BTEC NC - Electronic and Electrical Principles - Single Phase AC Circuit Theory

Brendan Burr BTEC National Certificate in ElectronicsSingle Phase AC Circuit Theory

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Task 11.1 Describe the following terms:

R.M.S Value – The RMS value of a sinusoidal waveform is the equivalent value that is equal to the DC value of the AC waveform. It is used to calculate power for an ac circuit.Frequency - The Frequency of a waveform is determined by how many cycles occur in one second. E.g. if a waveform has 342cycles a second its frequency is 342Hz.Periodic Time - Periodic Time is the amount of time it takes the waveform to complete one complete cycle. Peak Value – Peak Value is also known as the amplitude. It is the maximum value of voltage on the waveform.Peak To Peak Value – Peak To Peak Value is the difference between the lowest voltage and the highest voltage, i.e. 2 x Peak.

1.2 t = 4mS, Vmax = 10V, f = 50Hz, л 4

v = Vmax . Sin (ωt -

v = 10 . Sin (ω . 0.004 – л) 4

v = 10 . Sin (2 . л . 50 . 0.004 – л) 4

v = 10 . Sin (3л) 20

v = 4.5399004997 Radians

1.3 a) Peak Value:

PK = √2 x 415PK = 586.8986284VPK = 587V

b) Peak To Peak Value:

PK-PK= Peak Value x 2

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Angular Frequency (in Radians) :

ω = 2 . л . fω = 2 . л . 50ω = 100 л


PK-PK= 586.8986284 x 2PK-PK= 1173.797857PK-PK= 1174V

1.4a) Frequency:

f = Number of CyclesTime (Sec)

f = . 40 . 50 x 10-3

f = 800Hz

b) Periodic Time:

PT = 1 f

PT = . 1 . 800

PT = 0.00125 SecPT = 1.25 x 10-3 Sec

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Task 22.1 Calculation of transformer measurements:

Voltage Primary = No. Of Turns Primary

Voltage Secondary No. Of Turns Secondary

VS = NS x VP

NP

a) 20:1

100 = 20 ? 1

VS = 1 x 10020

VS = 5V (Step Down)

b) 1:10

100 = . 1 . ? 10

VS = 10 x 100 1

VS = 1000V (Step Up)

c) 1:1

100 = . 1 . ? 1

VS = 1 x 100 1

VS = 100V (Isolation Purposes)

d) 50:1

100 = 50 ? 1

VS = 1 x 10050

VS = 2V (Step Down)

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2.1 Transformer measurements using Crocodile clips simulation software:

20:1

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1:10

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1:1

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50:1

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Task 3

3.1a)

b) Maximum Voltage4.99V

c) Periodic Time[-0.003 + 0.017] = 0.02

d) Frequency of v = 3 Sin 100л t:

ω = 100л2лf = 100л

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2f = 100f = 100/2f = 50Hz

Periodic Time:

PT =

PT =

PT = 0.02 Secs

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Using a phasor diagram:

a)V2 = V1

2 + V22

V2 = 9 +16V2 = 25V = 5

b)Phase angle between V1

and V

So

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Task 4a) i) Increasing the input AC voltage to 240V (r.m.s)

By increasing the RMS voltage to 240 volts from 230 you are able to see that the sinusoidal waveform, on the top graph, has a higher peak value. The resultant of this is the dc waveform, the bottom graph, also has a higher voltage. The peak of the sinusoidal waveform ac voltage is around 34 volts and the peak of the dc voltage is around 32.5 volts.The input of voltage is 339.4 volts this allows the RMS value to be 240 volts. This is determined by the square root of 2 multiplied by the required voltage, in this instance 240.

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a) ii) Decreasing the input AC voltage to 220V (r.m.s)

By decreasing the RMS voltage to 220 volts, you can see that the peak of the sinusoidal waveform is lowered to 30 volt. The dc equivalent, on the bottom graph, has also clearly lowered in value. At 240 volts RMS the dc equivalent was around 32.5 volts and at 220 volts the dc equivalent is around 29.7 volts.Due to the frequency staying the same on both power sources, the periodic time is not affected, this is determined on the graphs as both waveforms had a periodic cycle of 20mS.

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b) i) Increasing the smoothing capacitor C to 10000uF

As you can see on the bottom graph the waveform appears much smoother that the previous two. This is because the smoothing capacitor has been replaced from 1000uF to 10000uF. The extra storage capacity of the capacitor allows for a much slower loss of charge, this makes it difficult to observe the degradation as the smallest value on the voltage axis is to 0.1 of a volt. I tried increasing the sample rate to improve the amount of information available, hoping this would allow me to zoom in slightly more, but it failed to do so.The only degradation of the capacitor that is visible is at around 7.4mS there is a slight dip in the pink trace suggesting that the dc voltage has dropped.

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b) ii) Decreasing the smoothing capacitor C to 100uF

When decreasing the size of the smoothing capacitor, it is easily visible that the capacitor does not hold its charge. The dc voltage drops from around 31.19 volts to 31.14 volts, this could cause an appliance to fail as it may require a steady dc input, it is however unlikely as it is such a small voltage drop.By comparing both graphs together, it is clear to see that the capacitor is charged at the same point as the peaks and troughs of the stepped down voltage input, signalling that the two waveforms are in phase with one another.

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c) i) Increasing the transformation ratio N to 20:1

The signal generator is giving an output of 230 volts RMS (325.269 volts) at a frequency of 50Hz.By stepping the transformer up to a 20:1 ratio, the voltage peak of the ac voltage is identifiably decreased exactly half, this labels the transformer as a step down device. This is because the original transformer was a 10:1 ratio which is half of the windings of 20:1. This tells me that the current has to travel twice as much in the 20:1 and is therefore halved.Because the ac input is halved the eventual dc voltage is also halved. This is shown in the bottom waveform above, where the dc is around 14.86 volts whereas before it was 31.19.

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c) ii) Decreasing the transformation ratio N to 5:1

The closer the ratio of the transformer gets to 1:1, the closer the voltage peak of the sinusoidal waveform gets to original source waveform. The voltage starts as 325.269 volts so a 1:1 would give 325.269 volts on the other winding. The 5:1 step down lowers the ac voltage to around 65 volts. This is ironically 325.269 divided by 5.The dc voltage is at around 63.62 volts, this can be observed from the bottom graph above.

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d) i) Increasing the load resistor R to 1MΩ

The relationship between the capacitor and the resistor determine the amount of discharge there is. The higher the resistor the less discharge there should be. With the graphs above it is clear that the dc voltage doesn’t drop much during discharging. Above the values are around 0.02 of a volt drop, which is with a 1 Mega Ohm resistor.

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d) ii) Decreasing the load resistor R to 10KΩ

As explained on the previous graphs I explained that the higher the resistor the less discharge there should be. When the resistor value is lowered to 10 Kilo Ohms the voltage drop is much more, in the above graph it is around 0.05 of a volt which is much more than the 0.02 voltage drop as shown on the previous page.This is because the positive of the capacitor bleeds away through the resistor. The higher the value the less the stored energy can discharge.The sinusoidal waveform has stayed exactly the same throughout changing the values of the load resistor. It has maintained a peak value of around 32 volts.

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BTEC NC - Electronic and Electrical Principles - Single Phase AC Circuit Theory

Documents

btec national

input ac voltage

dc voltage

ac voltage

dc equivalent

brendan burr

periodic time

bottom graph