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An uncharged 220 µF capacitor is connected in a circuit as shown - Scottish Highers Physics - Question 13 - 2017

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An uncharged 220 µF capacitor is connected in a circuit as shown. The 12V battery has negligible internal resistance. (a) Switch $S_1$ is closed and the capacitor ... show full transcript

Worked Solution & Example Answer:An uncharged 220 µF capacitor is connected in a circuit as shown - Scottish Highers Physics - Question 13 - 2017

Step 1

Calculate the initial charging current.

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Answer

To find the initial charging current when the switch S1S_1 is closed, we can apply Ohm’s Law, which states that:

V=IRV = IR

Here, we have:

  • Voltage, V=12VV = 12V
  • Resistance, R=6800ΩR = 6800 \Omega

Rearranging the formula to find current II gives: I=VRI = \frac{V}{R}

Substituting the known values: I=126800I = \frac{12}{6800}

Calculating this yields: I1.76×103AI \approx 1.76 \times 10^{-3} A

Thus, the initial charging current is approximately 1.76 mA1.76 \ mA.

Step 2

Explain why the time for the capacitor to fully charge is less than in part (a).

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Answer

When switch S1S_1 is closed and the capacitor is charged, the initial current is at its maximum because the capacitor behaves like a short circuit at that moment.

Once the capacitor is at a voltage close to the supply voltage, the charging current decreases significantly.

However, when switch S2S_2 is closed immediately after discharging the capacitor, the equivalent resistance of the circuit becomes less than when charging through a 6800 Ω resistance alone.

This lesser resistance results in a greater initial charging current when switch S1S_1 is closed again. Since the current is greater, the capacitor charges faster, leading to a shorter time to fully charge than in part (a).

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