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An emf source with \(\mathcal{E}=120 \mathrm{V},\) a resistor with \(R=\) \(80.0 \Omega,\) and a capacitor with \(C=4.00 \mu F\) are connected in series. As the capacitor charges, when the current in the resistor is 0.900 \(\mathrm{A}\) . what is the magnitude of the charge on each plate of the capacitor?

Short Answer

Expert verified
The charge on each plate of the capacitor is 19.2 \(\mu \text{C}\).

Step by step solution

01

Understand the Context

The problem involves a series circuit consisting of an emf source, a resistor, and a capacitor. As the capacitor is charging, we need to find the charge on the capacitor plates when the current is given.
02

Use the Charging Circuit Formula

For a charging circuit with a resistor and capacitor in series, the current at any time, given by \(I(t)\), is \(I(t) = \frac{\mathcal{E}}{R} e^{-\frac{t}{RC}}\). In this problem, \(I(t) = 0.900\, \text{A}\), \(\mathcal{E} = 120\, \text{V}\), and \(R = 80.0\, \Omega\).
03

Rearrange and Solve for Time

Rearrange the formula to solve for time \(t\): \(0.900 = \frac{120}{80.0} e^{-\frac{t}{80 \cdot 4\times10^{-6}}}\)Simplifying:\(0.900 = 1.5 e^{-\frac{t}{320\times10^{-6}}}\)\(e^{-\frac{t}{320\times10^{-6}}} = \frac{0.900}{1.5} = 0.6\),Take the natural logarithm:\(-\frac{t}{320\times10^{-6}} = \ln(0.6)\) \(t = -320\times10^{-6} \times \ln(0.6)\)
04

Calculate the Time

Solve for \(t\):\(t = -320\times10^{-6} \times (-0.5108)\approx 163.5 \times 10^{-6} \text{s}\) or \(163.5\, \mu \text{s}\).
05

Use the Formula for Charge on a Charging Capacitor

For the charge \(Q(t)\) on a charging capacitor, the formula is:\(Q(t) = C \mathcal{E} (1 - e^{-\frac{t}{RC}})\), where \(C = 4.00 \mu F\).
06

Calculate the Charge

Substitute for \(t\) into the charge formula:\[Q(t) = 4 \times 10^{-6} \times 120 \times (1 - e^{-\frac{163.5 \times 10^{-6}}{320 \times 10^{-6}}})\]First, evaluate the exponent:\(e^{-\frac{163.5 \times 10^{-6}}{320 \times 10^{-6}}} = e^{-0.5108} = 0.6\)Substitute back:\(Q(t) = 4 \times 10^{-6} \times 120 \times (1 - 0.6)\)\(Q(t) = 4 \times 10^{-6} \times 120 \times 0.4\)\(Q(t) = 19.2 \times 10^{-6} \text{C}\) or \(19.2 \mu \text{C}\).

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Key Concepts

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

Charging Capacitor
In an RC circuit, the charging of a capacitor is an important concept. When you connect a capacitor to a voltage source through a resistor, it starts charging. The voltage across the capacitor increases gradually.
This is because the resistor controls how fast the charge can flow to the capacitor. Let's consider an electromagnetic force (emf) source, a resistor, and a capacitor in series. The voltage builds up and approaches its maximum value slowly over time.
The charge, denoted as \(Q\), on the capacitor increases according to the formula:
  • \(Q(t) = C \mathcal{E} (1 - e^{-\frac{t}{RC}})\)
This equation shows that initially, the charge on the capacitor is zero, and it reaches a maximum value \(C \mathcal{E}\) as time progresses toward infinity.
Current in Resistor
The behavior of the current through the resistor, \(I(t)\), during the charging process is crucial. Initially, the current is at its maximum value because the voltage difference is highest.
As the capacitor charges, the current decreases. This is due to the buildup of voltage across the capacitor which acts against the emf source. In our given problem, this current is found using the exponential decay formula:
  • \(I(t) = \frac{\mathcal{E}}{R} e^{-\frac{t}{RC}}\)
This explains why, as time increases, \(I(t)\) approaches zero in a fully charged capacitor. At the moment given in the exercise, you see the current reduced from the maximum initial current.
Charge on Capacitor Plates
Finding the charge on the capacitor plates involves understanding how the charge accumulates over time. The charge on the plates is directly proportional to the capacitance \(C\) and the applied voltage \(\mathcal{E}\).
At any given time \(t\), the charge \(Q\) can be calculated using the formula mentioned earlier:
  • \(Q(t) = C \mathcal{E} (1 - e^{-\frac{t}{RC}})\)
Here, \(t\) is the precise moment when we need to know the charge. The term \(1 - e^{-\frac{t}{RC}}\) reflects how much of the total possible charge has been accumulated.
This is important because it shows the non-linear nature of capacitor charging, where the charge is gained most quickly at the beginning and slows down as \(t\) increases.
Time Constant
The time constant, denoted as \(\tau\), is defined as \(RC\) where \(R\) is resistance and \(C\) is capacitance. It is a measure of the speed at which the capacitor charges or discharges.
In simpler terms, the time constant gives us a sense of how fast the system reaches a certain charge level. After a time span equal to the time constant, the charge on the capacitor reaches approximately 63.2% of its maximum value.
Therefore, given a specific \(R\) and \(C\), the time constant indicates how quickly electrons are able to flow onto the capacitor’s plates, filling it up with charge.
This means the larger the resistor or capacitor, the slower the charge/discharge process, as more time is required to reach the full potential of the capacitor.

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Most popular questions from this chapter

A \(6.00-\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a \(5.00-\Omega\) resistor and an emf source with \(\mathcal{E}=50.0 \mathrm{V}\) and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of \(250 \mathrm{W},\) how much energy has been stored in the capacitor?

A \(1.50-\mu \mathrm{F}\) capacitor is charging through a \(12.0-\Omega\) resistor using a 10.0 -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 \(2.36-\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a \(5.86-\Omega\) resistor and an emf source with \(\mathcal{E}=120 \mathrm{V}\) and negligible internal resistance. (a) Just after the connection is made, what are (i) the rate at which electrical energy is being dissipated in the resistor; (ii) the rate at which the electrical energy stored in the capacitor is increasing; (iii) the electrical power output of the source? How do the answers to parts (i), (ii), and (iii) compare? (b) Answer the same questions as in part (a) at a long time after the connection is made. (c) Answer the same questions as in part (a) at the instant when the charge on the capacitor is one-half its final value.

A \(2.00-\mu \mathrm{F}\) capacitor that is initially uncharged is connected in series with a \(6.00-\mathrm{k} \Omega\) resistor and an emf source with \(\mathcal{E}=90.0 \mathrm{V}\) and negligible internal resistance. The circuit is completed at \(t=0\) . (a) Just after the circuit is completed, what is the rate at which electrical energy is being dissipated in the resistor? (b) At what value of \(t\) is the rate at which electrical energy is being dissipated in the resistor equal to the rate at which electrical energy is being stored in the capacitor? (c) At the time calculated in part (b), what is the rate at which electrical energy is being dissipated in the resistor?

Three resistors having resistances of 1.60 ? \(2,2.40 \Omega,\) and 4.80\(\Omega\) are connected in parallel to a 28.0 -V battery that has negligible internal resistance. Find (a) the equivalent resistance of the combination; (b) the current in each resistor; (c) the total current through the battery; (d) the voltage across each resistor; (e) the power dissipated in each resistor. (f) Which resistor dissipates the most power. the one with the greatest resistance or the least resistance? Explain why this should be.

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