A-Level Physics: Alternating Currents

Understanding alternating current (AC) is fundamental to explaining how electrical energy is generated, transformed, and distributed to our homes and industries. While direct current (DC) flows in one constant direction, alternating current (AC) periodically reverses direction, a characteristic that makes large-scale power transmission economically viable.

The Nature of Alternating Current and Voltage

The most common form of AC is sinusoidal, meaning the current or voltage varies with time according to a sine wave. This can be described mathematically. The instantaneous voltage $V$ at any time $t$ is given by the equation:

$$V=V_0\sin (2\pi ft)$$

Here, $V_0$ represents the peak voltage (or amplitude), which is the maximum value the voltage reaches. The symbol $f$ stands for the frequency, measured in hertz (Hz), which indicates how many complete cycles occur per second. In the UK, the mains frequency is 50 Hz. It’s crucial to distinguish this from the time period $T$, which is the time for one complete cycle; they are related by $f=1/T$. The continual oscillation between positive and negative peaks is what defines AC and underpins the operation of transformers.

Interpreting Oscilloscope Traces

An oscilloscope is the primary tool for visualizing AC waveforms. The screen is a grid where you can measure voltage on the vertical (y-) axis and time on the horizontal (x-) axis. To determine the peak voltage from a trace, you must know the y-gain or voltage sensitivity setting, measured in volts per division (V/div). If the trace spans 4 vertical divisions from the centre line to a peak and the y-gain is set to 5 V/div, the peak voltage is $4\times 5=20\text{ V}$.

Finding the frequency requires analysing the time base. The time-base setting is measured in seconds per division (s/div). You measure the horizontal distance for one complete cycle (the wavelength of the trace). For example, if one cycle occupies 5 horizontal divisions and the time-base is set to 5 ms/div, the time period $T=5\times 5\text{ ms}=25\text{ ms}=0.025\text{ s}$. The frequency is then $f=1/T=1/0.025=40\text{ Hz}$. Mastery of these measurements is essential for practical exam questions.

Root Mean Square (RMS) Values and Power

For a sinusoidal AC supply, the root mean square (RMS) voltage is the equivalent steady DC voltage that would deliver the same average power to a resistive load. It is not a simple average, which for a symmetrical AC wave is zero. The RMS value is derived from the process of squaring the instantaneous values (making them all positive), finding their mean, and then taking the square root.

For a sinusoidal waveform, the relationship between RMS and peak values is fixed:

$$V_{\text{rms}}=\frac{V_0}{\sqrt{2}}\text{and}{I}_{\text{rms}}=\frac{I_0}{\sqrt{2}}$$

Approximately, $V_{\text{rms}}\approx 0.707\times V_0$. This is critically important because all domestic AC voltage ratings (like 230 V in the UK) are RMS values. The peak mains voltage is therefore about $230\times \sqrt{2}\approx 325\text{ V}$. When calculating power in an AC circuit with a resistive component, you must use RMS values: $P=I_{\text{rms}}{V}_{\text{rms}}=I_{\text{rms}}^{2}R$.

Transformer Operation and the Turns Ratio

Transformers are devices that use electromagnetic induction to change (transform) the amplitude of an alternating voltage. They consist of a primary coil and a secondary coil wound around a laminated iron core. An alternating current in the primary coil produces a changing magnetic flux in the core, which induces an alternating voltage in the secondary coil.

For an ideal transformer (one with 100% efficiency, meaning no energy losses), the relationship between the voltages ($V_p$, $V_s$) and the number of turns on the coils ($N_p$, $N_s$) is given by the turns ratio equation:

$$\frac{V_p}{V_s}=\frac{N_p}{N_s}$$

If $N_s>N_p$, it is a step-up transformer ($V_s>V_p$). If $N_s<N_p$, it is a step-down transformer ($V_s<V_p$). For an ideal transformer, power is conserved, so $V_p{I}_p=V_s{I}_s$. Combining this with the turns ratio gives the current relationship: $\frac{I_p}{I_s}=\frac{N_s}{N_p}$. A step-up transformer increases voltage but decreases current proportionally, which is the key to efficient power transmission.

Power Transmission and Efficiency in the National Grid

The National Grid is the high-voltage network that distributes electricity across the country. Power is transmitted at very high voltages (up to 400 kV) and low currents. This is because the power loss in the transmission cables, due to their resistance $R$, is proportional to the square of the current: $P_{\text{loss}}=I^{2}R$. By reducing the current by a factor of 100 (via a step-up transformer), the power loss is reduced by a factor of 10,000.

Real transformers are not 100% efficient. Energy losses occur due to:

1. Resistance of the windings ($I^{2}R$ heating).
2. Eddy currents (induced currents in the iron core, minimized by lamination).
3. Hysteresis (energy needed to repeatedly realign magnetic domains in the core).
4. Flux leakage (not all magnetic flux links both coils).

Efficiency is calculated as: $\text{Efficiency}=\frac{\text{Useful Power Output}}{\text{Total Power Input}}\times 100\%=\frac{I_s{V}_s}{I_p{V}_p}\times 100\%$. Analysing these losses and efficiency calculations are common in exam questions about the grid system.

Common Pitfalls

1. Confusing Peak and RMS Values: A common error is using peak voltage in power calculations. Remember, for power and heating effects, you must use RMS values. The 230 V from a wall socket is an RMS value; the peak is significantly higher.
2. Misreading Oscilloscope Settings: Students often forget to multiply the number of divisions by the setting. If a peak-to-peak distance is 6 divisions at 10 V/div, the peak-to-peak voltage is 60 V, making the peak voltage 30 V (not 6 V or 10 V).
3. Misapplying the Transformer Equations: The turns ratio equation $\frac{V_p}{V_s}=\frac{N_p}{N_s}$ only applies directly to RMS voltages. Do not attempt to use it with peak values unless you are consistent. Also, remember that a transformer only works with AC; a DC supply will not induce a steady voltage in the secondary coil.
4. Ignoring Transformer Efficiency: Assuming a transformer is ideal when it is not will lead to incorrect calculations for secondary current or input power. Always check the context of the question to see if efficiency is 100% or if you need to account for losses using the efficiency formula.

Summary

• Alternating Current (AC) periodically reverses direction and is described by sinusoidal equations involving peak voltage $V_0$ and frequency $f$. Oscilloscope traces are analysed using the y-gain (for peak voltage) and time-base (for period and frequency) settings.
• The Root Mean Square (RMS) value ($V_{\text{rms}}=V_0/\sqrt{2}$) is the equivalent DC voltage for power delivery. All domestic voltage ratings and power calculations ($P=I_{\text{rms}}{V}_{\text{rms}}$) use RMS values.
• Transformers change AC voltage using the turns ratio: $\frac{V_p}{V_s}=\frac{N_p}{N_s}$. For an ideal transformer, input power equals output power ($V_p{I}_p=V_s{I}_s$).
• The National Grid transmits power at high voltage and low current to minimise energy losses ($P_{\text{loss}}=I^{2}R$) in transmission lines. Real transformers have losses due to winding resistance, eddy currents, and hysteresis.