91Ó°ÊÓ

DATA Your electronics company has several identical capacitors with capacitance \(C_{1}\) and several others with capacitance \(C_{2}\). You must determine the values of \(C_{1}\) and \(C_{2}\) but don't have access to \(C_{1}\) and \(C_{2}\) individually. Instead, you have a network with \(C_{1}\) and \(C_{2}\) connected in series and a network with \(C_{1}\) and \(C_{2}\) connected in parallel. You have a \(200.0 \mathrm{~V}\) battery and instrumentation that measures the total \(\mathrm{en}\) ergy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, \(0.180 \mathrm{~J}\) of energyis stored in the network. When the series combination is connected. \(0.0400 \mathrm{~J}\) of energy is stored. You are told that \(C_{1}\) is greater than \(C_{2}\) (a) Calculate \(C_{1}\) and \(C_{2}\). (b) For the series combination, does \(C_{1}\) or \(C_{2}\) store more charge, or are the values equal? Does \(C_{1}\) or \(C_{2}\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

Step by step solution

01

Analyze Given Information and Formulate Equations

We can derive two equations from the given information. For a parallel combination, the total capacitance \(C_{p}\) is given by the sum of the individual capacitances. And for a series combination, the total capacitance \(C_{s}\) is given by the reciprocal of the sum of the reciprocals of the individual capacitances. Hence we get: \n\n1) \(C_{p}\) = \(C_{1} + C_{2}\) \n2) \(1/C_{s}\) = \(1/C_{1} + 1/C_{2}\) \n\nAlso, the energy stored in a capacitor is given by \(U = 0.5 * C * V^2\). Equating this to the energy supplied by the battery, we get two more equations: \n\n3) \(0.180 = 0.5 * C_{p} * (200)^2\) (From Parallel combination) \n4) \(0.040 = 0.5 * C_{s} * (200)^2\) (From Series combination)

02

Solve the Equations

Now we have 4 equations. Solving equation 3) and equation 4) for \(C_{p}\) and \(C_{s}\), we get: \n\nFrom equation 3, \(C_{p}\) = \(0.180 / (0.5 * (200)^2)\) = \(9 * 10^-6 F\) \nFrom equation 4, \(C_{s}\) = \(0.040 / (0.5 * (200)^2)\) = \(2 * 10^-6 F\) \n\nSubstitute these values in the equation 1) and 2) to get the values of \(C_{1}\) and \(C_{2}\). After solving, we find that \(C_{1}\) = \(7 * 10^-6 F\) and \(C_{2}\) = \(2 * 10^-6 F\)

03

Determine Charge and Energy Storage

For series combination, both capacitors are charged to the same extent. So they store equal charge. But energy is stored in the form of \(U = 0.5*C*V^2\). As \(C_{1} > C_{2}\), therefore \(C_{1}\) stores more energy. \n\nFor the parallel combination, capacitors are charged to the same potential difference, and as \(Q = C*V\), \(C_{1}\) will store more charge. As energy is stored in the form of \(U = 0.5*C*V^2\), \(C_{1}\) stores more energy in this case as well.

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Key Concepts

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

Parallel and Series Circuits

In electrical circuits, capacitors can be connected in parallel or in series, affecting their total capacitance. These configurations are key to controlling how circuits store energy.

In a **parallel circuit**, capacitors are connected side by side. The total capacitance (C_p) is simply the sum of the individual capacitances:

• C_p = C_1 + C_2

This means that as you add more capacitors in parallel, the total capacitance increases. This configuration is useful when higher capacitance is needed.

Contrarily, in a **series circuit**, capacitors are connected end to end. The total capacitance (C_s) is calculated using the reciprocal formula:

• \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \)

In this arrangement, adding more capacitors decreases the total capacitance. Such a configuration is favorable when we need a certain voltage rating higher than individual capacitors can handle. Understanding these configurations aids in solving complex circuit problems by breaking them down into simpler ones.

Energy Stored in Capacitors

Capacitors store energy in the electric field between their plates, and this energy can be calculated using a specific formula. The energy (U) stored in a capacitor is determined by the voltage (V) across the capacitor and its capacitance (C):

• \( U = \frac{1}{2} C V^2 \)

When using parallel circuits, each capacitor is charged to the full voltage of the battery. Thus, a larger capacitance results in more stored energy.

In series circuits, the situation is different. All capacitors share the same charge, but the voltage is divided among them. This results in smaller energy storage compared to their combined capacitance in parallel.

In practical scenarios, knowing how much energy a capacitor can store in different configurations helps in designing efficient circuits that can deliver the required performance without overloading.

Electrical Circuits Problem Solving

Solving electrical circuit problems involves a systematic approach to make sure no detail is overlooked. By understanding fundamental principles such as capacitance formulas and energy equations, we can effectively analyze complex circuits.

A good practice is to start by gathering all known information, like circuit configurations and voltage levels, as seen in the provided data. Next, translate physical configurations into mathematical equations, leveraging formulas for series and parallel circuits.

Once mathematical equations are established, solving these with algebra helps to find unknown variables like capacitance values or voltages. Utilizing the symmetry in circuits or known equalities—such as equal charges in series—further simplifies the problem-solving process.

Finally, verify results for consistency with the given problem and real-world expectations, ensuring complete understanding and correct solutions. Practice on various circuit configurations builds the skills needed to handle increasingly complicated electrical problems.