Monitoring and measuring a.c.

Alternating Current in UK mains is 230 V with a frequency of 50 Hz. This means that 50 times every second the direction of the current flip flops backwards and forwards. A graph of how a.c. and d.c. compare can be seen below.

From the graph it is possible to see that at each instant the value of the potential difference is constantly changing both in direction and value each second.

As you can see on the graph the peak voltage is greater than 230 V. Why is this the case?

The value of 230 V is the root mean square (rms) value. This is the value which averages out all of the continuous changing voltage and current with a power equivalent to the same d.c. value rather than the peak value.

$$V_{\mathrm{rms}}=\frac{V_{\mathrm{peak}}}{\sqrt{2}}$$ $$\mathrm{230}=\frac{V_{\mathrm{peak}}}{\sqrt{2}}$$

V_peak = 325 V

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Determining the rms voltage

Attempt to solve the question yourself before clicking the solve button!

A circuit has a peak a.c. voltage of V. Determine the V_rms in this circuit.

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The choice of the rms value is choosen because the brightness of a lamp with 230V_rms a.c. is the same brightness as a lamp with 230 V d.c.

Determnation of frequency, peak voltage and r.m.s. values from an oscilloscope

We can use an oscilloscope to determine details of the peak voltage and frequency.
On the controls of an oscilloscope there are dials detailing the values for the size of each division horizontally and vertically. From these settings we can work out the relevant values.

Timebase

The horizontal control is used for time based measurements (timebase). In the diagram above the timebase setting is 5 ms/div. What is the frequency of the wave?
Periodic time for one full wave requires 4 horizontal divisions.
T = 4 x 5 x 10^-3 = 0.02 s $$T=\frac{1}{f}$$ $$\mathrm{0.02}=\frac{1}{f}$$

f = 50 Hz

Peak Voltage

The vertical divisions are in volts per division (V/div).

The oscilloscope is set on the y-axis to 2V/div. What is the peak voltage, and the r.m.s. voltage?

V_peak = 3.25 divisions
V_peak = 3.25 x 2 = 6.5 V $$V_{\mathrm{rms}}=\frac{V_{\mathrm{peak}}}{\sqrt{2}}$$ $$V_{\mathrm{rms}}=\frac{\mathrm{6.5}}{\sqrt{2}}$$ $$V_{\mathrm{rms}}=4.60V$$

The r.m.s. value can simply be calculated by 0.7 x peak value.

Root Mean Square Current

In any a.c. circuit the constantly changing magnitude of the voltage will result in a constantly changing magnitude of the current since V = IR and V is constantly changing. We know that the V_rms value gives us the same brightness of a lamp with the same d.c. voltage. The I_rms current is calculated in the same manner as the voltage and we can see this below:

$$I_{\mathrm{rms}}=\frac{I_{\mathrm{peak}}}{\sqrt{2}}$$

Calculate the I_rms in a series circuit where a 100 Ω resistor is supplied with a peak voltage of 70.7 V. In order to derermin the I_rms we need to first calculate the V_rms and then use Ohm's Law, V = IR, to determine the current.

$$V_{\mathrm{rms}}=\frac{V_{\mathrm{peak}}}{\sqrt{2}}$$ $$V_{\mathrm{rms}}=\frac{70.7}{\sqrt{2}}$$ $$V_{\mathrm{rms}}=50V$$

Now that we know the V_rms is 50 V we can substitute that into Ohm's Law to determine the I_rms.

$$V_{\mathrm{rms}}=I_{\mathrm{rms}}R$$ $$50=I_{\mathrm{rms}}\times 100$$ $$I_{\mathrm{rms}}=0.50A$$

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Determining an rms current

Attempt to solve the question yourself before clicking the solve button!

A series circuit has a peak a.c. voltage of V across a resistor of Ω. Determine the I_rms in this circuit.

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Frequency of supply

Attempt to solve the question yourself before clicking the solve button!

An oscilloscope has a timebase setting of ms and a y-gain setting of V. An ac supply shown on the oscilloscope has a wavelength of divisions and has a maximum peak height of divisions from the midpoint of the wave.
(a) Determine the frequency in this ac supply.
(b) Calculate the V_rms for this ac supply.

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