Problem with solution: Charge, Capacitance and Half-Life of a Capacitor When Discharging From 50V to 15V.

RC Circuit - discharging capacitor
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Considered circuit for discharging the capacitor.

A capacitor of unknown capacitance with a resistor \(R = 2 \, \mathrm{k\Omega} \) was charged to \( 50 \, \mathrm{V} \) and discharged to \( 15 \, \mathrm{V} \) within one millisecond.

  1. What is the capacitance \( C \) of the capacitor?

  2. What is the time constant of the capacitor?

  3. After what time \(t_{\mathrm h}\) have \( 50 \, \mathrm{V} \) decreased to half?

  4. How much charge \( Q \) did the capacitor carry before the discharge process?

Solution tips Use the formula of the voltage across the capacitor for the discharge process. And, to find the charge, look up how the charge and voltage on the capacitor are related.
Solution for (a)

To find out the capacitance \(C\) of the capacitor, we use the formula that describes the time dependence of the voltage across the capacitor during the discharge process:

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What we are looking for is the capacitance, so we rearrange Eq. 1 for \(C\):

Capacitance as a function of voltage, resistance and time
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The voltage \( U_{\mathrm C}(1\,\mathrm{ms}) = 15 \, \mathrm{V} \) after \( t = 1 \,\mathrm{ms} \) is given. The initial voltage \( U_0 = 50 \, \mathrm{V} \) and in series connected resistor \(R = 2000 \mathrm{\Omega} \) are also known. Inserting the concrete values yields the capacitance we are looking for:

Example: Calculate capacitance of a discharging capacitor
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Solution for (b)

The time constant \( \tau \) is given by:

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It describes the time after which the initial voltage has dropped to about 37%. Inserting the discharge resistance and the capacitance found in a) yields:

Example: Calculate time constant during capacitor discharge
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Solution for (c)

In this exercise we want to calculate the half-life \( t_{\mathrm h} \). The half-life indicates the time after which the initial voltage of \( U_0 = 50 \, \mathrm{V} \) has dropped to half, that is to \( U_{\mathrm C} = \frac{U_0}{2} = 25 \, \mathrm{V} \):

Equate initial voltage with discharge formula for half-life
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Rearranging for half-life \( t_{\mathrm h} \) yields:

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As you can see, the half-life is determined by the time constant \(R \, C \), which is simply multiplied by \( \ln(2) \). So we can simply use the time constant calculated in b) for the half-life:

Example: Calculate half-life during capacitor discharge
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Solution for (d)

Here we want to find out how much charge \(Q_0\) was on the capacitor before the capacitor was discharged. So we are looking for capacitor charge at the initial time \(t = 0\). To do this, we use the relationship between the voltage and charge on the capacitor:

Initial charge during the discharge process of the capacitor
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Of course we need the initial voltage \( U_0 \), because this was applied at the initial time \(t = 0\). Inserting the values results in:

Example: Calculate charge during capacitor discharge
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