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 = 0.4 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= 36 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 = 0.4 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= 36 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
Solution
To calculate the value of the capacitor C, we need to use the formula for the time constant τ (tau) of an RC (resistor-capacitor) circuit, which is τ = RC.
In this case, the time constant τ is equal to the half-life t1/2 divided by the natural logarithm of 2, because the current in an RC circuit drops to half its initial value in one half-life.
So, τ = t1/2 / ln(2) = 36E-6 / 0.6931 = 51.96E-6 seconds.
The total resistance R in the circuit is the sum of the internal resistance of the power supply R1 and the resistance of the resistor R2, so R = R1 + R2 = 0.4E3 + 0.7E3 = 1.1E3 ohms.
Now we can calculate the capacitance C using the formula for the time constant: C = τ / R = 51.96E-6 / 1.1E3 = 47.24E-9 farads.
So, the value of the capacitor C is approximately 47.2 nF to 3 significant figures.
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