Current in Circuits - National 5 Recap
All Current flow can be divided into two groups:-
1. Direct Current - Occurs in a circuit using a battery.
2. Alternating Current - Occurs in a circuit using Mains Electricity.
When a circuit is attached to a battery, Electrons are repelled by the negative terminal and attracted by the positive terminal. This causes a flow of Electrons from negative to positive. As all Electrons are moving in the same continuous direction, this is a Direct Current. On an oscilloscope, direct current is shown by a horizontal, flat line as shown below:-
When a circuit is attached to Mains Electricity, the direction of Electron flow changes back and forth many times a second, giving an Alternating Current. On an oscilloscope, Alternating Current is shown by a sinusoidal waveform:-
In the UK, Mains Electricity alternates at a frequency of 50 Hz and operates at a declared Voltage of 230 V.
However this presents a problem, as how can the Voltage have a constant value of 230 V, if it is alternating as shown above ?
The answer is that the Voltage is not constant. The declared value of 230 V is actually an average value, known as the Root-Mean-Square (r.m.s.) value.
The R.M.S. value is the value at which it is equivalent to a DC circuit. For example, a bulb will be equally bright when attached to a 6V DC supply or a 6Vrms AC supply.
As the Vrms value is an average value, it is not the maximum value that the Voltage reaches. In order to find the Peak Voltage or Peak Current (Vpeak , Ipeak), the following calculations can be performed:-
Vpeak = Vrms x √2
Ipeak = Irms x √2
Graphical A.C. Analysis
The Peak or R.M.S. values can also be found graphically. By using the value per division, and by counting divisions, the Voltage (Y-axis) and the Time (X-axis) can be found.
Example 1 -
The image below shows the display of an oscilloscope displaying a A.C. Signal:-
Use the image above to calculate:-
1. The Frequency of the signal.
2. The Vpeak of the signal.
3. The Vrms of the signal.
Timebase (X-axis) is set to 2ms/division. One complete wave is equal to 2 divisions, therefore:-
Period of Wave (T) = 1 x 2x10-3 s
Frequency of the signal = 1/T = 1/2x10-3 = 500 Hz
Vpeak of the signal:-
Y-gain control (Y-axis) is set to 4mV/division. The peak of the signal is 3 divisions up from the centre-line, therefore:-
Vpeak = 3 x 4 = 12 mV
The Vrms of the signal:-
Vpeak = Vrms x √2
Vrms = Vpeak / √2
Vrms = 12 / √2 = 8.5 mV
This A.C. signal can be described as a 8.5 mV 250 Hz signal.