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The circuit shows a 1 mF capacitor connected to an a.c - Edexcel - A-Level Physics - Question 18 - 2023 - Paper 1

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The circuit shows a 1 mF capacitor connected to an a.c. supply. The graph shows how the potential difference V varies with time t. (a) (i) Calculate the root-mean-s... show full transcript

Worked Solution & Example Answer:The circuit shows a 1 mF capacitor connected to an a.c - Edexcel - A-Level Physics - Question 18 - 2023 - Paper 1

Step 1

(i) Calculate the root-mean-square potential difference.

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Answer

To calculate the root-mean-square (RMS) potential difference for an AC voltage described by the equation V=V0imesextsin(heta)V = V_0 imes ext{sin}( heta), we use the formula:

Vrms=V02V_{rms} = \frac{V_0}{\sqrt{2}}

In this case, the maximum voltage V0V_0 is given as 5 V. Therefore, substituting this value:

Vrms=523.54 VV_{rms} = \frac{5}{\sqrt{2}} \approx 3.54 \text{ V}

Step 2

(ii) Explain why this formula leads to the graph above.

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Answer

The formula V=5sin(100πt)V = 5\sin(100\pi t) describes a sinusoidal wave with a maximum amplitude of 5 V. The term 100π100\pi indicates the angular frequency of the waveform, which determines how fast the wave oscillates. This results in a periodic graph, depicting the voltage as it varies over time.

  1. Amplitude and Periodicity: The graph shows that the voltage oscillates between +5 V and -5 V, reflecting the maximum amplitude given by the formula.
  2. Frequency: The frequency of the graph can be found from the coefficient of tt in the sine function. In this case, the angular frequency ω=100π\omega = 100\pi leads to a frequency of f=ω2π=50f = \frac{\omega}{2\pi} = 50 Hz, indicating the rapid oscillation.
  3. Negative Values: The sine function's nature allows for both positive and negative values, which is clearly illustrated in the graph as the voltage dips below 0 V, demonstrating the alternating nature of the current.

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