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A controller on an electronic arcade game consists of a variable resistor connected across the plates of a \(0.220 \mu \mathrm{F}\) capacitor. The capacitor is charged to \(5.00 \mathrm{~V}\), then discharged through the resistor. The time for the potential difference across the plates to decrease to \(0.800 \mathrm{~V}\) is measured by a clock inside the game. If the range of discharge times that can be handled effectively is from \(10.0 \mu \mathrm{s}\) to \(6.00 \mathrm{~ms}\), what should be the (a) lower value and (b) higher value of the resistance range of the resistor?

Short Answer

Expert verified
Calculate resistance for: lower time at 10.0 μs and higher at 6.00 ms with the formula.

Step by step solution

01

Understanding the Decay Formula

The voltage across a discharging capacitor decreases exponentially over time, which can be modeled using the equation: \[ V(t) = V_0 e^{-t/RC} \]Where \( V(t) \) is the voltage at time \( t \), \( V_0 \) is the initial voltage, \( R \) is the resistance, and \( C \) is the capacitance. Here, \( V_0 = 5.00 \text{ V} \) and \( C = 0.220 \mu \text{F} \).
02

Solve for Resistance (R) When Voltage is 0.800 V

We need to find \( R \) when the voltage decreases to \( 0.800 \text{ V} \). Substitute the known values into the decay formula:\[ 0.800 = 5.00 e^{-t/(R \times 0.220 \times 10^{-6})} \]Rearrange the equation to solve for \( R \):\[ e^{-t/RC} = \frac{0.800}{5.00} \]\[ -\frac{t}{RC} = \ln\left(\frac{0.800}{5.00}\right) \] \[ R = -\frac{t}{0.220 \times 10^{-6} \times \ln(0.160)} \]
03

Calculate Lower Value of Resistance

Substitute \( t = 10.0 \mu \text{s} = 10.0 \times 10^{-6} \text{ s} \) into the equation for \( R \):\[ R_{ ext{lower}} = -\frac{10.0 \times 10^{-6}}{0.220 \times 10^{-6} \times \ln(0.160)} \]Evaluate this expression to find the lower value of the resistance.
04

Calculate Higher Value of Resistance

Substitute \( t = 6.00 \text{ ms} = 6.00 \times 10^{-3} \text{ s} \) into the equation for \( R \):\[ R_{ ext{higher}} = -\frac{6.00 \times 10^{-3}}{0.220 \times 10^{-6} \times \ln(0.160)} \]Evaluate this expression to find the higher value of the resistance.

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Key Concepts

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

Exponential Decay
In physics, exponential decay describes how a quantity reduces over time in a manner that the rate of decrease is proportional to its current value. This is commonly seen in processes such as radioactive decay and discharging capacitors. In the case of an RC circuit, the exponential decay is represented by the equation
  • \[ V(t) = V_0 e^{-t/RC} \]
Here, \( V_0 \) is the initial voltage, and \( V(t) \) is the voltage at time \( t \). The term \( e^{-t/RC} \) describes how quickly the voltage decreases or 'decays'.
When the resistance \( R \) or capacitance \( C \) is large, the voltage decreases slowly, whereas a smaller \( R \) or \( C \) results in a faster decay.
Understanding this behavior is crucial for designing circuits that require precise timing, such as electronic controllers and timing circuits.
Capacitor Discharge
Capacitor discharge occurs when a charged capacitor releases its stored electrical energy through a resistor. This process can be observed when the voltage drops from its initial value to a lower value over time.
The time it takes for a capacitor to discharge is dependent on the resistance and capacitance in the circuit, often referred to as the time constant \( \tau \).
  • The time constant is given by the formula: \[ \tau = RC \]
A larger time constant means it takes the capacitor longer to discharge.
During discharge, the voltage across the capacitor does not drop linearly but rather follows the natural logarithm based on the exponential decay described earlier.
This behavior is key in applications requiring precise control over timing and energy release, such as camera flash units or signal processing.
Resistance Calculation
Calculating resistance in an RC circuit during a capacitor discharge involves understanding how the voltage decay formula is rearranged to find resistance.
In our problem, we need to calculate the resistance that allows the voltage to decrease to 0.800 V at both minimum and maximum discharge times.
  • Start by rearranging the formula: \[ R = -\frac{t}{C \cdot \ln(V(t)/V_0)} \]
  • Substitute \( V(t) = 0.800 \) V, \( V_0 = 5.00 \) V, and corresponding times into the formula, replacing \( \ln(V(t)/V_0) \) with \( \ln(0.160) \).
By substituting the different values of \( t \) into the formula, you can determine the range of resistance values required for the game controller to handle effectively.
Voltage Decay Formula
The voltage decay formula is fundamental in describing how the voltage across a capacitor decreases over time in an RC circuit.
  • The formula is expressed as: \[ V(t) = V_0 e^{-t/RC} \]
This equation shows that as time \( t \) increases, the value of \( e^{-t/RC} \) decreases, resulting in a lower voltage.
To solve a problem involving voltage decay, begin by identifying known values: initial voltage \( V_0 \), final voltage \( V(t) \), resistance \( R \), capacitance \( C \), and the time span \( t \).
By substituting these values into the formula, you can solve for the unknown quantity, whether it be \( R \) or \( t \).
This equation is crucial in electronics for designing circuits where controlling the rate of voltage drop is necessary.

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Most popular questions from this chapter

The current in a single-loop circuit with one resistance \(R\) is \(5.0 \mathrm{~A}\). When an additional resistance of \(2.0 \Omega\) is inserted in series with \(R\), the current drops to \(4.0 \mathrm{~A}\). What is \(R\) ?

The starting motor of a car is turning too slowly, and the mechanic has to decide whether to replace the motor, the cable, or the battery. The car's manual says that the \(12 \mathrm{~V}\) battery should have no more than \(0.020 \Omega\) internal resistance, the motor no more than \(0.200 \Omega\) resistance, and the cable no more than \(0.040 \Omega\) resistance. The mechanic turns on the motor and measures \(11.4 \mathrm{~V}\) across the battery, \(3.0 \mathrm{~V}\) across the cable, and a current of \(50 \mathrm{~A}\). Which part is defective?

A \(120 \mathrm{~V}\) power line is protected by a \(15 \mathrm{~A}\) fuse. What is the maximum number of \(500 \mathrm{~W}\) lamps that can be simultaneously operated in parallel on this line without "blowing" the fuse because of an excess of current?

A three-way \(120 \mathrm{~V}\) lamp bulb that contains two filaments is rated for \(100-200-300 \mathrm{~W}\). One filament burns out. Afterward, the bulb operates at the same intensity (dissipates energy at the same rate) on its lowest as on its highest switch positions but does not operate at all on the middle position. (a) How are the two filaments wired to the three switch positions? What are the (b) smaller and (c) larger values of the filament resistances?

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?

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