Question

Consider a series \(\mathrm{RC}\) circuit with \(R=10.0 \Omega\) \(C=10.0 \mu \mathrm{F}\) and \(V=10.0 \mathrm{~V}\) a) How much time, expressed as a multiple of the time constant, does it take for the capacitor to be charged to half of its maximum value? b) At this instant, what is the ratio of the energy stored in the capacitor to its maximum possible value? c) Now suppose the capacitor is fully charged. At time \(t=\) 0 , the original circuit is opened and the capacitor is allowed to discharge across another resistor, \(R^{\prime}=1.00 \Omega\), that is connected across the capacitor. What is the time constant for the discharging of the capacitor? d) How many seconds does it take for the capacitor to discharge half of its maximum stored charge, \(Q\) ?

Short answer

Question: Determine the time it takes for the capacitor to charge to half of its maximum value in terms of the time constant, the ratio of energy stored in the capacitor at that instant to its maximum possible value, the time constant for the discharging of the capacitor when connected to another resistor, and the time it takes for the capacitor to discharge half of its maximum stored charge. Answer: a) The time it takes for the capacitor to charge to half of its maximum value is \(\tau\ln(2)\). b) The ratio of energy stored in the capacitor to its maximum possible value at this instant is 1/4. c) The time constant for the discharging of the capacitor is \(10^{-5}\mathrm{s}\). d) The time it takes for the capacitor to discharge half of its maximum stored charge is \(10^{-5}\mathrm{s}\ln(2)\).

Step-by-step solution

1. Recall the charging equation and time constant

The charge on the capacitor as a function of time when charging in an RC circuit is given by: \(Q(t)=C(1-e^{-t/(\tau)})V\) Where \(Q(t)\) is the charge at time \(t\), \(V\) is the applied voltage, \(C\) is the capacitance, and \(\tau=RC\) is the time constant. Here, \(\tau = R \cdot C = 10.0 \Omega \cdot 10.0 \times 10^{-6} \mathrm{F}=10^{-4}\mathrm{s}\).

2. Solve for the time to reach half the maximum value

To find the time it takes for the capacitor to charge to half of its maximum value, we need to set \(Q(t)\) equal to half of the maximum charge: \(\frac{1}{2}Q_{\mathrm{max}}(t)=C(1-e^{-t/(\tau)})V\) Since \(Q_{\mathrm{max}} = CV\), we can write: \(\frac{1}{2}=1-e^{-t/(\tau)}\) Now, solve for \(t\): \(e^{-t/\tau} = \frac{1}{2}\) Taking the natural logarithm of both sides, we get: \(-t/\tau = \ln(\frac{1}{2})\) Thus: \(t = -\tau\ln(\frac{1}{2}) = \tau\ln(2)\)

3. Calculate the energy ratio

The energy stored in a capacitor is given by: \(U=\frac{1}{2}CV^2\) At half the charge, the voltage across the capacitor is half the maximum voltage (\(V/2\)). So, the energy ratio is: \(\frac{U_{\frac{1}{2}}}{U_{\mathrm{max}}}=\frac{\frac{1}{2}C(\frac{V}{2})^2}{\frac{1}{2}CV^2}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

4. Calculate the time constant for discharging

The time constant for discharging is given by: \(\tau' = R'C = 1.00\mathrm{\Omega} \cdot 10.0 \times 10^{-6}\mathrm{F} = 10^{-5}\mathrm{s}\)

5. Find the time to discharge half the maximum stored charge

The expression for the charge on the capacitor during discharge is given by: \(Q(t) = Q_{\mathrm{max}}e^{-t/\tau'}\) We want to find the time when the charge is half of its maximum value: \(\frac{1}{2}Q_{\mathrm{max}}=Q_{\mathrm{max}}e^{-t/\tau'}\) Canceling \(Q_{\mathrm{max}}\) on both sides and solving for \(t\), we get: \(t=-\tau'\ln(\frac{1}{2})=\tau'\ln(2)= 10^{-5}\mathrm{s}\ln(2)\) To summarize the answers: a) The time it takes for the capacitor to charge to half of its maximum value is \(\tau\ln(2)\). b) The ratio of energy stored in the capacitor to its maximum possible value at this instant is 1/4. c) The time constant for the discharging of the capacitor is \(10^{-5}\mathrm{s}\). d) The time it takes for the capacitor to discharge half of its maximum stored charge is \(10^{-5}\mathrm{s}\ln(2)\).

Key concepts

Time Constant

Understanding the time constant, denoted as \( \tau \), is essential when exploring RC circuits. It's a measure of the time required for a capacitor to charge to about 63.2% of the maximum charge or, conversely, the time it takes for the capacitor to discharge to approximately 36.8% of its maximum charge when connected to a resistor. The time constant is calculated by multiplying the resistance (\( R \)) by the capacitance (\( C \)), resulting in \( \tau = R \times C \).

The importance of the time constant lies in its universality; it is a constant for a given RC circuit regardless of the changes in time or voltage. This characteristic allows us to predict the behavior of the circuit over time. In our educational platform's exercise, the circuit has a resistance of 10.0 ohms and a capacitance of 10.0 microfarads, leading to a time constant of \( 10^{-4} \) seconds for the charging process. Conversely, when a 1.00-ohm resistor is used for discharging, the time constant becomes \( 10^{-5} \) seconds.

Capacitor Charge and Discharge

Capacitors have the ability to store and release energy in the form of an electrical charge. In the context of an RC circuit, the process of charging involves a capacitor accumulating charge over time until it reaches its maximum capacity, dictated by the supplied voltage and the circuit's time constant. The mathematical expression describing charge accumulation is \( Q(t) = C(1 - e^{-t / \tau})V \), which shows how charge \( Q(t) \) changes with time \( t \).

Discharging, on the other hand, entails the release of the stored charge from the capacitor through a resistor. The discharging behavior is represented by the equation \( Q(t) = Q_{\mathrm{max}}e^{-t / \tau'} \). The exercise requires recognizing that the time to reach half the maximum charge or discharge is a function of the time constant, specifically \( t = \tau \ln(2) \) for charging and \( t = \tau' \ln(2) \) for discharging. This relationship highlights the exponential nature of charge variation in capacitors, a key concept in understanding how they behave in circuits.

Exponential Decay

Exponential decay is a fundamental concept that describes how a quantity decreases rapidly at first, then more slowly over time. It is characterized by a decreasing rate that is proportional to the current value, producing a distinctive curve when plotted. In an RC circuit, the charge and voltage exhibit exponential decay during the discharging phase, as depicted in the equation \( Q(t) = Q_{\mathrm{max}}e^{-t / \tau'} \).

The base of the natural logarithm, \( e \), is pivotal in these expressions, highlighting the natural exponential function's role in the decay process. Real-world applications of exponential decay extend beyond RC circuits and are found in diverse fields such as radioactivity, pharmacokinetics, and even finance. In our specific RC circuit problem, the capacitor's charge halves in a duration given by \( \tau' \ln(2) \), showing the practical use of exponential decay in predicting how rapidly a capacitor will lose its charge over time.