during half of each cycle it is negative. However, Ｉ² is always positive (because the square of a negative number is positive). Notice that Ｉ² varies up and down, and that it has twice the frequency of the current.
Now, if we consider <Ｉ²>, the average (mean) value of Ｉ², we find that its value is half of the square of the peak current (because the graph is symmetrical). That is:
<Ｉ²> = 1/2 ×Ｉ○²
To find the r.m.s. value of I, we now take the square root of <Ｉ²>.
This gives:
Ｉᵣ.ₘ.ₛ. = √ <Ｉ²> = √ (1/2Ｉ○²)
or Ｉ○ = √2 × Ｉᵣ.ₘ.ₛ.
Summarising this process: to find the r.m.s. value of the current, we find the root of the mean of the square of the current – hence r.m.s.
Calculating Power
The importance of r.m.s. values is that they allow us to apply equations from our study of direct current to situations where the current is alternating.
So, to calculate the average power dissipated in a resistor, we can use the usual formulae for power:
P = Ｉ²R = ＩV = V²/R
Remember that it is essential to use the r.m.s. values of Ｉ and V. If you use peak values, your answer will be too great by a factor of 2. Where does this factor of 2 come from? Recall that r.m.s. and peak values are related by:
Ｉ○ = √2 × Ｉᵣ.ₘ.ₛ.
So, if you calculate Ｉ²R using Ｉ○ instead of Ｉᵣ.ₘ.ₛ., you will introduce a factor of or 2. The same is true if you calculate power using V○ instead of Vᵣ.ₘ.ₛ. It follows that, for a sinusoidal alternating current, peak power is twice average power.
Worked example 1 - A 20 Ω resistor is connected to an alternating supply. The voltage across the resistor has peak value 25 V. Calculate the average power dissipated in the resistor.
Step 1 - Calculate the r.m.s. value of the voltage.
Vᵣ.ₘ.ₛ. = V○/√2 = 25/2 = 17.7 V
Step 2 - Now calculate the average power dissipated. (Remember you must use the r.m.s. value, and not the peak value.)
P = V²/R = 17.7²/20 = 15.6 W
Note that, if we had used V○ rather than Vᵣ.ₘ.ₛ., we would have found:
P = 25²/20 = 31.3 W
which is double the correct answer.
Define Reactification
“The process of converting alternating current (a.c.) into direct current (d.c.).”
Why? As there are many appliances, such as electronic equipment, which require d.c. For these, the alternating mains voltage must be converted to direct voltage by the rectification.
Half-wave rectification
Use a diode, which is a component that will only allow current in only one direction.
An alternating input voltage is applied to a circuit with a diode and a resistor in series. The diode will only conduct during the positive cycles of the input voltage. Hence, there will be a current in the load resistor only during these positive cycles. The output voltage Vout across the resistor will fluctuate as shown in the Vout against time t graph. This graph is identical to the input alternating voltage, except the negative cycles have been ‘chopped-off’.
This type of rectification is known as half-wave rectification. For one-half of the time the voltage is zero, and this means that the power available from a half-wave rectified supply is reduced.
The bridge rectification
To overcome this problem of reduced power, a bridge rectifier circuit is used. This consists of four diodes connected across the input alternating voltage, as shown in Figure 27.12. The output voltage Vₒᵤₜ
is taken across the load resistor R. The resulting output voltage across the load resistor R is full-wave rectified.
The way in which this works is shown in Figure 27.13.
* During the positive cycles of the input voltage, A is positive and B is negative. The diodes 2 and 3 conduct because they are both in forward bias. The diodes 1 and 4 are in reverse bias, and therefore do not conduct. The current in the load resistor R will be downwards. Figure 27.13a shows the direction of the current.
* During the negative cycles of the input voltage, B is positive and A is negative. The diodes 4 and 1 conduct because they are now both in forward bias. The diodes 2 and 3 are in reverse bias, and therefore do not conduct. The current in the load resistor R will still be downwards. Figure 27.13b shows the direction of the current.
Note that in both positive and negative cycles, the current direction in the load resistor R is always the same (downwards). This means that the top end of R must always be positive.
Light-emitting diodes in bridge rectification
You can construct a bridge rectifier using light-emitting diodes (LEDs) that light up when current flows through them. By connecting this bridge to a slow a.c. supply (for instance, 1 Hz from a signal generator), you can see the sequence in which the diodes conduct during rectification.
