alternating voltages and currents
DESCRIPTION
Alternating voltages and currents. Alternating Potential Differences. a.c. supply. Most alternating voltages vary “sinusoidally”. Hence the wavy symbol!. The period T of the p.d. is the time taken for one complete wave or (cycle). p.d. time. The frequency of the supply is 1/T. - PowerPoint PPT PresentationTRANSCRIPT
Alternating Potential Differences
a.c. supply
Most alternating voltages vary “sinusoidally”. Hence the wavy symbol!
time
p.d
The period T of the p.d. is the time taken for one complete wave or (cycle).
The frequency of the supply is 1/T.
So the frequency (f) of the supply is equal to the number of cycles in a second.
The mains frequency in England is 50Hz .
time
Current
The size of the current follows a similar pattern to the p.d.
We need to be able to say what the “average” current is.
If we take a simple mean for the current we get a value zero!
0
Finding an average value for the a.c. current
Measuring an a.c. currentR R
RWe compare the magnitude of the current in an a.c. circuit with that in a d.c. circuit.
We do this in terms of the power dissipated through an exactly equal resistance.
Ia.c. Ia.c.
Id.c.
R R
resistor in a.c circuit Same resistor in d.c. circuit
When the power output from the two resistors is exactly the same we say that exactly the same current is flowing through the two resistors.
P =IV
The power formula tells us that power is the product of current and voltage
In this form this relationship doesn’t help much as I and V in the a.c. circuit are both varying continuously.
However substituting for V using V=IR yields
P= I2R
And the power through a given resistance can be seen to depend only on the current
R R
resistor in a.c circuit
resistor in d.c. circuit
P=I2RP=(Iac)2R
So the power through the resistor must be proportional to the average value of I2
R
This says that the power through the resistor is proportional to I2
II 2acIIaverage )( 2
This value is the
root mean square current
Root mean square value of the current
Current /I
Irms
It can be shown for a sine curve it can be shown that:
I0
00 707.02
II
I rms
Time t
Sinusoidal voltageA similar argument for sinusoidal voltage change gives a similar result
Root mean square value of the voltage
Voltage / V
Vrms
V0
Time t
00 707.02
VV
Vrms
Peak Voltage
Voltage / V
V0
Time t
Peak voltage
0
In this case the peak voltage is equal to the amplitude of the wave
Peak Voltage
Voltage / V
Time t
Peak voltage
0
In this case the peak voltage is not equal to the amplitude of the wave
This diagram illustrate the difference between peak voltage and amplitude.
The oscilloscope
• The oscilloscope is a simple means of producing p.d/ time graphical displays.
• It is an immesly powerful tool becausse it can be used to capure very small intervals of time and tiny voltages.
The time interval represented by the
horizontal distance across one square is (in this case)
0.2 s
The horizontal distance is referred to as the “time-base”.
It takes the spot 0.2 seconds to travel this distance at this setting
The p.d, between the input terminals of the oscilloscope is zero volts
The p.d. scale on the oscilloscope is +1V per
square
The p.d. input read by the oscilloscope is -1V
i.e the input terminals of the oscilloscope have been reversed.
An a.c voltage is represented by a sinusoidal wave.
On a time base of 1 ms with a voltage scale of 5V,
What is the frequency of the a.c. signal?
What is the peak voltage of the signal?
The potential of the mains electrical supply oscillates in this way at a frequency of 50 Hz.
This potential drives a current with the same frequency