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Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \({1 \over 675}\) s with an average light power output of 2.70 \(\times\) 10\(^5\) W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

Short Answer

Expert verified
(a) 421 J, (b) 53.8 mF.

Step by step solution

01

Calculate Energy Output as Light

First, we need to determine how much energy is emitted as light. The average power output is given as \(2.70 \times 10^5\) W and the flash duration is \(\frac{1}{675}\) s. Energy is calculated as \(E = P \cdot t\). So, the energy emitted as light is \(E_{light} = 2.70 \times 10^5 \times \frac{1}{675}\). Calculating this gives \(E_{light} = 400\) J.
02

Calculate Total Energy Stored in Capacitor

The light energy produced represents 95% of the total energy stored in the capacitor, due to the conversion efficiency. Let \(E_{total}\) be the total energy stored. Since \(95\%\) of \(E_{total}\) is \(400\) J, we have \(0.95E_{total} = 400\). Solving for \(E_{total}\), we get \(E_{total} = \frac{400}{0.95} = 421\) J.
03

Calculate Capacitance from Stored Energy

Now we have \(E_{total} = 421\) J and the potential difference \(V = 125\) V. The energy stored in a capacitor is given by the formula \(E = \frac{1}{2} C V^2\). Rearranging for \(C\), the capacitance, we have \(C = \frac{2E}{V^2}\). Substituting the values, \(C = \frac{2 \times 421}{125^2}\). This simplifies to \(C = 0.0538\) F or 53.8 mF.

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Key Concepts

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

Capacitors
Capacitors are crucial components in electronics, acting as energy storage devices. They are like small reserves of energy, holding charge in an electric field. These components are commonly found in camera flashes, providing the quick burst of power required to produce a flash. The amount of energy a capacitor can store is determined by its capacitance, denoted by the symbol "C," measured in Farads (F). The formula to calculate the energy stored in a capacitor is \(E = \frac{1}{2} C V^2\), where "E" is energy in Joules and "V" is the voltage across the capacitor's plates. This relationship shows that both capacitance and voltage play crucial roles in determining how much energy a capacitor can store.
Using capacitors, devices like camera flashes can efficiently release a large amount of energy in a very short time. They build up this energy over a longer duration, which can then be used in just a fraction of a second to create the flash. This is what makes them so useful in various applications where a quick, intense burst of power is needed.
Energy Conversion Efficiency
Energy conversion efficiency is a measure of how well a device converts energy from one form to another. In this context, we are looking at the conversion of electrical energy stored in a capacitor to light energy.
Every energy conversion process involves some losses, usually in the form of unwanted heat. For instance, in the camera flash unit, 95% of the electrical energy in the capacitor is converted efficiently to light. This conversion efficiency can be calculated using the ratio: \(Efficiency = \frac{E_{output}}{E_{input}}\). Here, the output energy is the energy that becomes light, and the input is the total energy stored in the capacitor.
High efficiency is desirable as it means more of the stored energy is being used for the intended purpose, in this case, emitting light, while minimal energy is wasted. Understanding and improving energy conversion efficiency can contribute to the design of more effective and sustainable electronic devices.
Electrical Energy Storage
Electrical energy storage is a key aspect of modern electronics, allowing devices to function efficiently and reliably. This process involves capturing energy in an electrical form and storing it for future use. Capacitors are a popular choice for short-term electrical energy storage due to their ability to rapidly charge and discharge.
The principle of storing energy in capacitors involves maintaining a difference in electrical potential between two conductors, or plates. When a voltage is applied across the capacitor, it stores electrical energy by accumulating charge on the plates. This stored energy can be quickly released when needed, which is why capacitors are preferred in environments requiring fast power delivery, such as camera flashes and other high-speed electronics.
Efficient electrical energy storage ensures that devices perform well and have a prolonged operational life, making capacitors an essential component in electronic circuit design.

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Most popular questions from this chapter

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 \(\mu\)C. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

A 20.0-\(\mu\)F capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged 10.0-\(\mu\)F capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

A parallel-plate capacitor has capacitance \(C\) = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm. The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 pC. (a) What is the dielectric constant \(K\) of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

An air capacitor is made by using two flat plates, each with area A, separated by a distance \(d\). Then a metal slab having thickness \(a\) (less than \(d\)) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (\(\textbf{Fig. P24.62}\)). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C\(_0\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

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