distinguish between root-mean-square (r.m.s.) and peak values and recall and use I r.m.s. = I0 / 2 and Vr.m.s. = V0 / 2 for a sinusoidal alternating current

Alternating Currents – Cambridge International AS & A Level Physics (9702)

Learning Objectives

• Identify period, frequency and peak (maximum) values of sinusoidal waveforms.
• Distinguish between peak, peak‑to‑peak and root‑mean‑square (r.m.s.) quantities.
• Derive and use the exact sinusoidal relationships

  \[

  I{\rm rms}= \frac{I{0}}{\sqrt{2}},\qquad

  V{\rm rms}= \frac{V{0}}{\sqrt{2}}

  \]

  and the exam‑style approximations

  \[

  I{\rm rms}= \frac{I{0}}{2},\qquad

  V{\rm rms}= \frac{V{0}}{2}.

  \]

• Calculate mean (average) power in resistive circuits and recognise the “½ \(P_{\max}\)” rule.
• Analyse half‑wave and full‑wave rectifiers and understand how a smoothing capacitor reduces ripple.
• Carry out a simple laboratory verification of the r.m.s.–peak relationship (AO3).

1. Prerequisite Concepts

• Instantaneous vs. average quantities.
• Sinusoidal function: \(x(t)=x_{0}\sin(\omega t+\phi)\) where \(\omega =2\pi f\).
• DC power: \(P=IV=I^{2}R=V^{2}/R\).
• Basic integration over one period \(T\) (used in the r.m.s. derivation).

2. Key Definitions

3. Why RMS is Important

• For a resistor \(R\), the average power delivered by an AC quantity is

  \[

  P{\rm avg}=I{\rm rms}^{2}R=\frac{V_{\rm rms}^{2}}{R}.

  \]

• Using r.m.s. values lets us apply the familiar DC power formulas to AC circuits without performing an integration each time.
• For a pure sinusoid the mean power is also

  \[

  P{\rm avg}= \frac{1}{2}P{\max},

  \]

  because the instantaneous power varies as \(\sin^{2}\) and its average over a period is \(1/2\).

4. Exact Derivation for a Sinusoid

Consider a sinusoidal current

\[

I(t)=I_{0}\sin(\omega t).

\]

The r.m.s. value is defined as

\[

I{\rm rms}= \sqrt{\frac{1}{T}\int{0}^{T} I^{2}(t)\,dt}.

\]

Substituting the sinusoid:

\[

\begin{aligned}

I{\rm rms}&= \sqrt{\frac{1}{T}\int{0}^{T} I_{0}^{2}\sin^{2}(\omega t)\,dt}\\

&= I{0}\sqrt{\frac{1}{T}\int{0}^{T}\sin^{2}(\omega t)\,dt}.

\end{aligned}

\]

Since \(\displaystyle\frac{1}{T}\int_{0}^{T}\sin^{2}(\omega t)\,dt = \frac12\),

\[

I{\rm rms}= \frac{I{0}}{\sqrt{2}}.

\]

Exactly the same steps give \(V{\rm rms}=V{0}/\sqrt{2}\) for a sinusoidal voltage.

5. The Exam‑Style “/2” Rule

• Cambridge exam questions often give the shortcut

  \[

  I{\rm rms}= \frac{I{0}}{2},\qquad V{\rm rms}= \frac{V{0}}{2},

  \]

  where \(I{0}\) (or \(V{0}\)) is the peak‑to‑peak amplitude.

  Because \(I{0(\text{p‑p})}=2I{0(\text{peak})}\), the exact relation becomes

  \[

  I{\rm rms}= \frac{I{\text{p‑p}}}{2\sqrt{2}}\approx\frac{I_{\text{p‑p}}}{2.83}.

  \]

• Use the “/2” form only when the question explicitly states it. Otherwise apply the exact \(\sqrt{2}\) factor.
• Always check whether the given “peak” value is a true peak or a peak‑to‑peak value before converting.

6. RMS Factors for Common Waveforms

7. Converting Between Peak and RMS Values

1. Identify the waveform type. (sinusoid, square, triangular, rectified, etc.)
2. Choose the appropriate factor \(k\). Use the table above or the exact \(\frac{1}{\sqrt{2}}\) for a pure sinusoid.
3. Apply the conversion

   \[

   X{\rm rms}=kX{0},\qquad X{0}= \frac{X{\rm rms}}{k}.

