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Magazine Chapter 18: Problem 38
A series RC circuit consisting of a 5.0-M\Omega resistor and a \(0.40-\mu \mathrm{F}\) capacitor is connected to a \(12-\mathrm{V}\) battery. If the capacitor is initially uncharged, (a) what is the change in voltage across it between \(t=2 \tau\) and \(t=4 \tau\) ? (b) By how much does the capacitor's stored energy change in the same time interval?
Short Answer
Expert verified
(a) 1.4008 V; (b) 0.00618 J
Step by step solution
01
Calculate the Time Constant
For an RC circuit, the time constant \( \tau \) is given by \( \tau = R \times C \). Here, \( R = 5.0 \times 10^{6} \; \Omega \) and \( C = 0.40 \times 10^{-6} \; F \). So, \( \tau = 5.0 \times 10^{6} \times 0.40 \times 10^{-6} = 2.0 \; \text{seconds} \).
02
Compute the Voltage Across the Capacitor at t=2蟿
The voltage across the capacitor at time \( t \) is \( V(t) = V_{0}(1 - e^{-t/\tau}) \), where \( V_{0} = 12 \; V \). At \( t = 2\tau \), the voltage is: \[V(2\tau) = 12 (1 - e^{-2}) \approx 12 (1 - 0.1353) \approx 10.3776 \; V.\]
03
Compute the Voltage Across the Capacitor at t=4蟿
Using the same formula \( V(t) = V_{0}(1 - e^{-t/\tau}) \). At \( t = 4\tau \), the voltage is: \[V(4\tau) = 12 (1 - e^{-4}) \approx 12 (1 - 0.0183) \approx 11.7784 \; V.\]
04
Determine the Change in Voltage
The change in voltage from \( t = 2\tau \) to \( t = 4\tau \) is the difference in voltages at these moments:\[\Delta V = V(4\tau) - V(2\tau) = 11.7784 - 10.3776 \approx 1.4008 \; V.\]
05
Calculate Capacitor Energy at t=2蟿
The energy stored in a capacitor is \( U = \frac{1}{2}CV^2 \). At \( t = 2\tau \), \[U_{2\tau} = \frac{1}{2} \times 0.40 \times 10^{-6} \times (10.3776)^2 \approx 0.02156 \; \text{Joules}.\]
06
Calculate Capacitor Energy at t=4蟿
At \( t = 4\tau \), the energy is:\[U_{4\tau} = \frac{1}{2} \times 0.40 \times 10^{-6} \times (11.7784)^2 \approx 0.02774 \; \text{Joules}.\]
07
Determine the Change in Stored Energy
The change in energy from \( t = 2\tau \) to \( t = 4\tau \) is:\[\Delta U = U_{4\tau} - U_{2\tau} = 0.02774 - 0.02156 \approx 0.00618 \; \text{Joules}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Time Constant
In an RC circuit, the time constant, represented by the symbol \( \tau \), is a fundamental parameter that helps to determine the charging and discharging rates of a capacitor through a resistor. The time constant is calculated as the product of resistance \( R \) and capacitance \( C \): \( \tau = R \times C \).
In the given exercise with a 5.0 M惟 resistor and a 0.40 \( \mu \text{F} \) capacitor, the time constant is:
The time constant tells us how quickly the capacitor charges or discharges to about 63% of the full voltage. With every increment of \( \tau \), the capacitor voltage climbs closer to the supply voltage, especially in the context of exponential charging.
In the given exercise with a 5.0 M惟 resistor and a 0.40 \( \mu \text{F} \) capacitor, the time constant is:
- \( \tau = 5.0 \times 10^6 \Omega \times 0.40 \times 10^{-6} \text{F} = 2.0 \text{ seconds} \)
The time constant tells us how quickly the capacitor charges or discharges to about 63% of the full voltage. With every increment of \( \tau \), the capacitor voltage climbs closer to the supply voltage, especially in the context of exponential charging.
Capacitor Voltage
Understanding how capacitor voltage changes over time in an RC circuit is vital for many electronic applications. The voltage \( V(t) \) across a capacitor in a charging process is given by the equation:
\[V(t) = V_0(1 - e^{-t/\tau})\]
where \( V_0 \) is the final voltage supplied, and \( e \) is the base of the natural logarithm.
In our exercise, the battery provides \( 12 \text{ V} \). Using this formula, you calculate:
Increasing the time from \( 2\tau \) to \( 4\tau \) effectively allows the capacitor to gain more voltage, as reflected in these calculations.
\[V(t) = V_0(1 - e^{-t/\tau})\]
where \( V_0 \) is the final voltage supplied, and \( e \) is the base of the natural logarithm.
In our exercise, the battery provides \( 12 \text{ V} \). Using this formula, you calculate:
- At \( t = 2\tau \), \( V(2\tau) = 12(1 - e^{-2}) \approx 10.38 \text{ V} \)
- At \( t = 4\tau \), \( V(4\tau) = 12(1 - e^{-4}) \approx 11.78 \text{ V} \)
Increasing the time from \( 2\tau \) to \( 4\tau \) effectively allows the capacitor to gain more voltage, as reflected in these calculations.
Stored Energy
The stored energy in a capacitor is critical for storing and releasing power in circuits. The energy \( U \) stored in a charged capacitor is calculated using the formula:
\[U = \frac{1}{2} C V^2\]
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.
For our given problem, let's find the stored energy at different times:
The change in energy indicates how the charging of the capacitor increases its power-storing capability over time.
\[U = \frac{1}{2} C V^2\]
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.
For our given problem, let's find the stored energy at different times:
- At \( t = 2\tau \), the stored energy is \( U_{2\tau} = \frac{1}{2} \times 0.40 \times 10^{-6} \times (10.38)^2 \approx 0.02156 \text{ Joules} \).
- At \( t = 4\tau \), \( U_{4\tau} = \frac{1}{2} \times 0.40 \times 10^{-6} \times (11.78)^2 \approx 0.02774 \text{ Joules} \).
The change in energy indicates how the charging of the capacitor increases its power-storing capability over time.
Exponential Charging
The concept of exponential charging is intrinsic to the behavior of RC circuits. When a capacitor in an RC circuit begins to charge, it does not move linearly; rather, it follows an exponential curve.
Initially, the charge enters the capacitor swiftly. Over successive time constants, the rate of charging slows as the capacitor voltage rises asymptotically towards the supply voltage.
The "exponential" nature of this charging implies the capacitor never fully reaches the applied voltage (theoretically), although it becomes negligible close beyond four or five \( \tau \). This property is utilized in timers, filters, and other critical components in electronic systems.
Initially, the charge enters the capacitor swiftly. Over successive time constants, the rate of charging slows as the capacitor voltage rises asymptotically towards the supply voltage.
- After one time constant \( (1\tau) \), approximately 63% of the voltage is reached.
- After two time constants \( (2\tau) \), the voltage reaches approximately 86.5% of the supply voltage.
- By four time constants \( (4\tau) \), it gets around 98% of \( V_0 \).
The "exponential" nature of this charging implies the capacitor never fully reaches the applied voltage (theoretically), although it becomes negligible close beyond four or five \( \tau \). This property is utilized in timers, filters, and other critical components in electronic systems.
