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A capacitor (with capacitance C) is placed in series in a circuit with a power supply and a resistor. The power supply has some internal resistance R1 = 0.2 kΩ and the resistor has value R2 = 0.7 kΩ.A square wave voltage input is supplied, with the voltage switching between constant positive and negative levels. The behaviour of the current through the circuit is observed on an oscilloscope. It is seen to drop exponentially with time after each voltage switch. The half time is measured to be t1/2= 30 µs.Calculate the value of the capacitor C. Give your answer to 3 significant figures.Note you need to use the total resistance in the circuit R = R1 + R2 in your calculation.Entering numbers in scientific notation: Example: 1.45 x 10-9 should be entered as 1.45E-9

Question

A capacitor (with capacitance C) is placed in series in a circuit with a power supply and a resistor. The power supply has some internal resistance R1 = 0.2 kΩ and the resistor has value R2 = 0.7 kΩ.A square wave voltage input is supplied, with the voltage switching between constant positive and negative levels. The behaviour of the current through the circuit is observed on an oscilloscope. It is seen to drop exponentially with time after each voltage switch. The half time is measured to be t1/2= 30 µs.Calculate the value of the capacitor C. Give your answer to 3 significant figures.Note you need to use the total resistance in the circuit R = R1 + R2 in your calculation.Entering numbers in scientific notation: Example: 1.45 x 10-9 should be entered as 1.45E-9

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Solution

The half-life of a capacitor discharging through a resistor is given by the formula t1/2 = 0.693RC.

In this case, we know that t1/2 = 30 µs and the total resistance R = R1 + R2 = 0.2 kΩ + 0.7 kΩ = 0.9 kΩ = 0.9E3 Ω.

We can rearrange the formula to solve for C:

C = t1/2 / (0.693*R)

Substituting the given values:

C = 30E-6 / (0.693*0.9E3)

This gives C = 4.87E-8 F or 48.7 nF to 3 significant figures.

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