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You conduct a series of experiments in which you connect the capacitors \(C_{1}\) and \(C_{2}>C_{1}\) to a battery in various ways. The experiments are as follows: \(A, C_{1}\) alone connected to the battery; \(\mathbf{B}, C_{2}\) alone connected to the battery; \(\mathrm{C}, \mathrm{C}_{1}\) and \(C_{2}\) connected to the battery in series; \(\mathbf{D}, C_{1}\) and \(C_{2}\) connected to the battery in parallel. Rank these four experiments in order of increasing equivalent capacitance. Indicate ties where appropriate.

Short Answer

Expert verified
C < A < B < D.

Step by step solution

01

Analyze Individual Capacitance Connections

In Experiment A, only capacitor \( C_1 \) is connected to the battery. The equivalent capacitance is simply \( C_1 \). In Experiment B, only capacitor \( C_2 \) is connected, resulting in an equivalent capacitance of \( C_2 \). Since \( C_2 > C_1 \), \( C_1 \) is less than \( C_2 \).
02

Evaluate Series Connection

For Experiment C, capacitors \( C_1 \) and \( C_2 \) are connected in series. The formula for equivalent capacitance \( C_s \) in a series is: \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \] This results in \( C_s < C_1 \) since the equivalent capacitance of a series of capacitors is always less than the smallest individual capacitance.
03

Evaluate Parallel Connection

In Experiment D, \( C_1 \) and \( C_2 \) are connected in parallel. The equivalent capacitance \( C_p \) is given by: \[ C_p = C_1 + C_2 \] In this case, \( C_p > C_2 \) and \( C_p > C_1 \), since the equivalent capacitance in parallel is the sum of the individual capacitances.
04

Order the Experiments by Equivalent Capacitance

From the above analysis, we have the following equivalent capacitances: \( C_1 \) in Experiment A, \( C_2 \) in Experiment B, \( C_s \) in Experiment C, and \( C_p \) in Experiment D. Since \( C_s < C_1 < C_2 < C_p \), the experiments are ranked as follows: Experiment C < Experiment A < Experiment B < Experiment D.

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Key Concepts

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

Series Circuit
In a series circuit, components are connected end-to-end, forming a single path for electric current to flow. When capacitors are connected in series, their overall or equivalent capacitance decreases. This happens because the charge has to pass through each capacitor sequentially, thereby increasing the overall opposition to charge buildup. The formula used to determine the equivalent capacitance \( C_s \) of capacitors in series is:
  • \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \)
This formula illustrates that the total reciprocal of the equivalent capacitance is the sum of the reciprocals of each individual capacitance. Therefore, the equivalent capacitance \( C_s \) is always less than the smallest capacitance in the series. In the given exercise, when capacitors \( C_1 \) and \( C_2 \) are connected in series, their combined effect is less than either of them alone, making it the smallest equivalent capacitance in the experiments provided.
This understanding is pivotal in designing circuits where reduced capacitance is required, often for timing and frequency applications in electronic devices.
Parallel Circuit
Parallel circuits are characterized by all components being connected across the same potential difference, or voltage, meaning they all share the same two electrical nodes or terminals. When capacitors are connected in parallel, their total or equivalent capacitance increases. This is due to the shared arrangement, which allows the combined storage ability of the capacitors to sum up. The formula for determining the equivalent capacitance \( C_p \) in a parallel connection is:
  • \( C_p = C_1 + C_2 \)
With this setup, the total capacitance is simply the sum of the individual capacitances. Thus, \( C_p \) is always greater than the largest single capacitance in the connection because the capacitors are effectively adding their charge-storing abilities together.
In the experiment context, connecting \( C_1 \) and \( C_2 \) in parallel results in the highest equivalent capacitance of all the configurations. This property is often employed in circuits needing a higher capacitance, for instance, in power supplies where large energy storage is beneficial.
Capacitor Connections
Capacitor connections, specifically in series and parallel, significantly influence how the overall capacitance behaves in an electric circuit. Understanding these connections can help determine how energy storage and voltage behavior within the circuit will work.
In series connections, the capacitors resist change more, reducing the system's capacitance and limiting the total charge stored. Conversely, parallel connections enhance the capacitance as they allow more storage capacity, combining the charge-storing capabilities directly.
  • Series connection results: Lower equivalent capacitance; offers greater voltage division.
  • Parallel connection results: Higher equivalent capacitance; benefits systems needing increased charge storage.
Both types of connections have their unique uses depending on requirements such as frequency filtering, smoothing, and stabilizing voltages. For example, in the original exercise, we learned that series connections result in an equivalent lower than individual capacitors, whereas parallel connections provide an equivalent greater than individual capacitors, impacting their respective rankings in the experiment setup.

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Most popular questions from this chapter

Consider a rectangular block of metal of height \(A\), width \(B\), and length \(C,\) as shown in Figure \(21-28\). If a potential difference \(V\) is maintained between the two \(A \times B\) faces of the block, a current \(l_{A B}\) is observed to flow. Find the current that flows if the same potential difference \(V\) is applied between the two \(B \times C\) faces of the block. Give your answer in terms of \(l_{A B}\)

The capacitor in an \(R C\) circuit \((R=120 \Omega, C=45 \mu F)\) is initially uncharged. Find (a) the charge on the capacitor and (b) the current in the circuit one time constant \((\tau=R C)\) after the circuit is connected to a \(9.0-\mathrm{V}\) battery.

Two lightbulbs operate on the same potential difference. Bulb A has four times the power output of bulb B. (a) Which bulb has the greater current passing through it? Explain. (b) What is the ratio of the current in bulb A to the current in bulb B?

Predict/Explain Two capacitors are connected in series. (a) If a third capacitor is now connected in series with the original two, does the equivalent capacitance increase, decrease, or remain the same? (b) Choose the best explaration from among the follow ing: I. Adding a capacitor generally tends to increase the capacitance, but putting it in series tends to decrease the capacitance; therefore, the net result is no change. II. Adding a capacitor in series will increase the total amount of charge stored, and hence increase the equivalent capacitance. III. Adding a capacitor in series decreases the equivalent capacitance since each capacitor now has less voltage across it, and hence stores less charge.

BIO Current Through a Cell Membrane A typical cell membrane is \(8.0 \mathrm{nm}\) thick and has an electrical resistivity of \(1.3 \times 10^{7} \Omega \cdot \mathrm{m} \cdot\) (a) If the potential difference between the inner and outer surfaces of a cell membrane is \(75 \mathrm{mV}\), how much current flows through a square area of membrane \(1.0 \mu \mathrm{m}\) on a side? (b) Suppose the thickness of the membrane is doubled, but the resistivity and potential difference remain the same. Does the current increase or decrease? By what factor?

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