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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 k... 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.
Answer
To find the time constant T, we use the formula: where R is the resistance and C is the capacitance. In this case, R = 500 k\Omega = 500,000 \Omega \text{ and } C = 30 \mu F = 30 \times 10^{-6} F.$$
Calculating: egin{align*} T & = 500,000 \Omega \times 30 \times 10^{-6} F \ & = 15 \text{ ms}. \end{align*} Thus, this statement is correct.
Step 2
B Q_0 is 12 µC.
Answer
The initial charge Q_0 on the capacitor can be calculated using: where V is the voltage. Here, V is 4.0 V.
Substituting the known values: egin{align*} Q_0 & = 30 \times 10^{-6} F \times 4.0 V \ & = 120 \times 10^{-6} C \ & = 12 \mu C. \end{align*} Thus, this statement is correct.
Step 3
C After a time T the pd across the capacitor is 1.5 V.
Answer
The pd V after a time T can be found using the formula: Substituting t = T and V_0 = 4.0 V: egin{align*} V(T) & = 4.0 V imes e^{-1} \ & = 4.0 V imes \frac{1}{e} \ & \approx 4.0 V \times 0.3679 \approx 1.47 V. \end{align*} This indicates that after time T, the pd across the capacitor is approximately 1.47 V, which rounds to 1.5 V. Thus, this statement is considered correct.
Step 4