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Magazine Chapter 20: Problem 101
A \(7.0-\mu \mathrm{F}\) and a \(3.0-\mu \mathrm{F}\) capacitor are connected in series across a \(24-\mathrm{V}\) battery. What voltage is required to charge a parallel combination of the two capacitors to the same total energy?
Short Answer
Expert verified
The required voltage is approximately 11 V.
Step by step solution
01
Understanding Series Capacitor Connections
When capacitors are connected in series, the total capacitance, \(C_s\), is given by: \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \]For a 7.0-\(\mu\mathrm{F}\) and a 3.0-\(\mu\mathrm{F}\) capacitor, compute the total capacitance: \[ \frac{1}{C_s} = \frac{1}{7.0\times10^{-6}} + \frac{1}{3.0\times10^{-6}} \] \[ C_s = \frac{1}{\frac{1}{7.0} + \frac{1}{3.0}} \times 10^{-6} \] \[ C_s = 2.1 \times 10^{-6} \mathrm{F} = 2.1 \mu\mathrm{F} \].
02
Calculate Total Energy in Series Connection
The energy stored in capacitors in series is given by: \[ U_s = \frac{1}{2} C_s V^2 \]Plug in the values of \(C_s\) and \(V = 24\,\mathrm{V}\): \[ U_s = \frac{1}{2} \times 2.1 \times 10^{-6} \times 24^2 \]\[ U_s = 0.0006048 \mathrm{J} = 604.8 \mu\mathrm{J} \].
03
Understanding Parallel Capacitor Connections
In a parallel connection, the total capacitance \(C_p\) is the sum of the individual capacitances: \[ C_p = C_1 + C_2 \]\[ C_p = 7.0 \times 10^{-6} + 3.0 \times 10^{-6} \]\[ C_p = 10.0 \times 10^{-6} = 10 \mu\mathrm{F} \].
04
Calculate Required Voltage for Parallel Energy Equivalence
For capacitors in parallel, energy stored \(U_p\) is given by: \[ U_p = \frac{1}{2} C_p V_p^2 \]We want \(U_p = U_s = 604.8 \mu\mathrm{J} \):\[ 604.8 \mu\mathrm{J} = \frac{1}{2} \times 10 \times 10^{-6} V_p^2 \]Solve for \( V_p \): \[ 1.2096 = 10^{-5} V_p^2 \]\[ V_p^2 = 1.2096 \times 10^5 \]\[ V_p = \sqrt{1.2096 \times 10^5} \]\[ V_p \approx 11 \mathrm{V} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Series and Parallel Circuits
Understanding how capacitors function in series and parallel circuits is essential for solving problems involving these components. When capacitors are connected in series, each capacitor shares the same charge, but the voltage across each capacitor may differ. Here鈥檚 a quick breakdown of how series circuits work:
- In a series circuit, the reciprocal of the total capacitance \(C_s\) is equal to the sum of the reciprocals of each individual capacitance. This can be expressed as: \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots \]
- Each capacitor in the series shares the same amount of electric charge, but the overall capacitance is less than the smallest individual capacitor in the series.
- In parallel, the total capacitance \(C_p\) is the sum of each individual capacitance: \[ C_p = C_1 + C_2 + \ldots\]
- All capacitors have the same voltage across them, unlike in series where the voltage can be different for each capacitor.
Energy Stored in Capacitors
Capacitors store electrical energy, and the energy storage capability differs depending on whether the capacitors are in series or in parallel. The energy stored in a capacitor is given by the formula: \[ U = \frac{1}{2} C V^2 \]
This represents the energy \(U\) stored, where \(C\) is the total capacitance and \(V\) is the voltage across the capacitor.For capacitors in a series circuit:
This represents the energy \(U\) stored, where \(C\) is the total capacitance and \(V\) is the voltage across the capacitor.For capacitors in a series circuit:
- The energy stored is a function of the reduced total capacitance \(C_s\) resulting in less energy storage for the same voltage compared to a parallel circuit.
- Using a 24-volt battery, as in the example, the energy stored \(U_s\) would be calculated based on the total series capacitance.
- The overall energy stored is higher because the total capacitance \(C_p\) increases, allowing for more energy storage assuming the same voltage.
- Capacitors in parallel can store more energy since they effectively behave as a single capacitor with a larger capacity.
Capacitance Calculations
Performing capacitance calculations involves using specific formulas to assess how capacitors behave when connected in different configurations. Let's explore these concepts further.In a series circuit, the total capacitance \(C_s\) is found by adding the reciprocals of each capacitor's capacitance and then taking the reciprocal of the result:
- Example: For two capacitors with capacitances \(7.0 \, \mu F \) and \(3.0 \, \mu F\), the total capacitance is calculated as: \[ \frac{1}{C_s} = \frac{1}{7.0} + \frac{1}{3.0} \]\[ C_s = \frac{1}{\frac{1}{7.0} + \frac{1}{3.0}} \]
- This gives a lower overall capacitance, specifically \(2.1 \, \mu F\) for the series described.
- For the same capacitors: \[ C_p = 7.0 + 3.0 = 10 \, \mu F \]
- This total is used when solving for different electrical properties, such as voltage required to achieve a certain energy storage.
