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A 500.0\(\Omega\) resistor is connected in series with a capacitor. What must be the capacitance of the capacitor to produce a time constant of 2.00 \(\mathrm{s} ?\)

Short Answer

Expert verified
The capacitance should be 4.00 mF.

Step by step solution

01

Understand the Time Constant Formula

The time constant \( \tau \) for an RC (resistor-capacitor) circuit is given by the formula \( \tau = R \times C \), where \( R \) is the resistance and \( C \) is the capacitance. We are given \( \tau = 2.00 \) seconds and \( R = 500.0 \) ohms.
02

Rearrange the Formula to Find Capacitance

We need to solve for \( C \) in the equation \( \tau = R \times C \). Rearranging the equation gives \( C = \frac{\tau}{R} \).
03

Substitute Known Values

Substitute the known values into the equation: \( C = \frac{2.00}{500.0} \).
04

Calculate the Capacitance

Perform the division to find the capacitance: \( C = \frac{2.00}{500.0} = 0.004 \) Farads or 4.00 milliseconds (4.00 mF).

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Key Concepts

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

Time Constant
The time constant in an RC circuit is a crucial concept that describes how quickly the circuit responds to changes. It is labeled as \( \tau \) and defined as the time it takes for voltage across the capacitor to either charge up to about 63.2% of its maximum value or discharge to about 36.8% of its initial value. In mathematical terms, the time constant \( \tau \) is given by the product of resistance \( R \) and capacitance \( C \), which is expressed as \( \tau = R \times C \).
The time constant gives us an intuitive sense of how fast or slow the circuit is: a larger time constant means a slower response, while a smaller one means a faster response. This concept is significant for applications like timing and filtering in electronics, where precise control over charge and discharge rates is essential.
Capacitance Calculation
Calculating capacitance in an RC circuit involves using the relationship between time constant, resistance, and capacitance. If you are given the time constant \( \tau \) and resistance \( R \), you can find capacitance \( C \) using the formula \( C = \frac{\tau}{R} \).
In our example, with a time constant of 2.00 seconds and a resistance of 500.0 ohms, the capacitance can be calculated as follows:
  • Start with the equation: \( C = \frac{\tau}{R} \)
  • Substitute the values: \( C = \frac{2.00}{500.0} \)
  • Perform the division: \( C = 0.004 \) Farads or 4.00 millifarads (mF)
This simple rearrangement and substitution process helps solve for unknown values and understand how these circuit components interact. Understanding this calculation helps in designing circuits with precise timing and charge retention.
Series Circuit
In a series circuit, components are connected end-to-end in a single path for the current to flow. This means each component shares the same current, but the voltage across the components adds up to the total voltage of the circuit. In our scenario, the resistor and capacitor are connected in series, which affects how they accumulate charge and discharge over time.
Series circuits are straightforward in terms of components' behavior:
  • The total resistance in a series circuit is the sum of individual resistances: \( R_{total} = R_1 + R_2 + ... + R_n \).
  • In terms of voltage, it is distributed across each component proportional to their resistances.
  • Since all components share the same current, any alteration in one affects the entire circuit.
Understanding how series circuits function is critical for analyzing how the current and voltage interact, especially in RC circuits where timing and response are crucial.
Resistor
A resistor is a fundamental component in electronics that limits the flow of current in a circuit. In the context of an RC circuit, the resistor, alongside the capacitor, defines the time constant \( \tau \), influencing how quickly the circuit can charge or discharge.
Key roles of resistors in circuits include:
  • Controlling current flow to prevent damage to other components.
  • Creating voltage drops to supply other parts of the circuit with the proper voltage levels.
  • Working with capacitors to shape signals in timing applications, such as clock generation or pulse shaping.
In the given exercise, a 500.0 \( \Omega \) resistor plays a critical role in determining the time constant, as demonstrated by its inclusion in the formula \( \tau = R \times C \). Understanding the function of resistors enhances your ability to manipulate electronic circuits for desired operational characteristics.

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Most popular questions from this chapter

Three resistors having resistances of \(1.60 \Omega, 2.40 \Omega,\) and \(4.80 \Omega,\) respectively, are connected in parallel to a 28.0 \(\mathrm{V}\) bat- tery 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, and (e) the power dissipated in each resistor. (f) Which resistor dissipates the most power, the one with the greatest resistance or the one with the least resistance? Explain why this should be.

An idealized voltmeter is connected across the terminals of a 15.0 \(\mathrm{V}\) battery, and a 75.0\(\Omega\) appliance is also connected across its terminals. If the voltmeter reads \(11.3 \mathrm{V} :\) (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

Struck by lightning. Lightning strikes can involve currents as high as \(25,000\) A that last for about 40\(\mu\) s. If a person is struck by a bolt of lightning with these properties, the current will pass through his body. We shall assume that his mass is 75 kg, that he is wet (after all, he is in a rainstorm) and therefore has a resistance of \(1.0 \mathrm{k} \Omega,\) and that his body is all water (which is reasonable for a rough, but plausible, approximation).(a) By how many degrees Celsius would this lightning bolt increase the temperature of 75 kg of water? (b) Given that the internal body temperature is about \(37^{\circ} \mathrm{C}\) , would the person's temperature actually increase that much? Why not? What would happen first?

When a solid cylindrical rod is connected across a fixed potential difference, a current \(I\) flows through the rod. What would be the current (in terms of \(I\) ) if (a) the length were doubled, (b) the diameter were doubled, (c) both the length and the diameter were doubled?

A \(540-\mathrm{W}\) electric heater is designed to operate from 120 \(\mathrm{V}\) lines. (a) What is its resistance? (b) What current does it draw? (c) If the line voltage drops to \(110 \mathrm{V},\) what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.

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