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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
34.8 V is required.

Step by step solution

01

Calculate the Equivalent Capacitance for Series Connection

When capacitors are connected in series, the reciprocal of the equivalent capacitance \(C_{s}\) is the sum of the reciprocals of the individual capacitances:\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \]For the given capacitors, \( C_1 = 7.0 \ \mu\text{F} \) and \( C_2 = 3.0 \ \mu\text{F} \), so:\[ \frac{1}{C_s} = \frac{1}{7.0}+\frac{1}{3.0} \]Calculating this gives:\[ \frac{1}{C_s} = \frac{10}{21}\]\[ C_s \approx 2.1 \ \mu\text{F} \]
02

Calculate Total Energy Stored in Series Connection

The energy \(E\) stored in a capacitor is given by:\[ E = \frac{1}{2} C V^2 \]For series capacitors with \( C_s \approx 2.1 \ \mu\text{F} \) and voltage \(V = 24\) V:\[ E_s = \frac{1}{2} \times 2.1 \times 10^{-6} \times (24)^2 \]\[ E_s = \frac{1}{2} \times 2.1 \times 10^{-6} \times 576 \]\[ E_s \approx 6.05 \times 10^{-3} \ \text{J} \]
03

Calculate Equivalent Capacitance for Parallel Connection

When capacitors are connected in parallel, the total capacitance \(C_p\) is the sum of the individual capacitances:\[ C_p = C_1 + C_2 \]\[ C_p = 7.0 + 3.0 \]\[ C_p = 10.0 \ \mu\text{F} \]
04

Calculate Voltage Required for the Same Energy

Set the energy in the parallel capacitors \(E_p\) equal to the energy calculated for the series capacitors \(E_s\), and solve for the voltage \(V_p\):\[ E_p = \frac{1}{2} C_p V_p^2 \]Setting \(E_p = E_s\):\[ \frac{1}{2} \times 10.0 \times 10^{-6} \times V_p^2 = 6.05 \times 10^{-3} \]\[ 5.0 \times 10^{-6} \times V_p^2 = 6.05 \times 10^{-3} \]\[ V_p^2 = \frac{6.05 \times 10^{-3}}{5.0 \times 10^{-6}} \]\[ V_p^2 = 1210 \]\[ V_p \approx 34.8 \ \text{V} \]
05

Concluding the Solution

The voltage required to charge the parallel combination of the two capacitors to the same total energy as the series combination is approximately 34.8 V.

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Key Concepts

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

Equivalent Capacitance
When dealing with capacitors, understanding the concept of equivalent capacitance is essential, especially when they're connected in different configurations, like series or parallel. The equivalent capacitance of a set of capacitors tells us how a group of capacitors behaves as a single capacitor. This is particularly useful in simplifying circuits.
  • For capacitors in series, imagine them like a chain. Here, the total or equivalent capacitance is less than any individual capacitor's capacitance. We use the formula: \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n} \] If you solve this, you invert the sum to find the equivalent capacitance \(C_s\).
  • In contrast, for capacitors in parallel, think of the capacitors as lanes on a highway. More lanes mean more space. Here, equivalent capacitance is the sum of all capacitances: \[ C_p = C_1 + C_2 + \cdots + C_n \] The result is that \(C_p\) is greater than all individual capacitances.
Each configuration is like a tool in your toolkit, allowing you to manage different circuit needs efficiently!
Energy Stored in Capacitors
Capacitors store electrical energy just like a battery, but in a different way. The energy stored depends on two factors: the capacitance and the voltage applied across it. The energy \(E\) stored in a capacitor is calculated using the formula:\[ E = \frac{1}{2} C V^2 \]This equation tells us that energy is proportional to the capacitance \(C\) and the square of the voltage \(V\). In practical terms:
  • Higher capacitance means more energy can be stored for a given voltage.
  • Increasing the voltage across the capacitor also increases the energy stored, but significantly more because of the voltage squared term.
It's essential to balance these factors to achieve the desired energy storage without exceeding the capacitor's voltage rating.In circuit applications, such as the example where capacitors are configured differently, the total energy stored can help determine the necessary voltage to achieve equivalent energy in different setups.
Series and Parallel Circuits
Circuits often require capacitors to be configured in series or parallel to achieve particular electrical behaviors. Let's dive into each configuration's unique characteristics. In series circuits: - Capacitors share the same current, but the voltage is divided among them. - They act like a single capacitor with smaller equivalent capacitance. - This setup is useful for applications requiring lower capacitance or higher voltage ratings. For parallel circuits: - Each capacitor experiences the same voltage across it, but the overall current is shared. - Here, capacitors add up, effectively increasing the total capacitance. - Useful when high capacitance is needed, allowing more charge storage. Understanding these configurations allows for effective design and manipulation of electronic circuits to meet specific requirements, like balancing energy storage, space, and efficiency needs.

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Most popular questions from this chapter

An \(86-\Omega\) resistor and a \(67-\Omega\) resistor are connected in series across a battery. The voltage across the \(86-\Omega\) resistor is 27 V. What is the voltage across the \(67-\Omega\) resistor?

A 60.0-W lamp is placed in series with a resistor and a 120.0-V source. If the voltage across the lamp is 25 V, what is the resistance R of the resistor?

The coil of wire in a galvanometer has a resistance of \(R_{C}=60.0 \Omega\) . The galvanometer exhibits a full-scale deflection when the current through it is 0.400 \(\mathrm{mA}\) . A resistor is connected in series with this combination so as to produce a nondigital voltmeter. The voltmeter is to have a full-scale deflection when it measures a potential difference of 10.0 \(\mathrm{V}\) . What is the resistance of this resistor?

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