A-Level AQA Physics Work, Energy & Power: The mean power dissipated in a r

The mean power dissipated in a resistor is 47.5 µW when the root mean square (rms) voltage across the resistor is 150 mV - AQA - A-Level Physics - Question 23 - 2017 - Paper 2

Question 23

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The mean power dissipated in a resistor is 47.5 µW when the root mean square (rms) voltage across the resistor is 150 mV. What is the peak current in the resistor?

Worked Solution & Example Answer:The mean power dissipated in a resistor is 47.5 µW when the root mean square (rms) voltage across the resistor is 150 mV - AQA - A-Level Physics - Question 23 - 2017 - Paper 2

Step 1

Calculate the RMS Current

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Answer

To find the peak current, we first need to calculate the RMS current using the formula for power:

$P = I_{rms}^2 R$

Where:

$P$ is the power in watts (W)
$I_{rms}$ is the root mean square current in amperes (A)
$R$ is the resistance in ohms (Ω)

We rearrange this to find $I_{rms}$ :

$I_{rms} = \sqrt{\frac{P}{R}}$

For this problem, we need to find resistance using the RMS voltage:

$P = \frac{V_{rms}^2}{R}$

Rearranging gives:

$R = \frac{V_{rms}^2}{P}$

Using $P = 47.5 \times 10^{-6} W$ and $V_{rms} = 150 \times 10^{-3} V$ , we first find R.

Step 2

Determine the Resistance

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Answer

Calculating Resistance:

Compute $V_{rms}^2$ :

$V_{rms}^2 = (150 \times 10^{-3})^2 = 2.25 \times 10^{-4}$
Calculate Resistance $R$ :

$R = \frac{2.25 \times 10^{-4}}{47.5 \times 10^{-6}} \approx 4.7368 \Omega$

Step 3

Calculate the RMS Current

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Answer

Now substituting $R$ back to find $I_{rms}$ :

$I_{rms} = \sqrt{\frac{P}{R}} = \sqrt{\frac{47.5 \times 10^{-6}}{4.7368}} \approx 0.0032 A = 3.2 \times 10^{-3} A$

Step 4

Convert RMS Current to Peak Current

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Answer

The relationship between peak current ( $I_{peak}$ ) and RMS current is:

$I_{peak} = I_{rms} \sqrt{2}$

Thus:

$I_{peak} = 3.2 \times 10^{-3} \sqrt{2} \approx 4.52 \times 10^{-3} A = 4.5 \times 10^{-3} A$

Therefore, the peak current in the resistor is approximately $4.5 \times 10^{-3} A$ .

The mean power dissipated in a resistor is 47.5 µW when the root mean square (rms) voltage across the resistor is 150 mV - AQA - A-Level Physics - Question 23 - 2017 - Paper 2

Worked Solution & Example Answer:The mean power dissipated in a resistor is 47.5 µW when the root mean square (rms) voltage across the resistor is 150 mV - AQA - A-Level Physics - Question 23 - 2017 - Paper 2

Calculate the RMS Current

Determine the Resistance

Calculate the RMS Current

Convert RMS Current to Peak Current

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