Question

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

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

Step-by-step solution

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.

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.

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.

Key concepts

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.