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A \(1.00-\mu \mathrm{F}\) capacitor, initially charged to \(12 \mathrm{~V}\), discharges when it is connected in series with a resistor. (a) What resistance is necessary to cause the capacitor to have only \(37 \%\) of its initial charge 1.50 s after starting? (b) What is the voltage across the capacitor at \(t=3 \tau\) if the capacitor is instead charged by the same battery through the same resistor?

Short Answer

Expert verified
(a) Resistance \( R \approx 9.58 \times 10^{5} \Omega \). (b) Voltage at \( t=3\tau \) is approximately \( 11.6 \text{ V} \).

Step by step solution

01

Understanding the Problem and Given Values

We have a capacitor with capacitance \(C = 1.00 \mu F\) and initially charged to a voltage of \(V_0 = 12 \text{ V}\). We need to find the resistance \(R\) such that the charge on the capacitor is \(37\%\) of its initial charge after \( t = 1.50 \text{ s}\). Then, calculate the voltage across the capacitor at \( t = 3\tau \) where \( \tau = RC \) during charging.
02

Calculate the Time Constant for Desired Discharge

The formula for the charge at time \( t \) during discharge is \( Q(t) = Q_0 e^{-t/\tau} \), where \( Q_0 \) is the initial charge. Given \( Q(t) = 0.37 Q_0 \) and \( t = 1.50 \text{ s}\), we set up the equation: \[ 0.37 = e^{-1.50/\tau}. \] Solve for \( \tau \) as follows:\[ \ln(0.37) = -1.50/\tau \Rightarrow \tau = -\frac{1.50}{\ln(0.37)}. \]
03

Solve for Resistance

Substitute the capacitance \( C = 1.00 \mu F = 1.00 \times 10^{-6} \text{ F} \) to find \( R \): \[ \tau = RC \quad \Rightarrow \quad R = \frac{\tau}{C} = \frac{-1.50}{\ln(0.37) \times 1.00 \times 10^{-6}}. \] Calculate \( R \) to find the required resistance.
04

Calculate the Voltage at \(t=3\tau\) During Charging

When charging, the voltage across the capacitor at time \( t \) is given by \( V(t) = V_0 (1 - e^{-t/\tau}) \). For \( t = 3\tau \), \[ V(3\tau) = 12 (1 - e^{-3}). \] Calculate \( V(3\tau) \) to find the voltage at this time.

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Key Concepts

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

Capacitor Discharge
The phenomenon where a capacitor releases its stored energy is called **capacitor discharge**. When a charged capacitor is connected in a circuit with a resistor, the electrons in the conductive material flow from the negatively charged plate to the positively charged one. This flow of electrons causes the stored electrical energy to be gradually released.

Over time, the voltage across the capacitor decreases, eventually reaching zero if left for a long enough period. This is because the charge is transferred from one plate to the other. The speed and nature of this discharge depend on the product of resistance and capacitance, known as the time constant.

It's important to note that the discharge does not happen instantaneously. The rate of discharge is exponential, not linear, meaning it slows over time as less charge is available to move between the plates. This is described by the equation:
  • \( Q(t) = Q_0 e^{-t/\tau} \), where \( Q_0 \) is the initial charge, and \( \tau \) is the time constant.
This equation shows how the charge decreases over time, helping us understand the discharge behavior.
Time Constant
In an RC circuit, the **time constant** is a crucial concept for determining how fast a capacitor will charge or discharge. It is denoted by the Greek letter tau (\( \tau \)) and is defined as \( \tau = RC \), where \( R \) is the resistance and \( C \) is the capacitance.

Physically, the time constant represents the time it takes for the charge or voltage to decrease to about 37% of its initial value during discharge, or to rise to about 63% of its final value when charging.

A larger resistance or capacitance leads to a larger time constant:
  • If \( R \) or \( C \) increases, the capacitor discharges more slowly.
  • Conversely, if they decrease, discharge happens more quickly.
Understanding the time constant helps in predicting how long a capacitor will take to either charge up or discharge to a certain level, which is vital in designing circuits for specific time-dependent applications.
Resistor-Capacitor Relationship
The relationship between a resistor and a capacitor in an RC circuit determines how electrons flow, affecting the charging and discharging speed of the capacitor. This pair of components creates a configuration that can have various effects in electronic circuits.

The interaction between resistors and capacitors in a circuit can be visualized as:
  • Resistors limit the flow of electrons, acting like a dam that checks the current flow.
  • Capacitors store and release energy, much like a water tank that can hold and release water.
Together, they form a time-delay circuit where the resistor sets the rate of charge or discharge of the capacitor. Consequently, changes in either component affect the whole circuit's behavior.
  • For instance, increasing resistance results in slower capacitor charge/discharge rates.
  • Decreasing resistance will speed up these processes.
This relationship is especially useful in timing applications and filters within various electronic devices, where precise control over time-dependent activities is required.

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Most popular questions from this chapter

A \(10.0-\mu \mathrm{F}\) capacitor in a heart defibrillator unit is charged fully by a 10000 -V power supply. Each capacitor plate is connected to the chest of a patient by wires and flat "paddles," one on either side of the heart. The energy stored in the capacitor is delivered through an \(\mathrm{RC}\) circuit, where \(R\) is the resistance of the body between the two paddles. Data indicate that it takes \(75.1 \mathrm{~ms}\) for the voltage to drop to 20.0 V. (a) Find the time constant. (b) Determine the resistance, \(R\). (c) How much time does it take for the capacitor to lose \(90 \%\) of its stored energy? (d) If the paddles are left in place for many time constants, how much energy is delivered to the chest/heart area of the patient?

Two identical resistors \((R)\) are connected in parallel and then wired in series to a \(40-\Omega\) resistor. If the total equivalent resistance is \(55 \Omega,\) what is the value of \(R ?\)

A voltmeter has a resistance of \(30 \mathrm{k} \Omega\). (a) Sketch the circuit diagram and find the current in a \(10-\Omega\) resistor that is in series with a \(6.0 \mathrm{~V}\) ideal battery when the voltmeter is properly connected across that \(10-\Omega\) resistor. (b) Find the voltage across the \(10-\Omega\) resistor under the same conditions. (Express your answer to five significant figures to show how the current differs from \(0.60 \mathrm{~A}\) and the voltage differs from \(6.0 \mathrm{~V}\), which are the expected values when no voltmeter is in place.)

Two identical resistors (each with resistance \(R\) ) are connected together in series and then this combination is wired in parallel to a \(20-\Omega\) resistor. If the total equivalent resistance is \(10 \Omega,\) what is the value of \(R ?\)

You are given four \(5.00-\Omega\) resistors. (a) Show how to connect all the resistors so as to produce an effective total resistance of \(3.75 \Omega\). (b) If this network were then connected to a 12-V battery, determine the current in and voltage across each resistor.

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