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A capacitor (C) is placed in series in a circuit with a power supply and a resistor (R).A sine wave voltage input is supplied and the voltage across the resistor (proportional to the current) and across both the capacitor and resistor (voltage) observed on an oscilloscope.A student determines the capacitance by measuring the period of the waveforms and time difference between the waveforms for two different input frequencies on the function generator. The values obtained are:C1 = 0.05 µFC2= 0.12 µFAn average value of the capacitance (C) can now be found. Determine the uncertainty in this average value. Give your answer in SI units to two significant figures.

Question

A capacitor (C) is placed in series in a circuit with a power supply and a resistor (R).A sine wave voltage input is supplied and the voltage across the resistor (proportional to the current) and across both the capacitor and resistor (voltage) observed on an oscilloscope.A student determines the capacitance by measuring the period of the waveforms and time difference between the waveforms for two different input frequencies on the function generator. The values obtained are:C1 = 0.05 µFC2= 0.12 µFAn average value of the capacitance (C) can now be found. Determine the uncertainty in this average value. Give your answer in SI units to two significant figures.

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Solution

To calculate the average value of the capacitance, you would add the two values together and divide by 2:

C_avg = (C1 + C2) / 2 C_avg = (0.05 µF + 0.12 µF) / 2 C_avg = 0.085 µF

To find the uncertainty in this average value, you would subtract the smallest value from the largest value and divide by 2:

Uncertainty = (C2 - C1) / 2 Uncertainty = (0.12 µF - 0.05 µF) / 2 Uncertainty = 0.035 µF

So, the average value of the capacitance is 0.085 µF with an uncertainty of 0.035 µF.

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