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First exams 2025

Alternating Current & Voltage (CIE A Level Physics)

Revision Note

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Ashika

Last updated

27 October 2024

Properties of alternating current & voltage

An alternating current (a.c) is defined as:

A current which periodically varies from positive to negative and changes its magnitude continuously with time

This means the direction of an alternating current varies every half-cycle
The variation of current, or p.d., with time can be described as a sine curve ie. sinusoidal
- Therefore, the electrons in a wire carrying a.c. move back and forth with simple harmonic motion

As with SHM, the time period of an alternating current is defined as:

$T = \frac{1}{f}$

Where:
- T = time period (s)
- f = frequency (Hz)

Peak current (I₀), or peak voltage (V₀), is defined as:

The maximum value of the alternating current or voltage

Peak current, or voltage, can be determined from the amplitude of the graph

Time period and peak current on a current-time graph

Alternating current graph, downloadable AS & A Level Physics revision notes

Graph of alternating current against time with a time period of 20 ms and peak current of 2 A

Mains electricity is supplied as alternating current
- Power stations produce alternating current
- This is the type of current supplied when devices are plugged into sockets

Worked example

The variation with time t of the output voltage V of an alternating voltage supply is shown in the graph below.

Worked example voltage graph, downloadable AS & A Level Physics revision notes

Use the graph to calculate the frequency of the supply.

Answer:

Step 1: Write down the period-frequency relation

$f = \frac{1}{T}$

Step 2: Calculate the time period from the graph

The time period is the time taken for one complete cycle
From the graph, this is equal to 0.2 ms
Therefore, the time period is

T = 0.2 ms = 0.2 × 10^-3 s

Step 3: Substitute into frequency equation

$f = \frac{1}{0.2 \times 10^{- 3}} = 5000 Hz = 5 kHz$

Examiner Tip

Remember to double check the units on the alternating current and voltage graphs. These are often shown in the range of milli-seconds (ms) instead of seconds (s) on the x axis.

Using sinusoidal representations

The equation representing alternating current which gives the value of the current I at any time t is:

$I = I_{0} \sin (ω t)$

Where:
- I = current (A)
- I₀ = peak current (A)
- ⍵ = angular frequency of the supply (rad s^-1)
- t = time (s)

Note: this a sine function since the alternative current graph is sinusoidal
A similar equation can be used for representing alternating voltage:

$V = V_{0} \sin (ω t)$

Where:
- V = voltage (V)
- V₀ = peak voltage (V)

Recall the relation between the equation for angular frequency ⍵:

$ω = \frac{2 π}{T} = 2 π f$

Where:
- f = frequency of the supply (Hz)

Worked example

An alternating current I varies with time t as shown in the graph below.

21-1-1-we-alternating-current--cie-new

Using the graph and the equation for alternating current, calculate the value of the current at a time 0.48 s.

Answer:

Step 1: Write out the equation for alternating current

$I = I_{0} \sin (ω t)$

Step 2: Write out the equation for angular frequency

$ω = \frac{2 π}{T}$

Step 3: Measure the time period T and peak current I₀ from the graph

The time period is the time taken for one full cycle, T = 0.10 s
Peak current (amplitude), I₀ = 17 A

Step 4: Substitute values into alternating current equation at time t

Using the time given in the question, t = 0.48 s

$I = I_{0} \sin (ω t) = I_{0} \sin (\frac{2 π t}{T})$

$I = 17 \sin (\frac{2 π (0.48)}{0.1}) = - 16.168 = - 16 A (2 s . f)$

Examiner Tip

These equations are written as $x = x_{0} \sin (ω t)$ , the x is replaced with I or V, depending on the question.

Remember to check that your calculator is in radians mode when using any of these equations. This is because the angular frequency ⍵ is measured in rad s^-1.

Current can also be positive or negative, depending on its direction. This is why the answer in the worked example is negative. At t = 0.48 s the current will be in the negative direction. The mark schemes in the exams are normally not too picky about this, but the size of the current (16 A in the worked example) is what will get the mark.

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