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A 2.00- and a \(7.50-\mu \mathrm{F}\) capacitor can be connected in series or parallel, as can a 25.0- and a \(100-\mathrm{k} \Omega\) resistor. Calculate the four \(R C\) time constants possible from connecting the resulting capacitance and resistance in series.

Short Answer

Expert verified
The four RC time constants are calculated by multiplying the series or parallel resistance with the series or parallel capacitance.

Step by step solution

01

Calculate the Equivalent Capacitance in Series

When capacitors are connected in series, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances. Use the formula: \( \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} \) where \(C_1 = 2.00 \mu\mathrm{F}\) and \(C_2 = 7.50 \mu\mathrm{F}\).
02

Calculate the Equivalent Capacitance in Parallel

For capacitors in parallel, the total capacitance is the sum of the individual capacitances. Use the formula: \( C_{\text{parallel}} = C_1 + C_2 \) where \(C_1 = 2.00 \mu\mathrm{F}\) and \(C_2 = 7.50 \mu\mathrm{F}\).
03

Calculate the Equivalent Resistance in Series

Resistors in series simply add together to produce the total resistance. Use the formula: \( R_{\text{series}} = R_1 + R_2 \) where \(R_1 = 25.0 ~\mathrm{k}\Omega\) and \(R_2 = 100 ~\mathrm{k}\Omega\).
04

Calculate the Equivalent Resistance in Parallel

For resistors in parallel, the reciprocal of the total resistance is the sum of the reciprocals of each resistance. Use the formula: \( \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \) where \(R_1 = 25.0 ~\mathrm{k}\Omega\) and \(R_2 = 100 ~\mathrm{k}\Omega\).
05

Calculate the Four RC Time Constants

The RC time constant \( \tau \) is the product of the resistance and the capacitance. Calculate the four possible time constants using the combinations of equivalent series and parallel resistances and capacitances found in the previous steps: \( \tau_{\text{series-series}} = R_{\text{series}} \times C_{\text{series}} \), \( \tau_{\text{series-parallel}} = R_{\text{series}} \times C_{\text{parallel}} \), \( \tau_{\text{parallel-series}} = R_{\text{parallel}} \times C_{\text{series}} \), and \( \tau_{\text{parallel-parallel}} = R_{\text{parallel}} \times C_{\text{parallel}} \).

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Key Concepts

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

Capacitors in Series and Parallel
When we combine capacitors in different configurations, the overall capacitance changes. In a series configuration, the cumulative effect is that the equivalent capacitance will always be less than the smallest individual capacitor in the circuit. Imagine each capacitor acting as a bottleneck, reducing the total 'storage space' for charge.

Conversely, in a parallel setup, capacitors are like additional storage units that add up. Here, the total or equivalent capacitance is the sum of all capacitances because each one provides its own separate charge storage. Thus, the overall capacitance is increased.

The formulas used for these calculations are fundamental in circuit analysis. For series:
\[ \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n} \]
For parallel:
\[ C_{\text{parallel}} = C_1 + C_2 + \cdots + C_n \]
Reminding ourselves about these principles is key in understanding how different capacitor configurations influence circuit behavior.
Resistors in Series and Parallel
Similar to capacitors, resistors can also be connected in series or parallel, altering the total or equivalent resistance of a circuit.

With resistors in series, they add straight up. Each resistor adds its resistance to the total, creating what is effectively a single, larger resistor that impedes electron flow. The overall resistance increases because the current has to travel through each resistor in turn. The formula for this cumulative resistance is simple addition:
\[ R_{\text{series}} = R_1 + R_2 + \cdots + R_n \]

In a parallel arrangement, resistors provide multiple paths for current to flow, each path reducing the overall resistance and making it easier for current to pass. The total resistance is then always less than the smallest individual resistor's resistance. The calculation for equivalent resistance in parallel uses reciprocals:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} \]
These rules are essential for predicting how a circuit will respond to different resistor configurations.
Equivalent Capacitance
The concept of equivalent capacitance is a means to simplify a complex network of capacitors. It is the total capacitance of the circuit that would have the same effect as all the combined capacitors. When dealing with series and parallel combinations, it's crucial to apply the right formula to obtain the overall capacitance.

In series: we use the reciprocal approach to find the equivalent capacitance as previously shown. In parallel: it's a sum of all capacitances. This idea of simplifying a network of capacitors into an equivalent capacitance makes it much simpler to analyze and understand how a complex circuit will behave electrically.
Equivalent Resistance
Likewise, equivalent resistance is a way to consolidate the total resistance of a network of resistors into a singular value. This single resistance value can replace the network of resistors without affecting the overall current and voltage in the circuit.

In series configurations, you add respective values to reach the equivalent resistance. In parallel, you must take the reciprocal of the sum of reciprocals, contrary to the intuitive addition you might assume at first glance. The resulting equivalent resistance allows for streamlined calculations in circuit analysis and helps in understanding circuit functionality, especially when combined with the laws of Ohm and Kirchhoff.
Time Constant Calculation
The RC time constant, symbolized as \( \tau \), is a measure of the time it takes for a capacitor to either charge or discharge through a resistor by approximately 63.2% of the difference between its initial value and its final value. In mathematical terms, the time constant is the product of the equivalent resistance and the equivalent capacitance in the circuit:

\[ \tau = R_{\text{equiv}} \times C_{\text{equiv}} \]
For a given circuit, there can be multiple time constants depending on the combination of capacitors and resistors. As such, calculating each possible time constant allows us to predict how quickly a system will respond to changes, which is vital for the design and understanding of filters, timers, and many other electronic applications. Calculating these time constants is an exercise in applying our understanding of equivalent resistances and capacitances in context.

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Most popular questions from this chapter

A \(500-\Omega\) resistor, an uncharged \(1.50-\mu \mathrm{F}\) capacitor, and a \(6.16-\mathrm{V}\) emf are connected in series. (a) What is the initial current? (b) What is the \(R C\) time constant? (c) What is the current after one time constant? (d) What is the voltage on the capacitor after one time constant?

There is a voltage across an open switch, such as in Figure. Why, then, is the power dissipated by the open switch small?

Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on \(120 \mathrm{~V}\) and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on \(120 \mathrm{~V}\) and has 39 remaining identical bulbs, what is then the operating voltage of each?

Calculate the emf \(_{\mathrm{x}}\) of a dry cell for which a potentiometer is balanced when \(R_{\mathrm{x}}=1.200 \Omega\), while an alkaline standard cell with an emf of \(1.600 \mathrm{~V}\) requires \(R_{\mathrm{s}}=1.247 \Omega\) to balance the potentiometer.

While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the angular acceleration produced given the mass lifted is \(10.0 \mathrm{kg}\) at a distance of \(28.0 \mathrm{cm}\) from the knee joint, the moment of inertia of the lower leg is \(0.900 \mathrm{kg} \cdot \mathrm{m}^{2}\), the muscle force is 1500 N, and its effective perpendicular lever arm is \(3.00 \mathrm{cm}\). (b) How much work is done if the leg rotates through an angle of \(20.0^{\circ}\) with a constant force exerted by the muscle?

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