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A student investigates the charging of a capacitor - Scottish Highers Physics - Question 13 - 2019

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A student investigates the charging of a capacitor. The student sets up the circuit as shown using a 47 μF capacitor. The capacitor is initially uncharged. The swit... show full transcript

Worked Solution & Example Answer:A student investigates the charging of a capacitor - Scottish Highers Physics - Question 13 - 2019

Step 1

Calculate the maximum charge stored by the capacitor.

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Answer

The maximum charge ( Q ) stored by a capacitor can be calculated using the formula:

Q=C×VQ = C \times V

where:

  • C = capacitance in farads (F)
  • V = voltage in volts (V)

Given:

  • C = 47 μF = 47 \times 10^{-6} F
  • V = 6 V

Substituting the values:

Q=47×106F×6V=2.82×105CQ = 47 \times 10^{-6} F \times 6 V = 2.82 \times 10^{-5} C

Thus, the maximum charge stored by the capacitor is approximately 28.2 µC.

Step 2

Add a line to the graph to show how the current now varies with time as the capacitor charges.

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Answer

To show the effect of a larger resistance on the current, the line drawn on the graph should start at a lower initial current compared to the original.

  • The line should rise more slowly and approach the x-axis over a longer time period, indicating a longer charging time due to the increased resistance.

Step 3

Suggest an alteration to increase the maximum energy stored by the 47 μF capacitor.

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Answer

To increase the maximum energy stored in the capacitor, the student could increase the supply voltage.

This is because the energy ( E ) stored in a capacitor can be calculated by:

E=12CV2E = \frac{1}{2} C V^2

Increasing the voltage supplies more potential energy to the capacitor, consequently increasing the energy stored.

Step 4

Comment on the analogy between vehicles and the charging of a capacitor.

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Answer

The analogy of vehicles going over a speed bump helps illustrate the concept of charging a capacitor. Just as vehicles experience a gradual increase in speed and then stabilize after the bump, a capacitor charges gradually, initially drawing a lot of current which tapers off over time.

Both scenarios depict a delayed response where an initial burst of energy decreases, representing the charging behavior of a capacitor as it approaches its maximum voltage.

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