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A \(12.0 \mu \mathrm{F}\) capacitor is charged to a potential of \(50.0 \mathrm{~V}\) and then discharged through a \(225 \Omega\) resistor. How long does it take the capacitor to lose (a) half of its charge and (b) half of its stored energy?

Short Answer

Expert verified
It takes approximately 1.9 ms for the capacitor to lose half of its charge and 2.7 ms for it to lose half of its stored energy.

Step by step solution

01

Calculate time to lose half charge

We are given that the capacitance \(C = 12.0 \mu F\), the voltage \(V=50.0V\) and the resistance \(R = 225 \Omega \). Since it's asked for the time it takes for the capacitor to lose half of its charge, we can use the formula for the time constant of an RC circuit: \(t=RCln2\). Plugging the values in gives us \(t = 225 \Omega * 12.0 \mu F * ln(2) = 1.9 ms\).
02

Calculate time to lose half energy

The energy stored in a capacitor is given by the formula \(U=\frac{1}{2}CV^{2}\). To lose half of this energy, the voltage must decrease by a factor of \(\sqrt{2}\) because energy is directly proportional to the square of the voltage. To calculate the corresponding time for this to happen, we can use the voltage decay in a discharging capacitor formula \(V=V_{0}e^{-t/RC}\). Isolating t, we get \(t=-RCln[V/V_{0}]\). Substituting \(V_{0} = 50.0 V\), \(V = 50.0 V/\sqrt{2}\), and \(R = 225 \Omega\), \(C = 12.0 \mu F\) gives us \(t = 225 \Omega * 12.0 \mu F(-ln(1/\sqrt{2})) = 2.7 ms\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitor Discharge Time
Understanding how a capacitor discharges is a fundamental concept in electronics. When connected to a resistor, the charge dissipates over time. The discharge time is especially relevant when calculating how long it takes for a capacitor to release a certain percentage of its charge. In our exercise, to determine the time it takes for the capacitor to lose half its initial charge, we use the formula \( t = RC \ln(2) \), where \( R \) is the resistance, \( C \) is the capacitance, and \( \ln \) denotes the natural logarithm. The result is a measure of time, typically in seconds, revealing the duration required for the capacitor's charge to drop to 50% of its original amount.

It's important to grasp that this discharge occurs exponentially, not linearly, meaning the rate of charge loss decreases over time. This property will become clearer as we discuss the concept of exponential decay in the context of RC circuits.
Energy Stored in Capacitor
Capacitors are like tiny energy storage devices. The energy stored in a capacitor can be calculated using the formula \( U = \frac{1}{2}CV^2 \), with \( U \) representing the energy in joules, \( C \) the capacitance in farads, and \( V \) the voltage across the capacitor. If we think of the capacitor as holding this energy in an 'electrical state', it’s easier to imagine that once the capacitor starts discharging through a resistor, this energy begins to decrease.

In the given exercise, when the capacitor loses half of its stored energy, the voltage across its plates needs to reduce by the square root of two. Understanding how energy correlates with voltage and charge in a capacitor is critical for analyzing and designing electronic circuits where energy management and timing are essential.
Time Constant of RC Circuit
The time constant, symbolized by \( \tau \) (tau), of an RC circuit is a measure that encapsulates how quickly an RC circuit charges or discharges. It's defined as the product of the resistance \( R \) and the capacitance \( C \) of the circuit, denoted as \( \tau = RC \). This time constant represents the time it takes for the voltage across the capacitor to either charge up to roughly 63% of its maximum value or discharge to about 37% during discharge.

For our example, the time constant also helps determine the specific moments in time where the charge and energy in the capacitor have reached certain fractions of their initial values. The concept of the time constant is pivotal because it's directly related to how fast an electronic circuit can respond to changes, impacting everything from filters to timers.
Exponential Decay in Circuits
The concept of exponential decay is essential when analyzing RC circuits. As the capacitor discharges, the amount of charge and the voltage do not decrease linearly over time, but instead follow an exponential pattern. This means that the rate of decline is rapid at first but then slows down as the values approach zero.

The formula for the voltage in a discharging RC circuit is \( V = V_0e^{-t/RC} \), where \( V_0 \) is the initial voltage, \( e \) is Euler's number (approx. 2.71828), \( t \) is time, and \( RC \) is the time constant. By understanding exponential decay, students can predict how quickly a capacitor will lose its charge over time and design circuits with precise timing functions.

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Most popular questions from this chapter

A \(1.50 \mu \mathrm{F}\) capacitor is charging through a \(12.0 \Omega\) resistor using a \(10.0 \mathrm{~V}\) battery. What will be the current when the capacitor has acquired \(\frac{1}{4}\) of its maximum charge? Will it be \(\frac{1}{4}\) of the maximum current?

A capacitor with \(C=6.00 \mu \mathrm{F}\) is fully charged by connecting it to a battery that has emf \(50.0 \mathrm{~V}\). The capacitor is disconnected from the battery. A resistor of resistance \(R=185 \Omega\) is connected across the capacitor, and the capacitor discharges through the resistor. (a) What is the charge \(q\) on the capacitor when the current in the resistor is \(0.180 \mathrm{~A} ?\) (b) If the connection to the resistor is completed at time \(t=0,\) what is the value of \(t\) when the current has the value specified in part (a)?

A \(42 \Omega\) resistor and a \(20 \Omega\) resistor are connected in parallel, and the combination is connected across a \(240 \mathrm{~V}\) de line. (a) What is the resistance of the parallel combination? (b) What is the total current through the parallel combination? (c) What is the current through each resistor?

A \(4.60 \mu F\) capacitor that is initially uncharged is connected in series with a \(7.50 \mathrm{k} \Omega\) resistor and an \(\mathrm{emf}\) source with \(\mathcal{E}=245 \mathrm{~V}\) and negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor; (b) the voltage drop across the resistor; (c) the charge on the capacitor; (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a)-(d)?

Assume that a typical open ion channel spanning an axon’s membrane has a resistance of \(1 \times 10^{11} \Omega .\) We can model this ion channel, with its pore, as a 12-nm-long cylinder of radius \(0.3 \mathrm{nm}\). What is the resistivity of the fluid in the pore? (a) \(10 \Omega \cdot \mathrm{m}\) (b) \(6 \Omega \cdot \mathrm{m}\) (c) \(2 \Omega \cdot \mathrm{m} ;\) (d) \(1 \Omega \cdot \mathrm{m}\)

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