Physics » Magnetism and Faraday's Law » Alternating Current

Power

Power

If the current and voltage are functions of time, so they are always changing, then so will the power that is dissipated in any circuit element. In circuits which contain only ohmic resistors the average power dissipated in any component can be calculated in terms of the rms values.

\begin{align*} P_{av} &=I_{rms}V_{rms} \\ &=\frac{I_{\max}}{\sqrt{\text{2}}}\frac{V_{\max}}{\sqrt{\text{2}}}\\ &=\frac{\text{1}}{\text{2}}I_{\max}V_{\max} \end{align*}

You might ask why we don’t need to use an rms value for power. In an AC circuit both the current and voltage have the same sign so they are both either positive or negative. This means that power, the product of the two is always positive. If power is always positive then the average value won’t be zero as in the case of current or voltage in AC circuits.

Example: Laptop Transformer Power

Question

The transformer for a laptop has the following information:

• INPUT: $$\text{100}$$-$$\text{240}$$ $$\text{V}$$; $$\text{1.5}$$ $$\text{A}$$; $$\text{50}$$/$$\text{60}$$ $$\text{Hz}$$
• OUTPUT:$$\text{20}$$ $$\text{V}$$; $$\text{3.25}$$ $$\text{A}$$

Using the input values and assuming $$\text{240}$$ $$\text{V}$$, what is the average power dissipated in the transformer?

Step 1: RMS values

As calculated previously:

\begin{align*} V_{\text{rms}} & = \frac{V_{\text{max}}}{\sqrt{\text{2}}} \\ V_{\text{rms}} & = \frac{\text{240}\text{ V}}{\sqrt{\text{2}}} \\ & = \text{169.71}\text{ V} \end{align*}

Therefore $$V_{\text{rms}}=\text{169.71}\text{ V}$$

\begin{align*} I_{\text{rms}} & = \frac{I_{\text{max}}}{\sqrt{\text{2}}} \\ I_{\text{rms}} & = \frac{\text{1.5}\text{ A}}{\sqrt{\text{2}}} \\ & = \text{1.06}\text{ A} \end{align*}

Therefore $$I_{\text{rms}}=\text{1.06}\text{ A}$$

Step 2: Average power

\begin{align*} P_{av} &=I_{rms}V_{rms} \\ &= \text{1.06}\text{ A} \cdot \text{169.71}\text{ V}\\ & = \text{179.89}\text{ W} \end{align*}

Example: Motors and Generators (Exam Question)

Question

Diesel-electric trains make use of electric motors as well as generators.

1. The table below compares a motor and a generator in terms of the type of energy conversion and the underlying principle on which each operates. Complete the table by writing down only the question number (11.1.1–11.1.4) in the ANSWER BOOK and next to each number the answer.

 TYPE OF ENERGY CONVERSION PRINCIPLE OF OPERATION Motor 11.1.1 11.1.3 Generator 11.1.2 11.1.4

(4 marks)

2. The simplified diagram below represents an electric motor.

Give a reason why the section of the coil labelled BC in the above diagram does not experience a magnetic force whilst the coil is in the position as shown.

(2 marks)

3. Graphs of the current and potential difference outputs of an AC generator are shown below.

Calculate the average power output of this generator.

(6 marks)

[TOTAL: 12 marks]

1

1. Electrical (energy) to mechanical (kinetic) energy
2. Mechanical (kinetic) energy to electrical (energy)
3. Motor effect
4. Electromagnetic induction

(4 marks)

Question 2

BC (conductor) is parallel to the magnetic field.

OR

Open switch, no current

(2 marks)

Question 3

Option 1:

\begin{align*} P_{\text{ave}} &= V_{\text{rms}} I_{\text{rms}} \\ &= \frac{V_{\text{max}}}{\sqrt{2}} \cdot \frac{I_{\text{max}}}{\sqrt{2}} \\ &= \frac{(311)(21)}{2} \\ &= \text{3265.5 W} \end{align*}

OR

\begin{align*} P_{\text{max}} &= V_{\text{max}}I_{\text{max}} \\ & = (311)(21) \\ & = \text{6531 W} \\ \therefore P_{\text{ave}} & = \frac{P_{\text{max}}}{2} \\ & = \frac{\text{6 531}}{2} \\ & = \text{3265.5 W} \end{align*}

Option 2:

\begin{align*} V_{\text{rms}} &= \frac{V_{\text{max}}}{\sqrt{2}} \\ &= \frac{311}{\sqrt{2}} \\ &= \text{219.91}\text{ V} \end{align*}\begin{align*} I_{\text{rms}} &= \frac{I_{\text{max}}}{\sqrt{2}} \\ &= \frac{21}{\sqrt{2}} \\ &= \text{14.85}\text{ A} \end{align*}\begin{align*} P_{\text{ave}} & =V_{\text{rms}}I_{\text{rms}}\\ & =(\text{219.91})(\text{14.85})\\ & = \text{3 265.66}\text{ W} \end{align*}

Option 3:

\begin{align*} R & = \frac{V}{I} \\ & = \frac{\text{311}}{\text{21}}\\ & = \text{14.81}\ \Omega \end{align*}\begin{align*} I_{\text{rms}} &= \frac{I_{\text{max}}}{\sqrt{2}} \\ &= \frac{21}{\sqrt{2}} \\ &= \text{14.85}\text{ A} \end{align*}\begin{align*} P_{\text{ave}} &= I_{\text{rms}}^{2}R \\ &= (\text{14.85})^{2}(\text{14.81}) \\ &= \text{3 265.83}\text{ W} \end{align*}

Option 4:

\begin{align*} R & = \frac{V}{I} \\ & = \frac{\text{311}}{\text{21}}\\ & = \text{14.81}\ \Omega \end{align*}\begin{align*} V_{\text{rms}} &= \frac{V_{\text{max}}}{\sqrt{2}} \\ &= \frac{311}{\sqrt{2}} \\ &= \text{219.91}\text{ V} \end{align*}\begin{align*} P_{\text{ave}} &= \frac{V_{\text{rms}}^{2}}{R} \\ &= \frac{(\text{219.41})^2}{(\text{14.81})} \\ &= \text{3 265.83}\text{ W} \end{align*}

(6 marks)

[TOTAL: 12 marks]

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