recall and use the fact that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current

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

1. Syllabus Mapping – Topic 21.1 (Characteristics of Alternating Currents)

2. Learning Objectives (aligned to Assessment Objectives)

• AO1 – Knowledge & Understanding: State the sinusoidal expressions for voltage and current, define instantaneous, mean (average) and RMS quantities, and recall that for a purely resistive load the mean power equals one‑half the peak power.
• AO2 – Application & Analysis: Derive RMS values from first principles, use them to obtain the mean‑power expression (including the power‑factor term), and solve numerical problems for resistive and simple R‑L circuits.
• AO3 – Practical Skills: Design and carry out a simple experiment to verify \(\overline{P}= \tfrac12 P_{\max}\) using a resistor, a true‑RMS voltmeter/ammeter (or power meter) and an oscilloscope; evaluate uncertainties and systematic errors.

3. Key Concepts

• A sinusoidal alternating current can be written as

  \[

  i(t)=I_{\max}\sin(\omega t),\qquad \omega=2\pi f,\qquad T=\frac{1}{f}.

  \]

  \(I_{\max}\) is the peak (maximum) current.

• For a pure resistor \(R\) the voltage is in phase:

  \[

  v(t)=V{\max}\sin(\omega t),\qquad V{\max}=I_{\max}R.

  \]

• Instantaneous power in the resistor:

  \[

  p(t)=v(t)i(t)=\frac{v^{2}(t)}{R}=i^{2}(t)R.

  \]

• Mean (average) power \(\overline{P}\) is the average of \(p(t)\) over one full period \(T\).

  RMS (root‑mean‑square) values are the square‑root of the mean of the square of a quantity.

4. Derivation of RMS Current and Voltage

For any periodic quantity \(x(t)\) the RMS value is defined as

\[

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

\]

Current:

\[

\begin{aligned}

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

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

\end{aligned}

\]

Using \(\sin^{2}x=\dfrac{1-\cos2x}{2}\) and noting that the integral of \(\cos2\omega t\) over a full cycle is zero,

\[

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

\]

Hence

\[

\boxed{ I{\text{rms}} = \frac{I{\max}}{\sqrt{2}} } .

\]

Because \(V{\max}=I{\max}R\), the RMS voltage follows directly:

\[

\boxed{ V{\text{rms}} = \frac{V{\max}}{\sqrt{2}} } .

\]

5. Derivation of Mean Power for a Purely Resistive Load

Instantaneous power:

\[

p(t)=V{\max}I{\max}\sin^{2}(\omega t).

\]

Apply the identity \(\sin^{2}x=\dfrac{1-\cos2x}{2}\):

\[

p(t)=\frac{V{\max}I{\max}}{2}\bigl[1-\cos(2\omega t)\bigr].

\]

Average over one period:

\[

\begin{aligned}

\overline{P}

&=\frac{1}{T}\int_{0}^{T}p(t)\,dt \\

&=\frac{V{\max}I{\max}}{2}\left[\frac{1}{T}\int{0}^{T}1\,dt-\frac{1}{T}\int{0}^{T}\cos(2\omega t)\,dt\right] \\

&=\frac{V{\max}I{\max}}{2}\;(1-0) \\

&=\frac{V{\max}I{\max}}{2}.

\end{aligned}

\]

The peak (maximum) power occurs when \(\sin^{2}(\omega t)=1\):

\[

P{\max}=V{\max}I_{\max}.

\]

Therefore, for a sinusoidal AC applied to a pure resistor,

\[

\boxed{\overline{P}= \frac{1}{2}\,P_{\max}} .

\]

Using RMS quantities the same result is obtained more compactly:

\[

\boxed{\overline{P}=I{\text{rms}}^{2}R=\frac{V{\text{rms}}^{2}}{R}=V{\text{rms}}I{\text{rms}} } .

\]

6. Phase‑Angle Box (1 minute read)

For any AC circuit the general average‑power formula is

\[

\boxed{\overline{P}=V{\text{rms}}I{\text{rms}}\cos\phi } .

\]

Example (series RL, \(\phi=30^{\circ}\)):

\[

\overline{P}=V{\text{rms}}I{\text{rms}}\cos30^{\circ}=0.866\,V{\text{rms}}I{\text{rms}} .

\]

7. Summary Table – Peak, RMS and Mean Power Relationships

8. Worked Example (Resistive Load)

Problem: A sinusoidal AC source supplies a resistor \(R=10\;\Omega\). The peak voltage across the resistor is \(V_{\max}=100\;\text{V}\). Find the mean (average) power dissipated in the resistor.

