You have an electric device containing a \(10.0-\mu \mathrm{F}\) capacitor, but an application requires an \(18.0-\mu \mathrm{F}\) capacitor. What modification can you make to your device to increase its capacitance to \(18.0-\mu \mathrm{F} ?\)

Short Answer

Expert verified
Answer: The additional capacitor should have a capacitance of 8.0 μF.

Step by step solution

01

STEP 1: Determine the approach

As mentioned in the analysis, we need to connect an additional capacitor in parallel with the original capacitor to increase the capacitance. For parallel connection, the total capacitance is the sum of the individual capacitances: \(C_T = C_1 + C_2\).
02

STEP 2: Setup the equation

We need to find the capacitance of the additional capacitor \(C_2\) that must be connected in parallel with the original capacitor with capacitance \(C_1 = 10.0\,\mu\mathrm{F}\) to get the total capacitance \(C_T = 18.0\,\mu\mathrm{F}\). Using the formula for parallel capacitors, we get: $$C_T = C_1 + C_2.$$
03

STEP 3: Solve for the unknown

Substituting the known values of \(C_T\) and \(C_1\), we have: $$18.0\,\mu\mathrm{F} = 10.0\,\mu\mathrm{F} + C_2.$$ Solve for \(C_2\): $$C_2 = 18.0\,\mu\mathrm{F} - 10.0\,\mu\mathrm{F} = 8.0\,\mu\mathrm{F}.$$
04

STEP 4: State the modification

To increase the capacitance of the device to \(18.0\,\mu\mathrm{F}\), an additional capacitor with a capacitance of \(8.0\,\mu\mathrm{F}\) should be connected in parallel to the original capacitor.

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Key Concepts

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

Electric Capacitance
Electric capacitance is a fundamental property of capacitors that determines their ability to store electrical energy. It is measured in farads (F), where one farad is the capacitance of a capacitor that can store one coulomb of charge with an electric potential difference of one volt across its plates.

In layman's terms, you can think of capacitance as a container for electrical charge; the larger the capacitance, the more charge it can hold for a given voltage. The capacitance of a capacitor is usually determined by its physical characteristics, such as the area of the plates, the distance between them, and the material of the dielectric (the insulating layer) between the plates. Understanding this concept is crucial when working with electronic circuits and in designing systems that depend on energy storage and timing applications.
Parallel Connection of Capacitors
When capacitors are connected in parallel, their total capacitance is the sum of their individual capacitances. This connection is akin to adding more storage bins side by side — the overall storage capacity increases because you now have more containers to store charges.

The formula for the total capacitance in a parallel circuit is expressed as:
\[C_{\text{T}} = C_1 + C_2 + C_3 + \ldots + C_n\]
where \(C_{\text{T}}\) is the total capacitance, and \(C_1\), \(C_2\), \(C_3\), ..., \(C_n\) are the capacitances of individual capacitors. One major consequence of this is that the voltage across each capacitor in parallel is the same and is equal to the voltage of the power source. In practical applications, connecting capacitors in parallel is an effective method to achieve the desired capacitance without having to manufacture a custom capacitor.
Solving for Unknown in Electric Circuits
To solve for an unknown component in an electric circuit, such as a missing capacitor value, we use the established relationships between different elements in the circuit—Ohm’s law for resistors, Kirchhoff’s rules for circuits, and the parallel formula for capacitors, as illustrated in the given exercise.

In our case, we addressed the unknown by setting up an equation based on the formula for capacitors in parallel. By rearranging the equation to isolate the unknown variable, we could solve for it arithmetically. This process of problem-solving is a fundamental aspect of electrical engineering and physics, providing a systematic approach to diagnose and design circuitry for various applications. It's crucial for students not only to memorize these equations but also to understand the principles behind them to apply them effectively to real-world scenarios.

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