A capacitor of capacitance 120 μF is charged and then discharged through a 20 kΩ resistor - AQA - A-Level Physics - Question 21 - 2017 - Paper 2
Question 21
A capacitor of capacitance 120 μF is charged and then discharged through a 20 kΩ resistor.
What fraction of the original charge remains on the capacitor 4.8 s after... show full transcript
Worked Solution & Example Answer:A capacitor of capacitance 120 μF is charged and then discharged through a 20 kΩ resistor - AQA - A-Level Physics - Question 21 - 2017 - Paper 2
Step 1
Calculate the Time Constant
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Answer
The time constant τ (tau) for an RC circuit is given by the formula:
τ=RimesC
Here, the resistance R = 20 kΩ = 20,000 Ω and the capacitance C = 120 μF = 120 × 10^{-6} F. Therefore:
τ=20,000imes120imes10−6=2.4s
Step 2
Determine the Charge Remaining
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Answer
The fraction of the charge remaining after time t can be calculated using:
Q(t)=Q0e−t/τ
Where Q(t) is the charge at time t, Q_0 is the initial charge, and e is Euler's number.
At t = 4.8 s:
Q(4.8)=Q0e−4.8/2.4=Q0e−2
To find the fraction of the original charge:
Fraction remaining=e−2≈0.1353
Step 3
Find the Closest Answer Choice
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Answer
The calculated fraction of approximately 0.1353 rounds to 0.14. Therefore, the correct answer is:
