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A \(1.0 \mu \mathrm{F}\) capacitor is being charged by a \(9.0 \mathrm{~V}\) battery through a \(10 \mathrm{M} \Omega\) resistor. Determine the potential across the capacitor at times (a) \(t=1.0 \mathrm{~s},\) (b) \(t=5.0 \mathrm{~s},\) and (c) \(t=20 \mathrm{~s}\)

Short Answer

Expert verified
(a) 0.855 V, (b) 3.262 V, (c) 7.782 V.

Step by step solution

01

Identify the Given Parameters

We have a capacitor with capacitance \( C = 1.0 \mu F = 1.0 \times 10^{-6} F \), a resistor with resistance \( R = 10 M\Omega = 10 \times 10^6 \Omega \), and an initial voltage source \( V_0 = 9.0 V \). We need to find the potential across the capacitor at specified times.
02

Understand the Charging Formula

The potential \( V(t) \) across a charging capacitor in an RC circuit is given by the formula: \[ V(t) = V_0 \left( 1 - e^{-t/RC} \right) \] where \( V_0 \) is the initial voltage, \( R \) is the resistance, \( C \) is the capacitance, and \( t \) is time.
03

Calculate the Time Constant

The time constant of an RC circuit is given by \( \tau = RC \). Substituting in our values:\[ \tau = (10 \times 10^6 \Omega)(1.0 \times 10^{-6} F) = 10 s \]
04

Compute Potential at \( t = 1.0 \, s \)

Using the formula \( V(t) = V_0 \left( 1 - e^{-t/RC} \right) \), we substitute \( t = 1.0 \, s \):\[ V(1.0) = 9.0 \left( 1 - e^{-1.0/10} \right) \approx 9.0 (1 - 0.9048) \approx 0.855 \, V \]
05

Compute Potential at \( t = 5.0 \, s \)

Now substitute \( t = 5.0 \, s \):\[ V(5.0) = 9.0 \left( 1 - e^{-5.0/10} \right) \approx 9.0 (1 - 0.6065) \approx 3.262 \, V \]
06

Compute Potential at \( t = 20.0 \, s \)

Finally, substitute \( t = 20.0 \, s \):\[ V(20.0) = 9.0 \left( 1 - e^{-20.0/10} \right) \approx 9.0 (1 - 0.1353) \approx 7.782 \, V \]
07

Summarize Results

The potential across the capacitor at \( t = 1.0 \, s \) is approximately \( 0.855 \, V \), at \( t = 5.0 \, s \) it is approximately \( 3.262 \, V \), and at \( t = 20.0 \, s \) it is approximately \( 7.782 \, V \).

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Key Concepts

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

Capacitor Charging
The process of capacitor charging involves connecting a capacitor to a voltage source, such as a battery, which causes the capacitor to store electrical energy. In an RC circuit, which consists of a resistor and a capacitor connected in series, the charging happens gradually. The battery drives electrons from one plate of the capacitor to the other, creating an electric field.
As the capacitor charges, the current in the circuit decreases because the increasing potential difference across the capacitor opposes the battery voltage. Ultimately, the charging stops once the capacitor voltage equals the battery voltage.
To understand how the potential difference across a capacitor changes over time, you'll often use the formula: \[ V(t) = V_0 (1 - e^{-t/RC}) \] where:
  • \( V(t) \) is the potential difference at time \( t \).
  • \( V_0 \) is the initial voltage from the source.
  • \( R \) is the resistance.
  • \( C \) is the capacitance.
  • \( e \) represents the base of the natural logarithm.
This exponential equation describes how the potential difference across the capacitor grows over time as it charges.
Time Constant
The time constant of an RC circuit, denoted as \( \tau \), is the product of the resistance \( R \) and the capacitance \( C \) of the circuit. It is a key parameter that governs how quickly the capacitor charges. The formula for the time constant is: \[ \tau = RC \] In the context of the exercise, with \( R = 10 \mathrm{M} \Omega \) and \( C = 1.0 \mu \mathrm{F} \), the time constant is calculated as: \[ \tau = (10 \times 10^6 \Omega) \cdot (1.0 \times 10^{-6} \mathrm{F}) = 10 \mathrm{s} \]
This means that, after a time span equal to \( \tau \), the capacitor will charge up to approximately 63.2% of the final voltage \( V_0 \).
The concept of the time constant is crucial because it indicates how fast or slow the capacitor charges, giving you a sense of the response time of the RC circuit.
Exponential Growth
In the context of charging a capacitor, exponential growth refers to the way the potential difference across the capacitor increases over time when connected in an RC circuit. Initially, the change is rapid, but it slowly decreases as the capacitor approaches its full voltage \( V_0 \).
The characteristic shape of exponential growth can be understood by examining the formula for potential \( V(t)\):
\[ V(t) = V_0 (1 - e^{-t/RC}) \] You will see that at \( t = 0 \), the potential difference \( V(t) \) is zero, and it asymptotically approaches \( V_0 \) as \( t \) becomes large.
  • Initial rapid growth: At the beginning of the charging process, the potential difference rises quickly.
  • Decreasing growth rate: As time continues, the growth rate slows down.
  • Approaching final value: Eventually, the potential difference gets very close to \( V_0 \), but never truly reaches it within a finite time span.
This concept helps to grasp how the potential difference changes in a non-linear fashion during the capacitor's charging.
Potential Difference Across Capacitor
The potential difference across a capacitor in an RC circuit is the voltage that develops between its plates over time as it charges. Capacitors store energy in the form of electric fields, and the potential difference is crucial in understanding this energy storage.
In our scenario with a 9V battery, as the capacitor charges through the 10Mω resistor, the potential difference across it at various moments can be calculated using the formula:
\[ V(t) = V_0 (1 - e^{-t/RC}) \] This helps determine the voltage across the capacitor at specific times, such as:
  • At \( t = 1 \mathrm{s} \), the potential difference is approximately \( 0.855 \mathrm{V} \).
  • At \( t = 5 \mathrm{s} \), it increases to around \( 3.262 \mathrm{V} \).
  • At \( t = 20 \mathrm{s} \), it reaches about \( 7.782 \mathrm{V} \).
Knowing the potential difference allows you to predict how the capacitor behaves over time, offering insight into energy storage efficiency and circuit performance.

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Most popular questions from this chapter

A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at \(25 \mathrm{~V}\) for a very short time, usually about \(3.0 \mathrm{~ms}\). (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to \(500 \mathrm{~V}\) and produce currents of \(80 \mathrm{~mA}\) (or even larger). A typical pulse lasts for \(10 \mathrm{~ms}\). What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

The power rating of a resistor is the maximum power it can safely dissipate without being damaged by overheating. (a) If the power rating of a certain \(15 \mathrm{k} \Omega\) resistor is \(5.0 \mathrm{~W},\) what is the maximum current it can carry without damage? What is the greatest allowable potential difference across the terminals of this resistor? (b) If a \(9.0 \mathrm{k} \Omega\) resistor is to be connected across a \(120 \mathrm{~V}\) potential difference, what power rating is required for that resistor?

The power rating of a lightbulb is the power it consumes when connected across a \(120 \mathrm{~V}\) outlet. (a) If you put two \(100 \mathrm{~W}\) bulbs in series across a \(120 \mathrm{~V}\) outlet, how much power would each consume if its resistance were constant? (b) How much power does each one consume if you connect them in parallel across a \(120 \mathrm{~V}\) outlet?

An automobile starter motor is connected to a \(12.0 \mathrm{~V}\) battery. When the starter is activated it draws 150 A of current, and the battery voltage drops to \(7.0 \mathrm{~V}\). What is the battery's internal resistance?

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