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A fully charged capacitor of capacitance 2.0 mF discharges through a 15 kΩ resistor - AQA - A-Level Physics - Question 21 - 2022 - Paper 2

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A fully charged capacitor of capacitance 2.0 mF discharges through a 15 kΩ resistor. What fraction of the stored energy remains after 1.0 minute?

Worked Solution & Example Answer:A fully charged capacitor of capacitance 2.0 mF discharges through a 15 kΩ resistor - AQA - A-Level Physics - Question 21 - 2022 - Paper 2

Step 1

What fraction of the stored energy remains after 1.0 minute?

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Answer

To find the fraction of the stored energy remaining after a certain time during the discharge of a capacitor, we use the formula:

E(t)=E0et/RCE(t) = E_0 e^{-t/RC}

where:

  • E(t)E(t) is the energy at time tt,
  • E0E_0 is the initial energy,
  • RR is the resistance, and
  • CC is the capacitance.

First, we calculate the time constant (au au) using:

au=RimesC au = R imes C

Given values:

  • Capacitance (CC) = 2.0 mF = 2.0 \times 10^{-3} F,
  • Resistance (RR) = 15 kΩ = 15 \times 10^{3} Ω.

Calculating the time constant:

au=(15×103)×(2.0×103)=30extseconds au = (15 \times 10^{3}) \times (2.0 \times 10^{-3}) = 30 ext{ seconds}

Next, we know that after 1.0 minute (60 seconds), the fraction of energy remaining is:

rac{E(t)}{E_0} = e^{-t/RC} = e^{-60/30} = e^{-2}

Therefore, the fraction of the stored energy remaining after 1.0 minute is:

1e2\frac{1}{e^{2}}

Thus, the correct choice is:

B) rac{1}{e^2}.

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