   \]

4. Insert the r.m.s. value into any power, energy or heating calculation.

8. Rectification and Smoothing

8.1 Half‑Wave Rectifier

• Only one half of the sinusoid is allowed to pass; the other half is blocked.
• Peak current through the load: \(I_{0}\).

  RMS current: \(I{\rm rms}=I{0}/2\).

• Average (DC) load current: \(\displaystyle I{\rm av}= \frac{I{0}}{\pi}\).

8.2 Full‑Wave Rectifier

• Both halves of the sinusoid are made positive (bridge or centre‑tap).
• RMS current: \(I{\rm rms}=I{0}/\sqrt{2}\) – the same as an un‑rectified sinusoid.
• Average (DC) load current: \(\displaystyle I{\rm av}= \frac{2I{0}}{\pi}\).

8.3 Smoothing with a Capacitor

After rectification a capacitor is placed across the load to reduce the ripple voltage.

• Ripple (peak‑to‑peak) voltage (approximation):

  \[

  V{\rm r} \approx \frac{I{\rm av}}{fC},

  \]

  where \(f\) is the ripple frequency (full‑wave: \(2f_{\text{source}}\)).

• Increasing the capacitance \(C\) or the load resistance reduces the ripple.
• For a well‑smoothed supply, the output voltage is close to the peak value minus the diode drop(s).

9. Worked Examples

Example 1 – Sinusoidal Current (Exact \(\sqrt{2}\) method)

Peak current \(I{0}=12\;\text{A}\). Find \(I{\rm rms}\) and the average power in a \(6\;\Omega\) resistor.

\[

I_{\rm rms}= \frac{12}{\sqrt{2}}=8.49\;\text{A}.

\]

\[

P{\rm avg}=I{\rm rms}^{2}R=(8.49)^{2}\times6\approx 432\;\text{W}.

\]

Example 2 – Using the “/2” Shortcut

Peak‑to‑peak current in a sinusoidal circuit is \(20\;\text{A}\). The question tells you to use the exam‑style rule.

\[

I_{0(\text{peak})}= \frac{20}{2}=10\;\text{A},\qquad

I{\rm rms}= \frac{I{0}}{2}=5\;\text{A}.

\]

Exact answer would be \(10/\sqrt{2}=7.07\;\text{A}\); the “/2” value is acceptable only because the question explicitly asks for it.

Example 3 – Full‑Wave Rectifier with Smoothing

A 240 V (r.m.s.) mains supply is fed into a full‑wave bridge. The diode drops are 0.7 V each. Find the approximate DC output voltage after a smoothing capacitor of \(1000\;\mu\text{F}\) supplying a 50 mA load.

1. Peak mains voltage: \(V_{0}= \sqrt{2}\times240=339\;\text{V}\).
2. After two diode drops: \(V_{\text{peak,load}} \approx 339-2\times0.7=337.6\;\text{V}\).
3. Ripple frequency for full‑wave: \(2f = 100\;\text{Hz}\).
4. Ripple voltage: \(\displaystyle V{r}\approx\frac{I{\rm av}}{f_{\rm ripple}C}

   =\frac{0.05}{100\times1000\times10^{-6}}=0.5\;\text{V}\).

5. DC output ≈ \(337.6\;\text{V} - \frac{V_{r}}{2}\approx 337.3\;\text{V}\).

10. Laboratory Verification of the RMS–Peak Relationship (AO3)

1. Set a function generator to produce a sinusoidal voltage of known peak‑to‑peak amplitude (e.g., 10 V p‑p).
2. Measure the true‑RMS voltage with a calibrated true‑RMS multimeter.
3. Calculate the expected r.m.s. value using both the exact \(\sqrt{2}\) formula and the “/2” shortcut.
4. Compare the measured value with the calculations; the discrepancy should be < 1 % for a true‑RMS meter.
5. Repeat with a square wave and a triangular wave to confirm the different RMS factors from the table.

11. Common Misconceptions & How to Avoid Them

• Peak = RMS. Only true for a square wave; for a sinusoid \(I{\rm rms}=0.707\,I{0}\).
• Confusing peak with peak‑to‑peak. Always read the wording; convert if necessary.
• Applying the “/2” rule to non‑sinusoidal waveforms. Use the appropriate factor from the waveform table.
• Using instantaneous values in power formulas. Power formulas \(P=I^{2}R\) or \(P=V^{2}/R\) require r.m.s. quantities, not instantaneous ones.

12. Quick Reference Summary – Exam Formulas

13. Summary Table – Peak ↔ RMS ↔ Power

14. Suggested Diagram (Insert in your notes)