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Two capacitors, \(C_{1}=25.0 \mu \mathrm{F}\) and \(C_{2}=5.00 \mu \mathrm{F}\) , are connected in parallel and charged with a \(100-\mathrm{V}\) power supply. (a) Draw a circuit diagram and calculate the total energy stored in the two capacitors. (b) What If? What potential difference would be required across the same two capacitors connected in series in order that the combination stores the same amount of energy as in (a)? Draw a circuit diagram of this circuit.

Short Answer

Expert verified
The total energy stored in the capacitors when connected in parallel is \(1.5\ J \). To store the same amount of energy with the capacitors in series, a potential difference of approximately \(316.2\ V\) is required.

Step by step solution

01

Understanding Parallel Capacitor Connection

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances, because the voltage across each capacitor is the same. For capacitors in parallel: \( C_{total} = C_1 + C_2 \).
02

Calculating Total Capacitance

Add the individual capacitances to find the total capacitance in parallel: \( C_{total} = 25.0 \mu F + 5.00 \mu F = 30.0 \mu F \).
03

Calculating Stored Energy in Parallel

The energy (\( E \)) stored in a capacitor is given by the formula \( E = \frac{1}{2}C V^2 \), where \( C \) is the capacitance and \( V \) is the voltage. Apply this formula using the total capacitance and given voltage to find the total energy: \( E = \frac{1}{2}(30.0 \times 10^{-6} \mathrm{F}) (100^2 \mathrm{V^2}) \).
04

Understanding Series Capacitor Connection

When capacitors are connected in series, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances. For capacitors in series: \( \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \).
05

Calculating Total Capacitance in Series

Use the formula for series connection to calculate the total capacitance: \( \frac{1}{C_{total}} = \frac{1}{25.0 \mu F} + \frac{1}{5.00 \mu F} \).
06

Calculating New Voltage for Same Energy in Series

Since we want to have the same energy as in the parallel case, set the energies equal and solve for the new voltage (\( V' \)): \( \frac{1}{2}C_{total} V'^2 = \frac{1}{2}(30.0 \times 10^{-6} F)(100^2 V^2) \). Calculate \( C_{total} \) using the result from step 5, and solve for \( V' \).

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Key Concepts

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

Capacitors in Parallel
When capacitors are connected side by side, or in parallel, they share the same voltage across them. This is similar to people holding hands in a chain – each person, like each capacitor, feels the same thing. So, for capacitors in parallel, their individual capacitance values simply add up to a larger total capacitance. Imagine if you had two water tanks, when connected side by side, they can hold more water together than each could alone.

To calculate the combined capacity of capacitors in parallel, it's like adding up the sizes of different water tanks to figure out how much water you can store altogether. In our problem, the total capacitance is the sum of the two given capacitances: 25.0μF and 5.00μF. It's a straightforward addition: the larger storage space (capacitance) you have, the more electrical 'water' (charge) you can store at the same 'pressure' (voltage).
Capacitors in Series
When capacitors are linked in a single line, or in series, they kind of act like runners in a relay race – they all experience the charge passing through them one by one. Unlike being in parallel, the voltage across them is divvied up. The formula for finding the total capacitance in series is a bit trickier. You have to take the reciprocal (think of it as the 'flip side') of each individual capacitor's capacitance, add those together, and then flip the result once more.

Isn't it a bit like filling balloons with air? If you connect balloons in a line using nozzles, the air you blow into the first one squeezes through to the next, and so on. The overall size isn't just the sum; it's constrained by the smallest balloon's size. In our textbook problem, you calculate the total capacitance in series by finding the reciprocal of each capacitance, summing them up, and then taking the reciprocal of that sum. This total determines how much electrical charge can be held across the complete set-up while maintaining an even distribution of voltage.
Electric Potential Energy
Electric potential energy in capacitors is like the energy stored in a stretched rubber band. Just as pulling back on a rubber band stores energy that can be released to propel an object, a charged capacitor holds electric potential energy that can be released to do work in a circuit. The formula\( E = \frac{1}{2}CV^2 \) lets us calculate how much energy is stored, with \( E \) representing the energy, \( C \) the capacitance, and \( V \) the voltage.

If we use our circuit scenario as an example, we place a certain amount of charge into the capacitor at a specific voltage, much like stretching a rubber band a certain distance. The result is stored energy ready to be released when the circuit is completed. Using our given values, the energy stored in the parallel configuration of the capacitors can be calculated, and it’s this energy that we want to match when we reconfigure the capacitors in series.
Circuit Diagrams
Circuit diagrams are the blueprints for electronic circuits. They use symbols to represent the different components, like resistors, batteries, and, of course, capacitors. It's a bit like a map, showing how all the parts are connected to make the full path for the electricity to travel. In our textbook exercise, drawing a circuit diagram helps visualize how the capacitors are connected, whether it's side by side in parallel or end-to-end in series.

Symbols are essential because they allow anyone, anywhere, to understand the design without needing a full description. Just as an architect's floor plan uses symbols for doors, windows, and walls, electrical diagrams have their own lingo. For instance, capacitors are often represented by two parallel lines, indicating the plates that store electric charge. In diagramming the exercise's parallel and series configurations, students better understand the flow of electric charge and the relationship between the capacitance and voltage in each setup.

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Most popular questions from this chapter

Capacitors \(C_{1}=6.00 \mu \mathrm{F}\) and \(C_{2}=2.00 \mu \mathrm{F}\) are charged as a parallel combination across a \(250-\mathrm{V}\) battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.

A wafer of titanium dioxide \((\kappa=173)\) of area 1.00 \(\mathrm{cm}^{2}\) has a thickness of 0.100 \(\mathrm{mm}\) . Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a \(12.0-\mathrm{V}\) battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part \((\mathrm{b}),\) what are the free and induced surface charge densities? (d) What is the magnitude of the electric field?

An air-filled capacitor consists of two parallel plates, each with an area of \(7.60 \mathrm{cm}^{2},\) separated by a distance of \(1.80 \mathrm{mm} .\) A \(20.0-\mathrm{V}\) potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled?

Two conductors having net charges of \(+10.0 \mu \mathrm{C}\) and \(-10.0 \mu \mathrm{C}\) have a potential difference of 10.0 \(\mathrm{V}\) between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to \(+100 \mu \mathrm{C}\) and \(-100 \mu \mathrm{C}\) ?

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