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An AC power supply of negligible internal resistance is connected to an 8.0 Ω resistor - Scottish Highers Physics - Question 19 - 2023

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An AC power supply of negligible internal resistance is connected to an 8.0 Ω resistor. The rms voltage of the power supply is 5.0 V. The peak power dissipated in th... show full transcript

Worked Solution & Example Answer:An AC power supply of negligible internal resistance is connected to an 8.0 Ω resistor - Scottish Highers Physics - Question 19 - 2023

Step 1

Calculate the Peak Voltage

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Answer

To find the peak voltage ( V_{peak} ), we use the formula:

Vpeak=Vrmsimesextsqrt(2)V_{peak} = V_{rms} imes ext{sqrt}(2)

Given that the rms voltage ( V_{rms} ) is 5.0 V:

Vpeak=5.0imesextsqrt(2) =5.0imes1.414 ≈7.07extVV_{peak} = 5.0 imes ext{sqrt}(2) \ = 5.0 imes 1.414 \ ≈ 7.07 ext{ V}

Step 2

Calculate the Peak Power

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Answer

The power dissipated in a resistor can be calculated using the formula:

P = rac{V_{peak}^2}{R}

Where:

P is the power

V_{peak} is the peak voltage

R is the resistance (8.0 Ω)

Substituting the values:

P = rac{(7.07)^2}{8.0} \ = rac{50.0049}{8.0} \ = 6.25 ext{ W}

The closest answer from the options provided is 6.3 W, which is E.

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