Refer to the image above containing a capacitor with an open switch. The capacitor is charged to some initial value Q. When the switch is closed, it slowly discharges throughout the resistor. If it takes 2 seconds for the capacitor to discharge to one-half of the initial value, how long does it take for it to discharge to one-quarter of the initial value?

To determine the time it takes for the capacitor to discharge to one-quarter of its initial value, we need to understand the behavior of a discharging capacitor in an RC circuit.

The voltage \( V(t) \) across a discharging capacitor as a function of time \( t \) is given by the equation:

\[ V(t) = V_0 e^{-\frac{t}{RC}} \]

where:
- \( V_0 \) is the initial voltage,
- \( R \) is the resistance,
- \( C \) is the capacitance,
- \( t \) is the time.

Since \( Q = CV \), the charge \( Q(t) \) on the capacitor as a function of time is similarly:

\[ Q(t) = Q_0 e^{-\frac{t}{RC}} \]

where \( Q_0 \) is the initial charge. 

We know it takes 2 seconds for the capacitor to discharge to one-half of its initial value. Therefore, if \( Q(t) = \frac{Q_0}{2} \) at \( t = 2 \) seconds:

\[ \frac{Q_0}{2} = Q_0 e^{-\frac{2}{RC}} \]

Dividing both sides by \( Q_0 \):

\[ \frac{1}{2} = e^{-\frac{2}{RC}} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{1}{2}\right) = -\frac{2}{RC} \]

\[ -\ln(2) = -\frac{2}{RC} \]

\[ \ln(2) = \frac{2}{RC} \]

\[ RC = \frac{2}{\ln(2)} \]

Next, we want to find the time \( t \) when \( Q(t) = \frac{Q_0}{4} \):

\[ \frac{Q_0}{4} = Q_0 e^{-\frac{t}{RC}} \]

Dividing both sides by \( Q_0 \):

\[ \frac{1}{4} = e^{-\frac{t}{RC}} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{1}{4}\right) = -\frac{t}{RC} \]

\[ \ln(4) = \frac{t}{RC} \]

Since \( 4 = 2^2 \):

\[ \ln(4) = \ln(2^2) = 2 \ln(2) \]

Substituting \( \ln(2) \) back in:

\[ \ln(4) = 2 \ln(2) = 2 \left(\frac{2}{RC}\right) \]

Since \( RC = \frac{2}{\ln(2)} \), we have:

\[ \ln(4) = 2 \cdot \ln(2) = 2 \cdot \left(\frac{2}{RC}\right) \]

\[ \frac{t}{RC} = 2 \]

\[ t = 2 RC \]

Given that \( RC = \frac{2}{\ln(2)} \):

\[ t = 2 \cdot \frac{2}{\ln(2)} \]

\[ t = 4 \cdot \frac{1}{\ln(2)} \]

\[ t = 4 \]

Therefore, it takes 4 seconds for the capacitor to discharge to one-quarter of its initial value.

Do you want more details or have any questions?

Here are five questions you might ask next:
1. What is the formula for the time constant in an RC circuit?
2. How does the resistance affect the discharge rate of a capacitor?
3. What happens to the voltage across a capacitor as it discharges?
4. How do you calculate the energy dissipated in a resistor during capacitor discharge?
5. What are the applications of RC circuits in real-life scenarios?

Tip: Understanding the exponential nature of capacitor discharge helps in analyzing the behavior of RC circuits in various applications.

Explore the solution to a capacitor discharge problem where it takes 2 seconds to reach one-half of its initial value. Learn how to calculate the time for the capacitor to discharge to one-quarter of its initial value using exponential decay in an RC circuit.

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