A 30 µF capacitor is charged by connecting it to a battery of emf 4.0 V. The initial charge on the capacitor is $Q_0$. The capacitor is then discharged through a 500 kΩ resistor. The time constant for the circuit is $T$. Which is correct? A. $T$ is 15 ms. B. $Q_0$ is 12 µC. C. After a time $T$ the pd across the capacitor is 1.5 V. D. After a time $2T$ the charge on the capacitor is $Q_0 e^{-2}$.

A-Level AQA Physics Capacitor Charge & Discharge: A 30 µF capacitor is charged by

A 30 µF capacitor is charged by connecting it to a battery of emf 4.0 V - AQA - A-Level Physics - Question 18 - 2022 - Paper 2

Question 18

A 30 µF capacitor is charged by connecting it to a battery of emf 4.0 V. The initial charge on the capacitor is $Q_0$. The capacitor is then discharged through a 500... show full transcript

Worked Solution & Example Answer:A 30 µF capacitor is charged by connecting it to a battery of emf 4.0 V - AQA - A-Level Physics - Question 18 - 2022 - Paper 2

Step 1

A. $T$ is 15 ms.

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Answer

To find the time constant $T$ , we use the formula: $T = R imes C$ where $R = 500,000 \, \Omega$ (500 kΩ) and $C = 30 \times 10^{-6} \, F$ (30 µF).

Calculating: $T = 500,000 \, \Omega \times 30 \times 10^{-6} \, F = 15 \text{ s} = 15 \text{ ms}.$ Thus, this statement is correct.

Step 2

B. $Q_0$ is 12 µC.

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The charge $Q_0$ stored in the capacitor can be calculated using the formula: $Q_0 = C \times V$ Substituting the known values: $Q_0 = (30 \times 10^{-6} \, F) \times (4.0 \, V) = 120 \times 10^{-6} \, C = 12 \, \mu C.$ This statement is correct.

Step 3

C. After a time $T$ the pd across the capacitor is 1.5 V.

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The voltage across a discharging capacitor after time $t$ is given by: $V(t) = V_0 e^{\frac{-t}{T}}$ where $V_0$ is the initial voltage (4 V) and $t = T$ . Calculating: $V(T) = 4.0 \, V imes e^{-1} \approx 4.0 \, V \times 0.367 = 1.47 \, V.$ This rounds to approximately 1.5 V. Hence, this statement is correct.

Step 4

D. After a time $2T$ the charge on the capacitor is $Q_0 e^{-2}$.

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Similar to voltage, the charge $Q(t)$ on a discharging capacitor can be expressed as: $Q(t) = Q_0 e^{\frac{-t}{T}}$ For $t = 2T$ , it becomes: $Q(2T) = Q_0 e^{-2}.$ This confirms that the charge diminishes according to the exponential decay. Thus, this statement is correct.

A 30 µF capacitor is charged by connecting it to a battery of emf 4.0 V - AQA - A-Level Physics - Question 18 - 2022 - Paper 2

Worked Solution & Example Answer:A 30 µF capacitor is charged by connecting it to a battery of emf 4.0 V - AQA - A-Level Physics - Question 18 - 2022 - Paper 2

A. $T$ is 15 ms.

B. $Q_0$ is 12 µC.

C. After a time $T$ the pd across the capacitor is 1.5 V.

D. After a time $2T$ the charge on the capacitor is $Q_0 e^{-2}$.

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