1. RMS voltage:

   \[

   V{\text{rms}}=\frac{V{\max}}{\sqrt{2}}=\frac{100}{\sqrt{2}}\;\text{V}\approx 70.7\;\text{V}.

   \]

2. Mean power using RMS values:

   \[

   \overline{P}= \frac{V_{\text{rms}}^{2}}{R}= \frac{(70.7)^{2}}{10}\;\text{W}\approx 500\;\text{W}.

   \]

3. Check with the half‑maximum rule:

   \[

   P{\max}= \frac{V{\max}^{2}}{R}= \frac{100^{2}}{10}=1000\;\text{W},\qquad

   \overline{P}= \frac{1}{2}P_{\max}=500\;\text{W}.

   \]

Both methods agree, confirming \(\overline{P}= \tfrac12 P_{\max}\) for a pure resistor.

9. Practical Activity – Verifying the Half‑Maximum Rule (AO3)

Objective: Measure the mean power dissipated in a known resistor and compare it with half the peak power calculated from the measured peak voltage.

1. Equipment: Function‑generator (sine‑wave), true‑RMS voltmeter, true‑RMS ammeter or a true‑RMS power meter, oscilloscope, \(R=10\;\Omega\) resistor (±1 % tolerance), connecting leads, safety resistors for the generator.
2. Procedure:

   • Set the function generator to 50 Hz sine‑wave. Adjust the amplitude so the oscilloscope shows a peak voltage of ≈100 V across the resistor.
   • Record the peak voltage \(V_{\max}\) from the oscilloscope (use the vertical‑scale division).
   • Measure the RMS voltage \(V_{\text{rms}}\) directly with the true‑RMS voltmeter.
   • Measure the RMS current \(I{\text{rms}}\) with the true‑RMS ammeter or compute it from \(I{\text{rms}}=V_{\text{rms}}/R\).
   • Calculate the mean power in two ways:

     1. \(\overline{P}=V{\text{rms}}I{\text{rms}}\) (direct RMS method).
     2. \(\displaystyle \frac{1}{2}P{\max}= \frac{1}{2}V{\max}I{\max}\) where \(I{\max}=V_{\max}/R\) (half‑maximum rule).

   • Compare the two results and comment on any discrepancy.

3. Data‑analysis questions (Paper 5 style):

   • Estimate the percentage uncertainty in \(\overline{P}\) arising from the resistor tolerance and the specifications of the RMS instruments.
   • Explain why the measured mean power may differ slightly from the theoretical \(\tfrac12 P_{\max}\). Discuss waveform distortion, instrument bandwidth, temperature‑coefficient of the resistor and any systematic offset in the function‑generator output.

10. Common Misconceptions

• Average vs. RMS: The arithmetic average of a sinusoidal voltage over a full cycle is zero; it cannot be used for power calculations. RMS is the appropriate “average” because power depends on the square of the quantity.
• Peak vs. RMS values: Confusing \(V{\max}\) (or \(I{\max}\)) with \(V{\text{rms}}\) (or \(I{\text{rms}}\)) leads to errors of a factor \(\sqrt{2}\) (≈1.414) in power calculations.
• Applicability of the half‑maximum rule: \(\overline{P}= \tfrac12 P_{\max}\) holds only for purely resistive loads where \(\phi=0^{\circ}\). For inductive or capacitive loads the factor \(\cos\phi\) reduces the mean power.
• Instantaneous power in reactive components: In a pure capacitor or inductor the instantaneous power oscillates between positive and negative values, giving a net mean power of zero even though energy is stored and returned each cycle.

11. Practice Questions (AO2)

1. For a sinusoidal current with \(I_{\max}=8\;\text{A}\) flowing through a \(5\;\Omega\) resistor, calculate:

   • Peak power \(P_{\max}\)
   • Mean power \(\overline{P}\)
   • RMS current \(I{\text{rms}}\) and RMS power \(P{\text{rms}}\)

2. A household appliance is rated at \(1200\;\text{W}\) mean power when connected to a \(240\;\text{V}\) RMS supply. Assuming the appliance behaves as a pure resistor, determine:

   • Its resistance \(R\)
   • The peak voltage across it

3. Explain why the mean power delivered to a purely capacitive load is zero, even though the instantaneous power is non‑zero during part of each cycle.
4. Design a brief experimental plan (max 150 words) to test the relationship \(\overline{P}= \tfrac12 P_{\max}\) using a function generator, resistor, and a true‑RMS power meter. Identify one possible source of systematic error and how you would minimise it.

12. Suggested Diagram (Figure 1)