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A \(2.00-\mu \mathrm{F}\) and a \(4.00-\mu \mathrm{F}\) capacitor are connected to a \(60.0-\mathrm{V}\) battery. What is the total charge supplied to the capacitors when they are wired (a) in parallel and (b) in series with each other?

Short Answer

Expert verified
360.0 μC in parallel, 80.0 μC in series.

Step by step solution

01

Understanding the Problem

The problem involves a circuit with capacitors connected both in parallel and in series to a voltage source. Our task is to find the total charge supplied to the capacitors in each configuration.
02

Determine Total Capacitance in Parallel

When capacitors are connected in parallel, the total capacitance is simply the sum of individual capacitances. Thus, \[ C_{parallel} = C_1 + C_2 = 2.00 \, \mu\mathrm{F} + 4.00 \, \mu\mathrm{F} = 6.00 \, \mu\mathrm{F} \]
03

Calculate Charge in Parallel

The formula for charge is \(Q = C \cdot V\), where \(Q\) is the charge, \(C\) is the capacitance, and \(V\) is the voltage. Using the total capacitance in parallel: \[ Q_{parallel} = C_{parallel} \cdot V = 6.00 \, \mu\mathrm{F} \cdot 60.0 \, \mathrm{V} = 360.0 \, \mu\mathrm{C} \]
04

Determine Total Capacitance in Series

For capacitors in series, the reciprocal of the total capacitance is equal to the sum of reciprocals of individual capacitances. Use the formula: \[ \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} \] Therefore, \[ \frac{1}{C_{series}} = \frac{1}{2.00 \, \mu\mathrm{F}} + \frac{1}{4.00 \, \mu\mathrm{F}} = \frac{1+0.5}{2.00 \, \mu\mathrm{F}} = \frac{3}{4} \] \[ C_{series} = \frac{4}{3} \, \mu\mathrm{F} \approx 1.33 \mu\mathrm{F} \]
05

Calculate Charge in Series

Using the formula for charge and the total capacitance in series, the total charge supplied is: \[ Q_{series} = C_{series} \cdot V = 1.33 \, \mu\mathrm{F} \cdot 60.0 \, \mathrm{V} = 80.0 \, \mu\mathrm{C} \]

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Key Concepts

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

Parallel Capacitors
When capacitors are connected in parallel, their total effect is like making one big capacitor that is just the sum of all smaller ones. This means adding up each capacitor's capacitance to find the total. If you have two capacitors, say one with a capacitance of 2.00 \( \mu\text{F} \)and another with a capacitance of 4.00 \( \mu\text{F} \),the total capacitance when connected in parallel will be:
  • \( C_{\text{parallel}} = 2.00 \mu\text{F} + 4.00 \mu\text{F} = 6.00 \mu\text{F} \)
In this configuration, each capacitor retains the same voltage that the electrical source provides. Thus, when you apply a 60.0 V battery, the voltage across each capacitor remains 60.0 V.
Having the total capacitance makes calculating the total charge supplied to the capacitors easy with the formula \( Q = C \cdot V \). For our example, it would be:
  • \( Q_{\text{parallel}} = 6.00 \mu\text{F} \cdot 60.0 \text{ V} = 360.0 \mu\text{C} \)
This setup is very useful in circuits where a higher capacitance is required.
Series Capacitors
For series connected capacitors, determining the total capacitance involves a slightly different approach. Instead of direct summation, we add the reciprocals of each capacitance. The formula to find the total capacitance in series looks like this:
  • \( \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} \)
Let's use our previous example. With capacitances of 2.00 \( \mu\text{F} \) and 4.00 \( \mu\text{F} \), the calculation is as follows:
  • \( \frac{1}{C_{\text{series}}} = \frac{1}{2.00 \mu\text{F}} + \frac{1}{4.00 \mu\text{F}} = \frac{1+0.5}{2.00 \mu\text{F}} = \frac{3}{4} \)
  • \( C_{\text{series}} = \frac{4}{3} \mu\text{F} \approx 1.33 \mu\text{F} \)
With capacitors in series, the same charge is experienced by each. Using the earlier formula, the total charge is calculated as:
  • \( Q_{\text{series}} = 1.33 \mu\text{F} \cdot 60.0 \text{ V} = 80.0 \mu\text{C} \)
This kind of connection is typical when you need a reduced total capacitance or specific voltage splits across components.
Total Capacitance Calculation
Calculating the total capacitance in circuits involving multiple capacitors is essential to understanding how these circuits store and deliver electric energy. The way capacitors are arranged — in series or parallel — greatly impacts this total capacitance.
In parallel arrangements, as discussed, the total capacitance is simply the sum of all individual capacitances. In contrast, for series, it involves summing the reciprocals:
  • Parallel: \( C_{\text{total}} = C_1 + C_2 + \dots + C_n \)
  • Series: \( \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n} \)
This mathematical diversity allows circuits to be designed to achieve different capacitance values, thus providing flexibility in energy storage according to the circuit's needs. Whether increasing total capacitance with a parallel arrangement or decreasing it with a series one, these arrangements let engineers tailor the circuit's properties for optimal performance. Understanding these basic concepts is crucial before tackling more complex circuit analyses.

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