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Chapter 26: Problem 61

An RC circuit has a time constant of 3.1 s. At \(t=0\), the process of charging the capacitor begins. At what time will the energy stored in the capacitor reach half of its maximum value?

Short Answer

Expert verified
Answer: The energy stored in the capacitor reaches half of its maximum value at approximately \(1.08\) seconds.

Step by step solution

01

Express the half energy voltage in terms of maximum voltage

We are given the energy at half the maximum capacity, and we need to find the corresponding voltage. Using the energy formula, we have: \(E_\frac{1}{2} = \frac{1}{2}CV^2_\text{half} = \frac{1}{4}CV^2_\text{max}\) Dividing both sides by \(\frac{1}{2}C\): \(V^2_\text{half} = \frac{1}{2}V^2_\text{max}\) Taking a square root of both sides: \(V_\text{half} = V_\text{max}\sqrt{\frac{1}{2}}\)
02

Write the charging equation for the half energy voltage

Plugging in the half voltage value into the charging equation: \(V_\text{max}\sqrt{\frac{1}{2}} = V_0(1 - e^{-\frac{t}{RC}})\) Where our goal is to find the time \(t\).
03

Solve for t in the charging equation

Now we will solve for \(t\) in the equation from Step 2: \(\sqrt{\frac{1}{2}} = 1 - e^{-\frac{t}{RC}}\) Rearrange the equation to isolate the exponential term: \(e^{-\frac{t}{RC}} = 1 - \sqrt{\frac{1}{2}}\) Take the natural logarithm of both sides: \(-\frac{t}{RC} = \ln(1 - \sqrt{\frac{1}{2}})\) Multiply both sides by \(-RC\) to find \(t\): \(t = -RC\ln(1 - \sqrt{\frac{1}{2}})\)
04

Plug the given time constant into the equation for t

We are given the time constant \(RC = 3.1\,\text{s}\), so we plug this value into the equation to find \(t\): \(t = -3.1\,\text{s}\ln(1 - \sqrt{\frac{1}{2}})\) Evaluate the expression to find the time: \(t \approx 1.08\,\text{s}\) So, at approximately \(1.08\) seconds, the energy stored in the capacitor will reach half of its maximum value.

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Most popular questions from this chapter

A \(12.0-V\) battery is attached to a \(2.00-\mathrm{mF}\) capacitor and a \(100 .-\Omega\) resistor. Once the capacitor is fully charged, what is the energy stored in it? What is the energy dissipated as heat by the resistor as the capacitor is charging?

A capacitor bank is designed to discharge 5.0 J of energy through a \(10.0-\mathrm{k} \Omega\) resistor array in under \(2.0 \mathrm{~ms}\) To what potential difference must the bank be charged, and what must the capacitance of the bank be?

A circuit consists of two \(1.00-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the current flowing through each resistor. b) A student trying to measure the current flowing through one of the resistors inadvertently connects an ammeter in parallel with that resistor rather than in series with it. How much current will flow through the ammeter, assuming that it has an internal resistance of \(1.0 \Omega ?\)

Two light bulbs for use at \(110 \mathrm{~V}\) are rated at \(60 \mathrm{~W}\) and \(100 \mathrm{~W}\), respectively. Which has the filament with lower resistance?

Voltmeters are always connected in parallel with a circuit component, and ammeters are always connected in series. Explain why.
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