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The potential difference between the plates of a leaky (meaning that charge leaks from one plate to the other) \(2.0 \mu \mathrm{F}\) capacitor drops to one-fourth its initial value in \(2.0 \mathrm{~s}\). What is the equivalent resistance between the capacitor plates?

Short Answer

Expert verified
The equivalent resistance is approximately 720,000 ohms.

Step by step solution

01

Understand the problem

We're asked to find the equivalent resistance (leakage resistance) between the plates of a leaky capacitor. The potential difference drops to one-fourth in 2 seconds for a 2.0 μF capacitor.
02

Use the formula for capacitive discharge

The potential difference across a discharging capacitor as a function of time is given by \[ V(t) = V_0 e^{-t/RC} \]where \( V_0 \) is the initial potential difference, \( R \) is the resistance, \( C \) is the capacitance, and \( t \) is time. Since \( V(t) = \frac{1}{4}V_0 \) at \( t = 2.0 \) s, we can set up the equation.
03

Set up the equation

Substitute the given values into the formula. We have:\[ \frac{1}{4}V_0 = V_0 e^{-2/RC} \]Dividing both sides by \( V_0 \) results in:\[ \frac{1}{4} = e^{-2/RC} \].
04

Solve for RC

Take the natural logarithm of both sides to remove the exponential function:\[ \ln\left(\frac{1}{4}\right) = -\frac{2}{RC} \].This simplifies to:\[ RC = -\frac{2}{\ln(1/4)} \].
05

Calculate the equivalent resistance R

Substitute known values to find \( R \). We know \( C = 2.0 \times 10^{-6} \) F:\[ R = \frac{-2}{\ln(1/4) \times 2.0 \times 10^{-6}} = \frac{2}{2.0 \times 10^{-6} \times \ln(4)} \].Calculate \( R \) using a calculator to find the value of \( \ln(4) \approx 1.386 \). This results in:\[ R \approx \frac{2}{2.0 \times 10^{-6} \times 1.386} \approx 720,000 \text{ ohms} \].

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Key Concepts

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

Leaky Capacitor
Imagine a capacitor that can't hold onto its charge very well; that's a leaky capacitor for you. Instead of keeping all the stored charge, some of it unexpectedly escapes, just like water leaking out of a broken bucket.
In the context of your exercise, the term "leaky" indicates that the charge slowly drifts from one plate of the capacitor to the other over time. This phenomenon is often seen as an analogy to a resistor being placed across the capacitor, allowing current to flow even when the external circuit is open. Hence, the leakage is characterized by an equivalent resistance that determines how quickly the charge leaks away.
For a clearer understanding:
  • The concept isn't about physical holes but rather poor insulation between the capacitor's plates, which allows charge to drift across.
  • Additional factors affecting leakage include the quality of the dielectric material between the plates and external conditions like temperature.
This understanding helps us explore the relationship between time and charge leakage in capacitors.
Equivalent Resistance
Equivalent resistance in the context of a leaky capacitor represents the resistance through which the charge leaks. It's as if there is an invisible resistor connected directly between the capacitor plates allowing the charge to seep through.
In circuits, equivalent resistance simplifies complex networks into a single resistor that has the same effect on the circuits' overall behavior in terms of current and voltage.
Here's why it matters in this problem:
  • Imagine we cut the capacitor off from the rest of the circuit; the equivalent resistance describes this internal leak rate.
  • Finding this resistance helps us understand how efficient or inefficient a capacitor is at holding charge.
  • By calculating this equivalent resistance, we gain insight into potential issues within electrical circuits or devices, possibly caused by time-dependent decay.
Understanding the equivalent resistance allows engineers to design more effective and reliable circuits by anticipating charge losses.
Exponential Decay
Exponential decay is a fundamental concept in physics that describes how quantities decrease at a rate proportional to their current value over time. It's like watching a parked car roll downhill; it starts slow but gains speed over time.
A leaky capacitor discharges following exponential decay principles: The rate at which it loses charge decreases over time because less charge remains to leak.
Why is this important for capacitors?
  • This principle helps predict the performance and lifespan of capacitors.
  • The exponential model is key to understanding just how fast a capacitor loses its stored energy.
  • It allows us to accurately model capacitor discharge in various applications such as timing circuits or analog signals.
In mathematical terms, the decay of the voltage across the capacitor follows the equation \[ V(t) = V_0 e^{-t/RC} \]where \( V(t) \) is the voltage at time \( t \), and \( V_0 \) is the initial voltage.
Logarithms in Physics
Logarithms might sound complicated, but they're just a unique way to simplify large numbers and solve equations where exponential decay is involved, turning multiplicative processes into additive ones.
They’re crucial in physics when dealing with exponential relationships, as they help to linearize exponential graphs, making them easier to understand and manipulate.
Here is why logarithms were used in solving your exercise:
  • The natural logarithm, \( \ln \), enabled us to navigate the exponential equation \( \frac{1}{4} = e^{-2/RC} \) by transforming it into a linear format.
  • This transformation isolates the time constant \( RC \), making direct calculation possible as seen when deriving \( RC = -\frac{2}{\ln(1/4)} \).
  • By applying logarithms, we can simplify complex relationships and solve for unknowns such as resistance or time constants in discharging circuits.
Ultimately, using logarithms is like having a universal translator that deciphers the language of exponential decay in capacitive discharge.

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Most popular questions from this chapter

A wire of resistance \(5.0 \Omega\) is connected to a battery whose emf \(\mathscr{E}\) is \(2.0 \mathrm{~V}\) and whose internal resistance is \(1.0 \Omega\). In \(2.0 \mathrm{~min}\), how much energy is (a) transferred from chemical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?

Nine copper wires of length \(l\) and diameter \(d\) are connected in parallel to form a single composite conductor of resistance \(R .\) What must be the diameter \(D\) of a single copper wire of length \(l\) if it is to have the same resistance?

A standard flashlight battery can deliver about \(2.0 \mathrm{~W} \cdot \mathrm{h}\) of energy before it runs down. (a) If a battery costs US\$0.80, what is the cost of operating a \(100 \mathrm{~W}\) lamp for \(8.0 \mathrm{~h}\) using batteries? (b) What is the cost if energy is provided at the rate of \(\mathrm{US} \$ 0.06\) per kilowatt-hour?

You are given a number of \(10 \Omega\) resistors, each capable of dissipating only \(1.0 \mathrm{~W}\) without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a \(10 \Omega\) resistance that is capable of dissipating at least \(5.0 \mathrm{~W} ?\)

When resistors 1 and 2 are connected in series, the equivalent resistance is \(16.0 \Omega\). When they are connected in parallel, the equivalent resistance is \(3.0 \Omega\). What are (a) the smaller resistance and (b) the larger resistance of these two resistors?

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